Chapter 14 Mathematical Reasoning

A statement is a sentence that is either true or false, but not both. We will examine each given sentence based on this definition.

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(i) 8 is less than 6.

This is a sentence.

We can determine its truth value. The sentence "$8$ is less than $6$" is false, as $8 > 6$.

Since it is a sentence that is definitely false, it has a definite truth value.

Therefore, it is a statement.

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(ii) Every set is a finite set.

This is a sentence.

We can determine its truth value. This sentence is false because there exist infinite sets, such as the set of all natural numbers ($\mathbb{N} = \{1, 2, 3, \dots\}$) or the set of all real numbers ($\mathbb{R}$).

Since it is a sentence that is definitely false, it has a definite truth value.

Therefore, it is a statement.

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(iii) The sun is a star.

This is a sentence.

We can determine its truth value based on scientific knowledge. The sun is indeed a star.

Since it is a sentence that is definitely true, it has a definite truth value.

Therefore, it is a statement.

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(iv) Mathematics is fun.

This is a sentence.

However, whether "Mathematics is fun" is true or false is subjective and varies from person to person. There is no universally agreed-upon truth value for this sentence.

Since its truth value is not definite (it is neither always true nor always false), it is not a statement.

Therefore, it is not a statement.

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(v) There is no rain without clouds.

This is a sentence.

Based on meteorological understanding, rain formation typically requires clouds. This sentence expresses a generally accepted scientific fact or observation, which is considered true in the context of typical reasoning. While there might be extremely rare or technical exceptions debated by meteorologists, in standard logic, such a widely accepted proposition is treated as having a definite truth value (true).

Since it is a sentence that has a definite truth value (generally considered true), it is a statement.

Therefore, it is a statement.

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(vi) How far is Chennai from here?

This is a sentence.

However, this sentence is a question. Questions do not assert anything; they ask for information. Therefore, they do not have a truth value (they are neither true nor false).

Since it does not have a truth value, it is not a statement.

Therefore, it is not a statement.

A statement is a sentence that is either true or false, but not both. We will examine each given sentence based on this definition.

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(i) There are 35 days in a month.

This is a sentence.

We can determine its truth value. Months in the standard calendar have 28, 29, 30, or 31 days. Thus, the sentence "$$There are 35 days in a month$$" is false.

Since it is a sentence that is definitely false, it has a definite truth value.

Therefore, it is a statement.

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(ii) Mathematics is difficult.

This is a sentence.

However, whether "Mathematics is difficult" is true or false is subjective and depends on the individual's perception, aptitude, and effort. There is no universally definite truth value for this sentence.

Since its truth value is not definite, it is not a statement.

Therefore, it is not a statement.

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(iii) The sum of 5 and 7 is greater than 10.

This is a sentence.

We can determine its truth value. The sum of 5 and 7 is $5 + 7 = 12$. The sentence states that 12 is greater than 10, which is true ($12 > 10$).

Since it is a sentence that is definitely true, it has a definite truth value.

Therefore, it is a statement.

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(iv) The square of a number is an even number.

This is a sentence.

We can determine its truth value. Consider the number 3. The square of 3 is $3^2 = 9$, which is an odd number. This provides a counterexample, showing the sentence is false.

Since it is a sentence that is definitely false, it has a definite truth value.

Therefore, it is a statement.

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(v) The sides of a quadrilateral have equal length.

This is a sentence.

We can determine its truth value. Consider a rectangle that is not a square. It is a quadrilateral, but its sides do not all have equal length. This shows the sentence is false.

Since it is a sentence that is definitely false, it has a definite truth value.

Therefore, it is a statement.

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(vi) Answer this question.

This is a sentence.

However, this sentence is an imperative sentence (a command). Imperative sentences do not assert anything; they direct or request an action. Therefore, they do not have a truth value (they are neither true nor false).

Since it does not have a truth value, it is not a statement.

Therefore, it is not a statement.

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(vii) The product of (–1) and 8 is 8.

This is a sentence.

We can determine its truth value. The product of -1 and 8 is $(-1) \times 8 = -8$. The sentence states that the product is 8, which is false ($-8 \neq 8$).

Since it is a sentence that is definitely false, it has a definite truth value.

Therefore, it is a statement.

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(viii) The sum of all interior angles of a triangle is 180°.

This is a sentence.

We can determine its truth value based on a fundamental theorem in Euclidean geometry. The sum of the interior angles of any triangle is indeed 180 degrees.

Since it is a sentence that is definitely true, it has a definite truth value.

Therefore, it is a statement.

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(ix) Today is a windy day.

This is a sentence.

However, the truth value of this sentence depends on the specific day it is uttered and the location where the speaker is. Its truth value changes from day to day and place to place. Without knowing the context (the specific "today" and "here"), we cannot assign a definite truth value.

Since its truth value is not definite, it is not a statement.

Therefore, it is not a statement.

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(x) All real numbers are complex numbers.

This is a sentence.

We can determine its truth value based on the definition of real and complex numbers. A complex number is typically written in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit ($\sqrt{-1}$). A real number $r$ can be written as $r + 0i$. Thus, every real number is a complex number with the imaginary part equal to zero.

Since it is a sentence that is definitely true, it has a definite truth value.

Therefore, it is a statement.

A sentence is not a statement if it cannot be definitively classified as either true or false.

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Here are three examples of sentences that are not statements:

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Example 1: What is your name?

Reason: This sentence is an interrogative sentence (a question). It is asking for information and does not make an assertion about something being true or false. Therefore, it does not have a truth value.

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Example 2: Please close the door.

Reason: This sentence is an imperative sentence (a command or request). It directs or requests an action and does not assert that something is true or false. Therefore, it does not have a truth value.

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Example 3: He is a good person.

Reason: This sentence contains a pronoun ("He") without specifying who "He" is. Its truth value depends on who "He" refers to, which is not specified within the sentence itself. Such sentences, whose truth value depends on context or undefined terms, are generally considered open sentences or not statements in logic unless the context is explicitly provided and fixes the reference.

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Other types of sentences that are generally not statements include exclamatory sentences (e.g., "Wow, that's amazing!"), subjective opinions (e.g., "Vanilla ice cream is the best."), and sentences containing vague or ambiguous terms without sufficient context.

The negation of a statement is a statement that is true when the original statement is false, and false when the original statement is true. The negation of a statement $P$ is typically written as "Not $P$", "It is not the case that $P$", or using the symbol $\sim P$ or $\neg P$.

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(i) Statement: Both the diagonals of a rectangle have the same length.

This statement asserts that a specific property (having the same length) holds for both diagonals of a rectangle.

To negate this, we need to state that this property does not hold. The simplest way is to say it is not true that both diagonals have the same length.

Negation: It is not the case that both the diagonals of a rectangle have the same length.

An equivalent and more natural way to phrase this is to state that at least one of the diagonals does not have the same length as the other. However, since there are only two diagonals, this is equivalent to saying they do not have the same length.

Equivalent Negation: The diagonals of a rectangle do not have the same length.

Note: In geometry, the original statement is true. Therefore, its negation is false.

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(ii) Statement: $\sqrt{7}$ is rational.

This statement asserts that the number $\sqrt{7}$ belongs to the set of rational numbers.

To negate this, we state that $\sqrt{7}$ does not belong to the set of rational numbers.

Negation: It is not the case that $\sqrt{7}$ is rational.

An equivalent and more common way to express the negation of "is rational" is "is irrational".

Equivalent Negation: $\sqrt{7}$ is irrational.

Note: In mathematics, the original statement is false (as $\sqrt{7}$ is indeed irrational). Therefore, its negation is true.

The negation of a statement $P$ is the statement "It is not the case that $P$", which has the opposite truth value of $P$.

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(i) Statement: Australia is a continent.

Negation: It is not the case that Australia is a continent.

An equivalent way to state the negation is: Australia is not a continent.

Check the truth value of the negation: The original statement "Australia is a continent" is true.

Therefore, its negation "Australia is not a continent" is false.

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(ii) Statement: There does not exist a quadrilateral which has all its sides equal.

This statement claims that no quadrilateral exists with all sides equal.

Negation: It is not the case that there does not exist a quadrilateral which has all its sides equal.

An equivalent and more natural way to state the negation is: There exists a quadrilateral which has all its sides equal.

Check the truth value of the negation: The original statement "There does not exist a quadrilateral which has all its sides equal" is false (e.g., a rhombus or a square is a quadrilateral with all sides equal).

Therefore, its negation "There exists a quadrilateral which has all its sides equal" is true (e.g., a square or a rhombus exists).

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(iii) Statement: Every natural number is greater than 0.

This statement is a universal quantification.

Negation: It is not the case that every natural number is greater than 0.

An equivalent way to state the negation is: There exists a natural number which is not greater than 0. (i.e., there exists a natural number less than or equal to 0).

Check the truth value of the negation: The original statement "Every natural number is greater than 0" is true (assuming the standard definition of natural numbers as $\{1, 2, 3, \dots\}$).

Therefore, its negation "There exists a natural number which is not greater than 0" is false.

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(iv) Statement: The sum of 3 and 4 is 9.

Negation: It is not the case that the sum of 3 and 4 is 9.

An equivalent way to state the negation is: The sum of 3 and 4 is not 9.

Check the truth value of the negation: The original statement "The sum of 3 and 4 is 9" is false (since $3+4=7$).

Therefore, its negation "The sum of 3 and 4 is not 9" is true.

A compound statement is formed by combining two or more simple statements using connecting words like "and", "or", "if...then", "if and only if", etc. The simple statements that make up a compound statement are called its component statements.

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(i) Compound Statement: The sky is blue and the grass is green.

The connecting word is "and". We can separate the two parts connected by "and" into individual sentences that are statements.

Component Statements:

$p$: The sky is blue.

$q$: The grass is green.

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(ii) Compound Statement: It is raining and it is cold.

The connecting word is "and".

Component Statements:

$p$: It is raining.

$q$: It is cold.

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(iii) Compound Statement: All rational numbers are real and all real numbers are complex.

The connecting word is "and".

Component Statements:

$p$: All rational numbers are real.

$q$: All real numbers are complex.

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(iv) Compound Statement: 0 is a positive number or a negative number.

The connecting word is "or".

Component Statements:

$p$: 0 is a positive number.

$q$: 0 is a negative number.

We need to identify the component statements in each compound statement and determine their truth value (True or False).

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(i) Compound Statement: A square is a quadrilateral and its four sides equal.

The connecting word is "and". The component statements are:

$p$: A square is a quadrilateral.

$q$: Its four sides equal. (The subject "its" refers to "a square").

Check Truth Value:

$p$: A square is a polygon with four sides, which is the definition of a quadrilateral. So, $p$ is True.

$q$: By definition, a square has four sides of equal length. So, $q$ is True.

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(ii) Compound Statement: All prime numbers are either even or odd.

The connecting word is "or". The structure is "Property A is either B or C". This can be broken down into two statements.

$p$: All prime numbers are even.

$q$: All prime numbers are odd.

Check Truth Value:

$p$: The only even prime number is 2. All other prime numbers (3, 5, 7, 11, ...) are odd. So, it is not true that *all* prime numbers are even. Thus, $p$ is False.

$q$: The number 2 is a prime number, and it is even, not odd. So, it is not true that *all* prime numbers are odd. Thus, $q$ is False.

Note: While the individual component statements are false, the original compound statement "All prime numbers are either even or odd" is true, because every integer (including every prime number) is indeed either even or odd.

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(iii) Compound Statement: A person who has taken Mathematics or Computer Science can go for MCA.

The connecting word is "or". The component statements are:

$p$: A person who has taken Mathematics can go for MCA.

$q$: A person who has taken Computer Science can go for MCA.

Check Truth Value:

Determining the truth value of these statements requires specific knowledge of the eligibility criteria for MCA (Master of Computer Applications) programs, which can vary between institutions. In general, a Bachelor's degree with a strong background in Mathematics or Computer Science is required. Assuming standard eligibility criteria:

$p$: A Bachelor's degree in Mathematics often satisfies the eligibility for MCA. So, $p$ is likely True (depending on specific program requirements).

$q$: A Bachelor's degree in Computer Science typically satisfies the eligibility for MCA. So, $q$ is likely True (depending on specific program requirements).

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(iv) Compound Statement: Chandigarh is the capital of Haryana and UP.

The connecting word is "and". The component statements are:

$p$: Chandigarh is the capital of Haryana.

$q$: Chandigarh is the capital of UP.

Check Truth Value:

$p$: Chandigarh is indeed the capital of Haryana. So, $p$ is True.

$q$: The capital of Uttar Pradesh (UP) is Lucknow, not Chandigarh. So, $q$ is False.

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(v) Compound Statement: $\sqrt{2}$ is a rational number or an irrational number.

The connecting word is "or". The component statements are:

$p$: $\sqrt{2}$ is a rational number.

$q$: $\sqrt{2}$ is an irrational number.

Check Truth Value:

$p$: A rational number can be expressed as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$. $\sqrt{2}$ cannot be expressed in this form. So, $p$ is False.

$q$: An irrational number is a real number that is not rational. Since $\sqrt{2}$ is a real number and is not rational, it is irrational. So, $q$ is True.

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(vi) Compound Statement: 24 is a multiple of 2, 4 and 8.

The connecting word is "and". This statement is a conjunction of three implicit component statements.

$p$: 24 is a multiple of 2.

$q$: 24 is a multiple of 4.

$r$: 24 is a multiple of 8.

Check Truth Value:

$p$: A multiple of 2 is an integer $k$ such that $k = 2 \times n$ for some integer $n$. $24 = 2 \times 12$. So, $p$ is True.

$q$: A multiple of 4 is an integer $k$ such that $k = 4 \times n$ for some integer $n$. $24 = 4 \times 6$. So, $q$ is True.

$r$: A multiple of 8 is an integer $k$ such that $k = 8 \times n$ for some integer $n$. $24 = 8 \times 3$. So, $r$ is True.

The negation of a statement $P$ is the statement "It is not the case that $P$".

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(i) Statement: Chennai is the capital of Tamil Nadu.

Negation: It is not the case that Chennai is the capital of Tamil Nadu.

Equivalent phrasing: Chennai is not the capital of Tamil Nadu.

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(ii) Statement: $\sqrt{2}$ is not a complex number.

This statement asserts that $\sqrt{2}$ does not belong to the set of complex numbers.

Negation: It is not the case that $\sqrt{2}$ is not a complex number.

The phrase "not not" cancels out. Equivalent phrasing: $\sqrt{2}$ is a complex number.

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(iii) Statement: All triangles are not equilateral triangle.

This statement is slightly ambiguously phrased, but it most likely means "It is not the case that all triangles are equilateral triangles" or "Not all triangles are equilateral triangles". Assuming this interpretation:

Assuming the original statement is $P$: "Not all triangles are equilateral triangles."

Negation of P: It is not the case that not all triangles are equilateral triangles.

Equivalent phrasing: All triangles are equilateral triangles.

If the original statement is interpreted as "For all triangles, it is not the case that the triangle is equilateral", then the negation would be "There exists a triangle such that it is equilateral". This leads to the same result.

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(iv) Statement: The number 2 is greater than 7.

Negation: It is not the case that the number 2 is greater than 7.

Equivalent phrasing: The number 2 is not greater than 7.

Alternatively, "not greater than" means "less than or equal to": The number 2 is less than or equal to 7.

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(v) Statement: Every natural number is an integer.

This is a universal quantification ("Every... is..."). The negation of "Every A is B" is "There exists an A that is not B".

Negation: It is not the case that every natural number is an integer.

Equivalent phrasing: There exists a natural number which is not an integer.

Two statements are negations of each other if and only if they always have opposite truth values. That is, if one is true, the other must be false, and if one is false, the other must be true.

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(i) Pair of Statements:

Statement 1: The number x is not a rational number.

Statement 2: The number x is not an irrational number.

Analysis:

Let $P$ be the statement "The number x is a rational number".

Statement 1 is the negation of $P$, which is "Not $P$". This means $x$ is irrational (assuming $x$ is a real number).

Let $Q$ be the statement "The number x is an irrational number".

Statement 2 is the negation of $Q$, which is "Not $Q$". This means $x$ is rational (assuming $x$ is a real number).

So, the pair of statements is effectively "x is irrational" and "x is rational".

If "x is irrational" is true, then "x is rational" is false. If "x is irrational" is false (meaning x is rational), then "x is rational" is true.

The two statements in the pair always have opposite truth values.

Conclusion:

Yes, the given pair of statements are negations of each other.

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(ii) Pair of Statements:

Statement 1: The number x is a rational number.

Statement 2: The number x is an irrational number.

Analysis:

Let $P$ be the statement "The number x is a rational number".

Let $Q$ be the statement "The number x is an irrational number".

For any real number $x$, $x$ is either rational or irrational, and it cannot be both. This is the definition of rational and irrational numbers within the real number system.

If Statement 1 ($P$) is true, then $x$ is rational, which means $x$ cannot be irrational. So Statement 2 ($Q$) must be false.

If Statement 2 ($Q$) is true, then $x$ is irrational, which means $x$ cannot be rational. So Statement 1 ($P$) must be false.

If Statement 1 ($P$) is false, then $x$ is not rational, which means $x$ must be irrational. So Statement 2 ($Q$) must be true.

If Statement 2 ($Q$) is false, then $x$ is not irrational, which means $x$ must be rational. So Statement 1 ($P$) must be true.

The two statements always have opposite truth values.

Conclusion:

Yes, the given pair of statements are negations of each other.

A compound statement is formed by combining simple statements using connectives like "and", "or", etc. The simple statements that form the compound statement are its component statements. We will identify the component statements and determine their truth values.

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(i) Compound Statement: Number 3 is prime or it is odd.

The connective is "or". The component statements are:

$p$: Number 3 is prime.

$q$: It is odd. (referring to the number 3)

Check Truth Value:

$p$: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The number 3 is divisible only by 1 and 3. So, $p$ is True.

$q$: An odd number is an integer that is not divisible by 2. The number 3 is not divisible by 2. So, $q$ is True.

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(ii) Compound Statement: All integers are positive or negative.

The connective is "or". The phrasing "All ... are either ... or ..." suggests properties applied to a set. The component statements are:

$p$: All integers are positive.

$q$: All integers are negative.

Check Truth Value:

$p$: Integers include negative numbers (e.g., -1, -5) and zero, which are not positive. So, $p$ is False.

$q$: Integers include positive numbers (e.g., 1, 5) and zero, which are not negative. So, $q$ is False.

Note that the original compound statement "All integers are positive or negative" is also false, because the integer 0 is neither positive nor negative.

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(iii) Compound Statement: 100 is divisible by 3, 11 and 5.

The connective is "and". This statement is a conjunction of three properties applied to the number 100. The component statements are:

$p$: 100 is divisible by 3.

$q$: 100 is divisible by 11.

$r$: 100 is divisible by 5.

Check Truth Value:

$p$: A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of digits of 100 is $1+0+0=1$, which is not divisible by 3. So, $p$ is False.

$q$: To check divisibility by 11, one method is alternating sum of digits. For 100: $0 - 0 + 1 = 1$, which is not divisible by 11. So, $q$ is False.

$r$: A number ending in 0 or 5 is divisible by 5. The number 100 ends in 0. So, $r$ is True.

A compound statement is true if and only if its component statements, connected by "and", are all true.

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(i) Compound Statement: A line is straight and extends indefinitely in both directions.

The connective is "and". The component statements are:

$p$: A line is straight.

$q$: A line extends indefinitely in both directions.

Check Truth Value of Component Statements:

$p$: According to the fundamental postulates of geometry, a line is a straight object. So, $p$ is True.

$q$: A key property of a line in geometry is that it extends infinitely in both directions. So, $q$ is True.

Check Truth Value of Compound Statement:

The compound statement is "$p$ and $q$". Since both $p$ and $q$ are true, the compound statement is True.

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(ii) Compound Statement: 0 is less than every positive integer and every negative integer.

The connective is "and". The component statements are:

$p$: 0 is less than every positive integer.

$q$: 0 is less than every negative integer.

Check Truth Value of Component Statements:

$p$: Positive integers are $1, 2, 3, \dots$. The statement "$0$ is less than every positive integer" ($0 < 1$, $0 < 2$, $0 < 3$, etc.) is true. So, $p$ is True.

$q$: Negative integers are $-1, -2, -3, \dots$. The statement "$0$ is less than every negative integer" ($0 < -1$, $0 < -2$, $0 < -3$, etc.) is false, because $0$ is greater than any negative number. So, $q$ is False.

Check Truth Value of Compound Statement:

The compound statement is "$p$ and $q$". Since $p$ is true and $q$ is false, the compound statement is False.

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(iii) Compound Statement: All living things have two legs and two eyes.

The connective is "and". This universal statement with "and" can be broken down based on the properties mentioned. The component statements are:

$p$: All living things have two legs.

$q$: All living things have two eyes.

Check Truth Value of Component Statements:

$p$: Consider living things like snakes (zero legs), spiders (eight legs), trees (zero legs). Not all living things have exactly two legs. So, $p$ is False.

$q$: Consider living things like single-celled organisms (no eyes), starfish (multiple eyes), many insects (compound eyes), or trees (no eyes). Not all living things have exactly two eyes. So, $q$ is False.

Check Truth Value of Compound Statement:

The compound statement is "$p$ and $q$". Since both $p$ and $q$ are false, the compound statement is False.

The word "or" is used in two senses: inclusive and exclusive.

An inclusive "or" means that the statement is true if at least one of the conditions is met (either one or both). $P \text{ or } Q$ (or both).

An exclusive "or" means that the statement is true if exactly one of the conditions is met (one but not the other). $P \text{ or } Q$ (but not both).

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(i) Statement: To enter a country, you need a passport or a voter registration card.

This sentence implies that having either a passport or a voter registration card is sufficient for entry. It doesn't typically exclude someone who possesses both documents. If having a passport is sufficient, having both a passport and a voter registration card would still allow entry. The "or" means "at least one of these".

Therefore, it is an inclusive "Or".

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(ii) Statement: The school is closed if it is a holiday or a Sunday.

The school is closed if it's a holiday. The school is closed if it's a Sunday. If it happens to be both a holiday and a Sunday, the school is still closed. The condition for the school being closed is satisfied if either or both reasons are true.

Therefore, it is an inclusive "Or".

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(iii) Statement: Two lines intersect at a point or are parallel.

In Euclidean geometry, two distinct lines in a plane can either intersect at exactly one point or be parallel (never intersect). They cannot simultaneously intersect at a point and be parallel. This means that exactly one of these possibilities must occur for any pair of distinct lines.

Therefore, it is an exclusive "Or".

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(iv) Statement: Students can take French or Sanskrit as their third language.

This sentence describes a choice for a single slot (the third language). A student typically selects one language *as* their third language from the options provided; they cannot simultaneously take both French and Sanskrit as *the* third language. The choices are mutually exclusive for that specific academic requirement.

Therefore, it is an exclusive "Or".

The word "or" can be used in an inclusive sense (at least one condition is true, possibly both) or an exclusive sense (exactly one condition is true, not both). We will determine which sense is intended based on the meaning and context of the statement, and then determine the truth value of the compound statement.

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(i) Statement: $\sqrt{2}$ is a rational number or an irrational number.

The two conditions are "$$\sqrt{2}$$ is a rational number" and "$$\sqrt{2}$$ is an irrational number". By the definition of rational and irrational numbers (within the real number system), a number cannot be both rational and irrational; it must be exactly one of them.

Type of "Or": Since the two conditions are mutually exclusive for the number $\sqrt{2}$, the "or" is used in the exclusive sense.

Truth value: Let $p$ be "$\sqrt{2}$ is a rational number" and $q$ be "$\sqrt{2}$ is an irrational number". We know that $p$ is false and $q$ is true. The compound statement is $p \text{ XOR } q$. The truth value of $p \text{ XOR } q$ is true if $p$ is true and $q$ is false, or if $p$ is false and $q$ is true. Since $p$ is false and $q$ is true, the compound statement is True.

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(ii) Statement: To enter into a public library children need an identity card from the school or a letter from the school authorities.

The two conditions are "children need an identity card from the school" and "children need a letter from the school authorities". In a typical scenario for needing identification or permission, if a child has both an identity card and a letter, they would likely be allowed entry. The rule means that possessing at least one of these documents is sufficient.

Type of "Or": Since possessing both items would typically satisfy the requirement, the "or" is used in the inclusive sense.

Truth value: This is a statement about a policy. Assuming this statement accurately reflects the policy of a particular public library where possessing either an identity card or a letter (or both) grants entry, the statement is True in that context.

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(iii) Statement: A rectangle is a quadrilateral or a 5-sided polygon.

The two conditions are "a rectangle is a quadrilateral" and "a rectangle is a 5-sided polygon". A rectangle is a geometric shape with specific properties. It is defined as a four-sided polygon. It cannot simultaneously be a four-sided polygon and a five-sided polygon.

Type of "Or": Since the properties "being a quadrilateral" and "being a 5-sided polygon" are mutually exclusive descriptions for a single shape like a rectangle, the "or" is used in the exclusive sense.

Truth value: Let $p$ be "A rectangle is a quadrilateral" and $q$ be "A rectangle is a 5-sided polygon". We know that $p$ is true and $q$ is false. The compound statement is $p \text{ XOR } q$. The truth value of $p \text{ XOR } q$ is true if $p$ is true and $q$ is false, or if $p$ is false and $q$ is true. Since $p$ is true and $q$ is false, the compound statement is True.

We will identify the connecting word(s) and then list the component statements for each compound statement.

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(i) Compound Statement: All rational numbers are real and all real numbers are not complex.

The connecting word is: and.

The component statements are:

$p$: All rational numbers are real.

$q$: All real numbers are not complex.

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(ii) Compound Statement: Square of an integer is positive or negative.

The connecting word is: or.

The component statements are:

$p$: Square of an integer is positive.

$q$: Square of an integer is negative.

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(iii) Compound Statement: The sand heats up quickly in the Sun and does not cool down fast at night.

The connecting word is: and.

The component statements are:

$p$: The sand heats up quickly in the Sun.

$q$: The sand does not cool down fast at night.

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(iv) Compound Statement: x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.

The connecting word is: and.

The component statements are:

$p$: x = 2 is a root of the equation $3x^2 - x - 10 = 0$.

$q$: x = 3 is a root of the equation $3x^2 - x - 10 = 0$.

A quantifier is a word or phrase that indicates the scope of a statement (e.g., "all", "every", "some", "there exists"). The negation of a quantified statement reverses the type of quantifier and negates the assertion.

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(i) Statement: There exists a number which is equal to its square.

The quantifier is: There exists (Existential Quantifier).

Let the statement be $P$. $P$: $\exists x, x = x^2$.

The negation of an existential statement "There exists an $x$ such that $P(x)$" is the universal statement "For every $x$, it is not the case that $P(x)$".

Negation: It is not the case that there exists a number which is equal to its square.

Equivalent phrasing: For every number, it is not equal to its square.

In symbols: $\neg (\exists x, x = x^2) \equiv \forall x, x \neq x^2$.

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(ii) Statement: For every real number x, x is less than x + 1.

The quantifier is: For every real number x (Universal Quantifier).

Let the statement be $P$. $P$: $\forall x \in \mathbb{R}, x < x + 1$.

The negation of a universal statement "For every $x$, $P(x)$ is true" is the existential statement "There exists an $x$ such that $P(x)$ is false".

Negation: It is not the case that for every real number x, x is less than x + 1.

Equivalent phrasing: There exists a real number x such that x is not less than x + 1.

The condition "x is not less than x + 1" is equivalent to "x is greater than or equal to x + 1" ($x \geq x+1$).

Another equivalent phrasing: There exists a real number x such that x $\ge$ x + 1.

In symbols: $\neg (\forall x \in \mathbb{R}, x < x + 1) \equiv \exists x \in \mathbb{R}, \neg (x < x + 1) \equiv \exists x \in \mathbb{R}, x \ge x + 1$.

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(iii) Statement: There exists a capital for every state in India.

This statement contains two quantifiers, although the "for every" applies to the states and "There exists a capital" applies to the existence of a capital for each state.

The quantifiers are: There exists (Existential) and for every (Universal).

The structure is "For every state s, there exists a capital c such that c is a capital for s". Let $P(s, c)$ be "c is a capital for state s". The statement is $\forall s, \exists c, P(s, c)$.

The negation of $\forall s, \exists c, P(s, c)$ is $\exists s, \neg (\exists c, P(s, c))$, which is equivalent to $\exists s, \forall c, \neg P(s, c)$.

Negation: It is not the case that there exists a capital for every state in India.

Equivalent phrasing: There exists a state in India for which there is no capital.

Another equivalent phrasing: There exists a state in India which has no capital.

In symbols: $\neg (\forall s, \exists c, P(s, c)) \equiv \exists s, \forall c, \neg P(s, c)$.

Two statements are negations of each other if and only if one is true whenever the other is false, and vice versa.

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Statement 1: $x + y = y + x$ is true for every real numbers x and y.

This statement is a universal quantification. It asserts that the property "$x + y = y + x$" holds for all possible pairs of real numbers $(x, y)$.

In mathematics, the property $x + y = y + x$ is the commutative property of addition for real numbers, which is a fundamental axiom (or theorem derived from axioms) of the real number system. This property is indeed true for every pair of real numbers.

Therefore, Statement 1 is True.

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Statement 2: There exists real numbers x and y for which $x + y = y + x$.

This statement is an existential quantification. It asserts that there is at least one pair of real numbers $(x, y)$ for which the property "$x + y = y + x$" holds.

Since the property $x + y = y + x$ is true for *every* pair of real numbers (as established above), it must necessarily be true for at least one pair of real numbers. For example, it is true for $x=1$ and $y=2$, since $1+2 = 3$ and $2+1 = 3$, so $1+2 = 2+1$.

Therefore, Statement 2 is also True.

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Since both Statement 1 and Statement 2 are true, they do not have opposite truth values. For statements to be negations of each other, one must be true when the other is false.

The negation of Statement 1 would be "There exist real numbers x and y for which $x + y \neq y + x$".

The negation of Statement 2 would be "For every real numbers x and y, $x + y \neq y + x$".

Neither of the original statements is the negation of the other.

Conclusion:

No, the given pair of statements are not negations of each other because both statements are true.

The word "or" is used in two senses: inclusive (at least one condition is true, possibly both) and exclusive (exactly one condition is true, not both).

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(i) Statement: Sun rises or Moon sets.

The two events described are the Sun rising and the Moon setting. These two events can happen at the same time (approximately, depending on the observer's location and the phase of the Moon). For example, during certain times of the month, the Moon sets shortly after sunrise.

Since it is possible for both events to occur simultaneously, the compound statement is true if at least one or both of the component events happen.

Therefore, it is an inclusive "Or".

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(ii) Statement: To apply for a driving licence, you should have a ration card or a passport.

This statement specifies conditions for eligibility for a driving licence. The requirement is usually satisfied by providing one acceptable document from a list. If a person possesses both a ration card and a passport, they would certainly meet the requirement. The intention is that having at least one of these documents is sufficient.

Therefore, it is an inclusive "Or".

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(iii) Statement: All integers are positive or negative.

This statement asserts a property about the set of all integers. The property is that each integer belongs to the set of positive numbers or the set of negative numbers.

For any single integer, it is either positive, negative, or zero. A number cannot be both positive and negative simultaneously. So, the properties "being positive" and "being negative" are mutually exclusive for any given non-zero integer.

However, the statement is a universal claim about *all* integers. It asserts that *every* integer falls into one of these two mutually exclusive categories. The existence of the integer 0, which is neither positive nor negative, makes this universal statement false. The "or" here is connecting two mutually exclusive properties, so it is primarily used in an exclusive sense when considering a single integer. But the structure of the sentence applies this disjunction universally.

The fundamental "or" between "positive" and "negative" for a single number is exclusive. The statement claims this exclusive disjunction holds for all integers. The presence of the integer 0, which satisfies neither property, makes the entire statement false and highlights the exclusive nature of the properties being combined.

Therefore, the "Or" used here is exclusive "Or", applied to the classification of individual integers into positive or negative categories.

A conditional statement has the form "If $P$, then $Q$". The contrapositive of this statement is "If not $Q$, then not $P$". A conditional statement and its contrapositive are logically equivalent; they always have the same truth value.

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(i) Statement: If a number is divisible by 9, then it is divisible by 3.

Let $P$: A number is divisible by 9.

Let $Q$: It (the number) is divisible by 3.

The statement is "If $P$, then $Q$".

We need to form "If not $Q$, then not $P$".

Not $Q$: It is not the case that the number is divisible by 3, which means the number is not divisible by 3.

Not $P$: It is not the case that the number is divisible by 9, which means the number is not divisible by 9.

Contrapositive: If a number is not divisible by 3, then it is not divisible by 9.

Note: The original statement is true. Its contrapositive is also true.

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(ii) Statement: If you are born in India, then you are a citizen of India.

Let $P$: You are born in India.

Let $Q$: You are a citizen of India.

The statement is "If $P$, then $Q$".

We need to form "If not $Q$, then not $P$".

Not $Q$: It is not the case that you are a citizen of India, which means you are not a citizen of India.

Not $P$: It is not the case that you are born in India, which means you are not born in India.

Contrapositive: If you are not a citizen of India, then you are not born in India.

Note: The original statement is generally true under many citizenship laws, although there can be exceptions (e.g., citizenship by descent, renouncing citizenship). Assuming a straightforward interpretation, if the original is true, the contrapositive is also true.

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(iii) Statement: If a triangle is equilateral, it is isosceles.

Let $P$: A triangle is equilateral.

Let $Q$: It (the triangle) is isosceles.

The statement is "If $P$, then $Q$".

We need to form "If not $Q$, then not $P$".

Not $Q$: It is not the case that the triangle is isosceles, which means the triangle is not isosceles.

Not $P$: It is not the case that the triangle is equilateral, which means the triangle is not equilateral.

Contrapositive: If a triangle is not isosceles, then it is not equilateral.

Note: The original statement is true because an equilateral triangle (all three sides equal) also satisfies the definition of an isosceles triangle (at least two sides equal). Therefore, its contrapositive is also true.

A conditional statement has the form "If $P$, then $Q$". The converse of this statement is "If $Q$, then $P$". The converse is not logically equivalent to the original statement; its truth value may be different.

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(i) Statement: If a number n is even, then n² is even.

Let $P$: A number $n$ is even.

Let $Q$: $n^2$ is even.

The statement is "If $P$, then $Q$".

The converse is "If $Q$, then $P$".

Converse: If n² is even, then the number n is even.

Note: The original statement is true. The converse is also true.

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(ii) Statement: If you do all the exercises in the book, you get an A grade in the class.

Let $P$: You do all the exercises in the book.

Let $Q$: You get an A grade in the class.

The statement is "If $P$, then $Q$".

The converse is "If $Q$, then $P$".

Converse: If you get an A grade in the class, then you did all the exercises in the book.

Note: The original statement implies that doing all exercises is a sufficient condition for an A grade. The converse implies that getting an A grade is a sufficient condition for having done all exercises. These are generally not the same; other factors like tests or participation might contribute to the grade, and one might get an A without doing all exercises, or do all exercises and not get an A.

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(iii) Statement: If two integers a and b are such that a > b, then a – b is always a positive integer.

Let $P$: Two integers $a$ and $b$ are such that $a > b$.

Let $Q$: $a - b$ is always a positive integer.

The statement is "If $P$, then $Q$".

The converse is "If $Q$, then $P$".

Converse: If two integers a and b are such that a – b is always a positive integer, then a > b.

Note: The original statement is true. The converse seems circular due to the phrasing "always a positive integer" in the premise $Q$. A more precise way to phrase $Q$ for the converse would be "for two integers a and b, a - b is a positive integer". Then the converse is "If for two integers a and b, a - b is a positive integer, then a > b". This converse is also true.

A conditional statement of the form "If $P$, then $Q$" consists of the hypothesis $P$ and the conclusion $Q$. We will identify these component statements and determine their truth values, and then determine the truth value of the entire conditional statement.

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(i) Compound Statement: If a triangle ABC is equilateral, then it is isosceles.

This is a conditional statement of the form "If $P$, then $Q$".

Component statement (Hypothesis): $P$: A triangle ABC is equilateral.

Component statement (Conclusion): $Q$: It (the triangle ABC) is isosceles.

Check Truth Value of Component Statements:

The truth value of $P$ and $Q$ depends on the specific triangle ABC being considered. However, the statement is a general geometric property. To evaluate the truth of the conditional statement "If $P$, then $Q$", we consider whether it is impossible for $P$ to be true while $Q$ is false.

A triangle is equilateral if all three of its sides are equal in length. A triangle is isosceles if at least two of its sides are equal in length.

If a triangle is equilateral, then all three sides are equal. If all three sides are equal, it necessarily has at least two sides equal. Thus, every equilateral triangle is also an isosceles triangle.

This means that whenever $P$ is true (a triangle is equilateral), $Q$ is also true (the triangle is isosceles). There is no case where $P$ is true and $Q$ is false.

Check Truth Value of Compound Statement:

A conditional statement "If $P$, then $Q$" is true unless $P$ is true and $Q$ is false. Since this scenario is impossible for these component statements, the compound statement is True.

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(ii) Compound Statement: If a and b are integers, then ab is a rational number.

This is a conditional statement of the form "If $P$, then $Q$".

Component statement (Hypothesis): $P$: a and b are integers.

Component statement (Conclusion): $Q$: ab is a rational number.

Check Truth Value of Component Statements:

The truth value of $P$ depends on the specific values of $a$ and $b$ being considered. However, the statement is a general claim about the properties of integers and rational numbers. We evaluate the truth of the conditional statement by considering whether it is impossible for $P$ to be true while $Q$ is false.

If $a$ and $b$ are integers (meaning $P$ is true), their product $ab$ is also an integer. An integer is a rational number because any integer $n$ can be written as $\frac{n}{1}$, which is in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. Thus, if $ab$ is an integer, it is also a rational number.

This means that whenever $P$ is true (a and b are integers), $Q$ is also true ($ab$ is a rational number). There is no case where $P$ is true and $Q$ is false.

Check Truth Value of Compound Statement:

A conditional statement "If $P$, then $Q$" is true unless $P$ is true and $Q$ is false. Since this scenario is impossible for these component statements, the compound statement is True.

The phrase "if and only if" (often abbreviated as "iff" or $\Leftrightarrow$) is used to combine two conditional statements, where one is the converse of the other. A statement "$P$ if and only if $Q$" is a biconditional statement that means both "If $P$, then $Q$" and "If $Q$, then $P$" are true.

In the given pairs, statement $q$ is the converse of statement $p$. So, combining them using "if and only if" creates a single statement that asserts the logical equivalence of the hypothesis and conclusion from statement $p$ (or $q$).

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(i) Statement p: If a rectangle is a square, then all its four sides are equal.

Let $A$: A rectangle is a square.

Let $B$: All its four sides are equal.

Statement p is "If $A$, then $B$".

Statement q: If all the four sides of a rectangle are equal, then the rectangle is a square.

Statement q is "If $B$, then $A$".

Combining "If $A$, then $B$" and "If $B$, then $A$" using "if and only if" gives "$A$ if and only if $B$".

Combined Statement: A rectangle is a square if and only if all its four sides are equal.

Alternatively, reversing the order: All four sides of a rectangle are equal if and only if the rectangle is a square.

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(ii) Statement p: If the sum of digits of a number is divisible by 3, then the number is divisible by 3.

Let $C$: The sum of digits of a number is divisible by 3.

Let $D$: The number is divisible by 3.

Statement p is "If $C$, then $D$".

Statement q: If a number is divisible by 3, then the sum of its digits is divisible by 3.

Statement q is "If $D$, then $C$".

Combining "If $C$, then $D$" and "If $D$, then $C$" using "if and only if" gives "$C$ if and only if $D$".

Combined Statement: The sum of digits of a number is divisible by 3 if and only if the number is divisible by 3.

Alternatively, reversing the order: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The given statement is a conditional statement of the form "If $P$, then $Q$", where:

$P$: A natural number is odd.

$Q$: Its square is also odd.

The meaning of "If $P$, then $Q$" is that $P$ implies $Q$, or $Q$ is a necessary condition for $P$, or $P$ is a sufficient condition for $Q$. We need to rephrase "If a natural number is odd, then its square is also odd" in five different ways.

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Here are five different ways to phrase the statement "If a natural number is odd, then its square is also odd" while conveying the same meaning:

1. Using "implies":

A natural number being odd implies that its square is also odd.

2. Using "unless" (as "P unless not Q" is equivalent to "If not (not Q), then not P", which simplifies to "If Q, then not P" - this is not the original statement. However, "P unless Q" can sometimes imply "If not Q then P". The phrasing "If P then Q" can be rewritten as "P is true unless Q is false"):

A natural number is odd unless its square is even. (This means if the square is NOT even, the number MUST be odd, which is the contrapositive and has the same meaning). More formally, "If P then Q" is equivalent to "Not Q implies Not P". So "P unless not Q" is equivalent to "If Not(not Q) then Not P", which is "If Q then Not P". This doesn't quite work directly. A simpler phrasing using 'unless' might be interpreted as 'if it's not the case that Q, then it must be that P'. Let's rephrase based on 'P is sufficient for Q' or 'Q is necessary for P'.

Let's use standard logical equivalences for "If P, then Q":

• $P \implies Q$
• $\neg P \vee Q$ (Not P or Q)
• $\neg Q \implies \neg P$ (Contrapositive)

Let's try rephrasing the statement in terms of necessary and sufficient conditions:

$P$: A natural number is odd.

$Q$: Its square is odd.

"If P, then Q" means P is sufficient for Q, and Q is necessary for P.

1. Using "implies":

A natural number being odd implies that its square is odd.

2. Using "sufficient":

A natural number being odd is a sufficient condition for its square to be odd.

3. Using "necessary":

A natural number's square being odd is a necessary condition for the natural number to be odd. (This is the converse of the contrapositive, equivalent to the original statement).

4. Using "whenever":

Whenever a natural number is odd, its square is also odd.

5. Using the contrapositive form ("If not Q, then not P"):

Not Q: Its square is not odd (i.e., its square is even).

Not P: A natural number is not odd (i.e., a natural number is even).

Contrapositive: If the square of a natural number is even, then the natural number is even.

Given a conditional statement "If $P$, then $Q$":

• The contrapositive is "If not $Q$, then not $P$".
• The converse is "If $Q$, then $P$".

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(i) Statement: If x is a prime number, then x is odd.

Let $P$: x is a prime number.

Let $Q$: x is odd.

Not $P$: x is not a prime number.

Not $Q$: x is not odd (i.e., x is even).

Contrapositive: If x is not odd, then x is not a prime number. (Or: If x is even, then x is not a prime number.)

Converse: If x is odd, then x is a prime number.

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(ii) Statement: If the two lines are parallel, then they do not intersect in the same plane.

Let $P$: The two lines are parallel.

Let $Q$: They (the two lines) do not intersect in the same plane.

Not $P$: The two lines are not parallel.

Not $Q$: It is not the case that they do not intersect in the same plane, which means they intersect in the same plane.

Contrapositive: If the two lines intersect in the same plane, then they are not parallel.

Converse: If the two lines do not intersect in the same plane, then they are parallel.

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(iii) Statement: Something is cold implies that it has low temperature.

Rewrite as "If something is cold, then it has low temperature."

Let $P$: Something is cold.

Let $Q$: It (something) has low temperature.

Not $P$: Something is not cold.

Not $Q$: It (something) does not have low temperature.

Contrapositive: If something does not have low temperature, then it is not cold.

Converse: If something has low temperature, then it is cold.

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(iv) Statement: You cannot comprehend geometry if you do not know how to reason deductively.

Rewrite as "If you do not know how to reason deductively, then you cannot comprehend geometry."

Let $P$: You do not know how to reason deductively.

Let $Q$: You cannot comprehend geometry.

Not $P$: It is not the case that you do not know how to reason deductively, which means you know how to reason deductively.

Not $Q$: It is not the case that you cannot comprehend geometry, which means you can comprehend geometry.

Contrapositive: If you can comprehend geometry, then you know how to reason deductively.

Converse: If you cannot comprehend geometry, then you do not know how to reason deductively.

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(v) Statement: x is an even number implies that x is divisible by 4.

Rewrite as "If x is an even number, then x is divisible by 4."

Let $P$: x is an even number.

Let $Q$: x is divisible by 4.

Not $P$: x is not an even number (i.e., x is odd).

Not $Q$: x is not divisible by 4.

Contrapositive: If x is not divisible by 4, then x is not an even number. (Or: If x is not divisible by 4, then x is odd.)

Converse: If x is divisible by 4, then x is an even number.

A statement in the form "If $P$, then $Q$" means that $P$ is the condition (hypothesis) and $Q$ is the result (conclusion). Other phrases can imply this structure, such as "$P$ implies $Q$", "$P$ whenever $Q$", "$Q$ if $P$", "$Q$ is a necessary condition for $P$", or "$P$ is a sufficient condition for $Q$".

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(i) Statement: You get a job implies that your credentials are good.

The word "implies" directly indicates the "if-then" structure. The part before "implies" is the hypothesis, and the part after is the conclusion.

Hypothesis ($P$): You get a job.

Conclusion ($Q$): Your credentials are good.

If-Then Form: If you get a job, then your credentials are good.

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(ii) Statement: The Banana trees will bloom if it stays warm for a month.

The word "if" introduces the condition (hypothesis), even though it appears later in the sentence. The part that follows "if" is the hypothesis, and the other part is the conclusion.

Hypothesis ($P$): It stays warm for a month.

Conclusion ($Q$): The Banana trees will bloom.

If-Then Form: If it stays warm for a month, then the Banana trees will bloom.

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(iii) Statement: A quadrilateral is a parallelogram if its diagonals bisect each other.

Again, the word "if" introduces the condition. The part that follows "if" is the hypothesis, and the other part is the conclusion.

Hypothesis ($P$): Its diagonals bisect each other. (The "its" refers to "A quadrilateral").

Conclusion ($Q$): A quadrilateral is a parallelogram.

If-Then Form: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

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(iv) Statement: To get an A+ in the class, it is necessary that you do all the exercises of the book.

The phrase "it is necessary that Q" means "If P, then Q". The necessary condition is the conclusion ($Q$), and the situation for which it is necessary is the hypothesis ($P$).

$Q$: You do all the exercises of the book.

$P$: You get an A+ in the class.

If-Then Form: If you get an A+ in the class, then you do all the exercises of the book.

Given a statement "If $P$, then $Q$":

• The contrapositive is "If not $Q$, then not $P$".
• The converse is "If $Q$, then $P$".

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(a) Original Statement: If you live in Delhi, then you have winter clothes.

Let $P$: You live in Delhi.

Let $Q$: You have winter clothes.

Not $P$: You do not live in Delhi.

Not $Q$: You do not have winter clothes.

Check the statements in (a):

(i) Statement (a)(i): If you do not have winter clothes, then you do not live in Delhi.

This statement is of the form "If not $Q$, then not $P$". This is the contrapositive of the original statement (a).

(ii) Statement (a)(ii): If you have winter clothes, then you live in Delhi.

This statement is of the form "If $Q$, then $P$". This is the converse of the original statement (a).

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(b) Original Statement: If a quadrilateral is a parallelogram, then its diagonals bisect each other.

Let $P$: A quadrilateral is a parallelogram.

Let $Q$: Its diagonals bisect each other. (The "its" refers to "a quadrilateral").

Not $P$: A quadrilateral is not a parallelogram.

Not $Q$: Its diagonals do not bisect each other.

Check the statements in (b):

(i) Statement (b)(i): If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.

This statement is of the form "If not $Q$, then not $P$". This is the contrapositive of the original statement (b).

(ii) Statement (b)(ii): If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

This statement is of the form "If $Q$, then $P$". This is the converse of the original statement (b).

Given Statement:

If x, y ∈ Z are such that x and y are odd, then xy is odd.

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To Check:

Whether the statement is true or false.

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Solution:

The statement is a conditional statement of the form "If $P$, then $Q$".

Hypothesis ($P$): x and y are integers and both x and y are odd.

Conclusion ($Q$): xy is odd.

To check if the conditional statement is true, we need to determine if it is impossible for the hypothesis $P$ to be true while the conclusion $Q$ is false.

Assume the hypothesis $P$ is true. This means $x$ and $y$ are odd integers.

An integer is odd if it can be written in the form $2k + 1$ for some integer $k$.

So, if $x$ is odd, we can write $x = 2k_1 + 1$ for some integer $k_1$.

If $y$ is odd, we can write $y = 2k_2 + 1$ for some integer $k_2$.

Now, consider the product $xy$:

$xy = (2k_1 + 1)(2k_2 + 1)$

Expand the product:

$xy = 2k_1(2k_2 + 1) + 1(2k_2 + 1)$

$xy = 4k_1k_2 + 2k_1 + 2k_2 + 1$

Factor out 2 from the first three terms:

$xy = 2(2k_1k_2 + k_1 + k_2) + 1$

Let $K = 2k_1k_2 + k_1 + k_2$. Since $k_1$ and $k_2$ are integers, $K$ is also an integer.

So, $xy = 2K + 1$, where $K$ is an integer.

This means that $xy$ has the form $2K + 1$, which is the definition of an odd integer.

Thus, whenever the hypothesis $P$ is true (x and y are odd integers), the conclusion $Q$ is also true (xy is odd).

There is no case where the hypothesis is true and the conclusion is false.

Therefore, the conditional statement "If x, y ∈ Z are such that x and y are odd, then xy is odd" is true.

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Conclusion:

The statement is True.

Given Statement:

If x, y ∈ Z such that xy is odd, then both x and y are odd.

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To Check:

Whether the statement is true or false by proving its contrapositive.

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Solution:

The given statement is a conditional statement of the form "If $P$, then $Q$".

Hypothesis ($P$): x, y ∈ Z and xy is odd.

Conclusion ($Q$): Both x and y are odd.

The contrapositive of "If $P$, then $Q$" is "If not $Q$, then not $P$".

Not $Q$: It is not the case that both x and y are odd. This means at least one of x or y is not odd (i.e., at least one of x or y is even).

Not $P$: It is not the case that (x, y ∈ Z and xy is odd). Since x, y ∈ Z is given in the premise, this part is assumed to be true. So, not $P$ means (x, y ∈ Z and it is not the case that xy is odd). Not $P$ simplifies to (x, y ∈ Z and xy is not odd), which means (x, y ∈ Z and xy is even).

So, the contrapositive statement is: If at least one of x or y is even, then xy is even.

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Proving the Contrapositive:

Assume the hypothesis of the contrapositive is true: At least one of x or y is even.

Case 1: x is even. (y can be any integer, even or odd).

If x is even, then x can be written in the form $2k$ for some integer $k$.

Consider the product xy: $xy = (2k)y = 2(ky)$.

Let $K = ky$. Since $k$ and $y$ are integers, $K$ is also an integer.

So, $xy = 2K$, which is the definition of an even integer.

Thus, if x is even, then xy is even.

Case 2: y is even. (x can be any integer, even or odd).

If y is even, then y can be written in the form $2m$ for some integer $m$.

Consider the product xy: $xy = x(2m) = 2(xm)$.

Let $M = xm$. Since $x$ and $m$ are integers, $M$ is also an integer.

So, $xy = 2M$, which is the definition of an even integer.

Thus, if y is even, then xy is even.

If at least one of x or y is even, then either Case 1 or Case 2 is true (or both, if both are even). In either case, we have shown that the product xy is even.

This proves the contrapositive statement: "If at least one of x or y is even, then xy is even".

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Conclusion:

Since the contrapositive statement "If at least one of x or y is even, then xy is even" is proven to be true, the original statement "If x, y ∈ Z such that xy is odd, then both x and y are odd" which is logically equivalent to its contrapositive, must also be true.

Therefore, the statement is True.

Statement to Prove (p): $\sqrt{7}$ is irrational.

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Method of Contradiction:

To prove a statement $P$ by contradiction, we assume the negation of $P$ ($\neg P$) is true and show that this assumption leads to a contradiction (a statement that is logically false). Since the assumption $\neg P$ leads to a contradiction, $\neg P$ must be false, which means $P$ must be true.

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Proof:

Assume the negation of statement p is true. The negation of "$\sqrt{7}$ is irrational" is "$\sqrt{7}$ is rational".

Assume $\sqrt{7}$ is rational.

By the definition of a rational number, if $\sqrt{7}$ is rational, it can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and the fraction $\frac{a}{b}$ is in its simplest form (i.e., $a$ and $b$ have no common factors other than 1; $\text{gcd}(a, b) = 1$).

where $a, b \in \mathbb{Z}$, $b \neq 0$, and $\text{gcd}(a, b) = 1$.

Square both sides of the equation (1):

$(\sqrt{7})^2 = \left(\frac{a}{b}\right)^2$

$7 = \frac{a^2}{b^2}$

Multiply both sides by $b^2$:

Equation (2) shows that $a^2$ is a multiple of 7. This means $a^2$ is divisible by 7.

A property of prime numbers (like 7) is that if a prime number $p$ divides a square ($a^2$), then $p$ must divide the base ($a$). Since 7 is a prime number and $7$ divides $a^2$, it must be that $7$ divides $a$.

If $7$ divides $a$, then $a$ can be written as $7k$ for some integer $k$.

Substitute equation (3) into equation (2):

$7b^2 = (7k)^2$

$7b^2 = 49k^2$

Divide both sides by 7:

Equation (4) shows that $b^2$ is a multiple of 7. This means $b^2$ is divisible by 7.

Again, since 7 is a prime number and $7$ divides $b^2$, it must be that $7$ divides $b$.

So, we have concluded that $7$ divides $a$ (from $a=7k$) and $7$ divides $b$ (from $b^2=7k^2$).

This means that $a$ and $b$ have a common factor of 7.

However, our initial assumption was that $\frac{a}{b}$ is in its simplest form, meaning $a$ and $b$ have no common factors other than 1 ($\text{gcd}(a, b) = 1$).

The conclusion that $a$ and $b$ have a common factor of 7 contradicts the assumption that $\text{gcd}(a, b) = 1$.

This contradiction arose from our initial assumption that $\sqrt{7}$ is rational.

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Conclusion:

Since the assumption that $\sqrt{7}$ is rational leads to a contradiction, the assumption must be false. Therefore, the original statement, "$\sqrt{7}$ is irrational", must be true.

Hence, $\sqrt{7}$ is irrational. The statement is True.

Given Statement:

If n is an odd integer, then n is prime.

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To Show:

The statement is false by giving a counterexample.

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Solution:

A counterexample to a conditional statement "If $P$, then $Q$" is a situation where the hypothesis $P$ is true, but the conclusion $Q$ is false.

In this statement:

Hypothesis ($P$): n is an odd integer.

Conclusion ($Q$): n is prime.

We need to find an odd integer $n$ such that $n$ is not prime.

Recall the definitions:

• An odd integer is an integer not divisible by 2.
• A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's consider some odd integers greater than 1 (since prime numbers must be greater than 1):

• If $n=3$: 3 is odd. Is 3 prime? Yes, its only positive divisors are 1 and 3. (This is consistent with the statement).
• If $n=5$: 5 is odd. Is 5 prime? Yes, its only positive divisors are 1 and 5. (This is consistent with the statement).
• If $n=7$: 7 is odd. Is 7 prime? Yes, its only positive divisors are 1 and 7. (This is consistent with the statement).
• If $n=9$: 9 is odd. Is 9 prime? No, its positive divisors are 1, 3, and 9. Since it has a divisor other than 1 and itself (namely 3), 9 is not prime.

We have found an integer, $n=9$, for which the hypothesis "n is an odd integer" is true (9 is odd), but the conclusion "n is prime" is false (9 is not prime).

This serves as a counterexample.

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Counterexample:

Let $n = 9$.

Check the hypothesis: Is $n=9$ an odd integer? Yes, 9 is not divisible by 2.

Check the conclusion: Is $n=9$ prime? No, because 9 is divisible by 3 (and $3 \neq 1$ and $3 \neq 9$).

Since there exists an odd integer (9) that is not prime, the statement "If n is an odd integer, then n is prime" is false.

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Conclusion:

The statement "If n is an odd integer, then n is prime" is False. A counterexample is $n = 9$.

Given Statement (p): If x is a real number such that $x^3 + 4x = 0$, then x is 0.

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This is a conditional statement of the form "If $P$, then $Q$".

Hypothesis ($P$): x is a real number such that $x^3 + 4x = 0$.

Conclusion ($Q$): x is 0.

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(i) Method: Direct Proof

Assume the hypothesis $P$ is true and logically deduce that the conclusion $Q$ must also be true.

Assume x is a real number such that $x^3 + 4x = 0$.

We need to show that this implies $x=0$.

Start with the given equation:

$x^3 + 4x = 0$

Factor out the common term $x$ from the left side:

$x(x^2 + 4) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

So, either $x = 0$ or $x^2 + 4 = 0$.

Consider the second possibility, $x^2 + 4 = 0$.

$x^2 = -4$

We are given that x is a real number. The square of any real number is always non-negative ($x^2 \ge 0$).

The equation $x^2 = -4$ has no real solutions, since $-4$ is negative.

Therefore, the only possibility is $x = 0$.

Thus, if $x$ is a real number such that $x^3 + 4x = 0$, it must be that $x = 0$.

The conclusion $Q$ ($x$ is 0) is logically deduced from the hypothesis $P$ ($x^3 + 4x = 0$).

Hence, the statement is true by the direct method.

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(ii) Method: Proof by Contradiction

Assume the negation of the entire statement is true. The negation of "If $P$, then $Q$" is "$P$ and not $Q$".

Assume the negation is true: x is a real number such that $x^3 + 4x = 0$ (Hypothesis $P$ is true) AND x is not 0 (Conclusion $Q$ is false).

So, we assume x is a real number, $x^3 + 4x = 0$, and $x \neq 0$.

From the equation $x^3 + 4x = 0$, we factor out $x$: $x(x^2 + 4) = 0$.

Since the product is zero, at least one factor must be zero. This means either $x = 0$ or $x^2 + 4 = 0$.

We assumed that $x \neq 0$.

This leaves only the possibility $x^2 + 4 = 0$.

$x^2 = -4$

We are given that x is a real number. The square of any real number is non-negative ($x^2 \ge 0$).

The statement $x^2 = -4$ implies that a non-negative number is equal to a negative number, which is a contradiction.

The contradiction arises from our assumption that the negation of the original statement is true (i.e., $P$ is true and $Q$ is false).

Since the assumption leads to a contradiction, the assumption must be false. Therefore, the original statement "If x is a real number such that $x^3 + 4x = 0$, then x is 0" must be true.

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(iii) Method: Proof by Contrapositive

The contrapositive of the statement "If $P$, then $Q$" is "If not $Q$, then not $P$". The original statement is true if and only if its contrapositive is true.

Original Statement: If x is a real number such that $x^3 + 4x = 0$ ($P$), then x is 0 ($Q$).

Not $Q$: x is not 0 (i.e., $x \neq 0$).

Not $P$: It is not the case that (x is a real number such that $x^3 + 4x = 0$). Since "x is a real number" is part of the universal set we are considering, Not P becomes "x is a real number and $x^3 + 4x \neq 0$".

Contrapositive Statement: If x is a real number such that x is not 0, then $x^3 + 4x \neq 0$.

Assume the hypothesis of the contrapositive is true: x is a real number and $x \neq 0$.

We need to show that this implies $x^3 + 4x \neq 0$.

Consider the expression $x^3 + 4x$. Factor out $x$: $x(x^2 + 4)$.

We are assuming $x \neq 0$.

Consider the second factor, $x^2 + 4$. Since x is a real number, $x^2 \ge 0$.

Therefore, $x^2 + 4 \ge 0 + 4 = 4$.

So, $x^2 + 4$ is always a positive number (greater than or equal to 4). In particular, $x^2 + 4 \neq 0$.

We have a product of two factors: $x$ and $(x^2 + 4)$.

We assumed $x \neq 0$.

We have shown that $x^2 + 4 \neq 0$ for any real number $x$.

Since neither factor is equal to zero, their product cannot be zero.

$x(x^2 + 4) \neq 0$

Thus, $x^3 + 4x \neq 0$.

The conclusion of the contrapositive ($x^3 + 4x \neq 0$) is logically deduced from the hypothesis of the contrapositive ($x \neq 0$).

Hence, the contrapositive statement is true. Since the original statement is logically equivalent to its contrapositive, the original statement is also true.

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In all three methods, we have shown that the given statement is true.

Given Statement:

For any real numbers a and b, if $a^2 = b^2$, then $a = b$.

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To Show:

The statement is not true by giving a counterexample.

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Solution:

The given statement is a universal conditional statement of the form "For any a and b, if $P(a, b)$, then $Q(a, b)$".

Hypothesis ($P(a, b)$): $a$ and $b$ are real numbers such that $a^2 = b^2$.

Conclusion ($Q(a, b)$): $a = b$.

To show that this universal statement is false, we need to find a specific instance (a pair of real numbers $a$ and $b$) for which the hypothesis $P(a, b)$ is true, but the conclusion $Q(a, b)$ is false.

We need to find real numbers $a$ and $b$ such that:

1. $a^2 = b^2$ (Hypothesis is true)

2. $a \neq b$ (Conclusion is false)

Let's consider some pairs of real numbers.

Consider $a=2$ and $b=2$. $a^2 = 2^2 = 4$, $b^2 = 2^2 = 4$. So $a^2 = b^2$ is true. Also, $a = b$ is true. This pair does not serve as a counterexample.

Consider $a=2$ and $b=-2$.

Check the hypothesis: $a^2 = (2)^2 = 4$. $b^2 = (-2)^2 = 4$. So, $a^2 = b^2$ is true.

Check the conclusion: Is $a = b$? $2 \neq -2$. So, $a = b$ is false.

We have found a pair of real numbers $a=2$ and $b=-2$ for which the hypothesis ($a^2 = b^2$) is true, but the conclusion ($a = b$) is false.

This pair serves as a counterexample to the statement.

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Counterexample:

Let $a = 2$ and $b = -2$.

These are real numbers.

Check the hypothesis: $a^2 = 2^2 = 4$ and $b^2 = (-2)^2 = 4$. Thus, $a^2 = b^2$ is true.

Check the conclusion: $a = 2$ and $b = -2$. Thus, $a \neq b$, so the conclusion $a = b$ is false.

Since we found a case where the hypothesis is true and the conclusion is false, the statement "For any real numbers a and b, $a^2 = b^2$ implies that $a = b$" is false.

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Conclusion:

The statement is False. A counterexample is $a = 2$, $b = -2$.

More generally, any pair of real numbers $a$ and $b$ such that $b = -a$ and $a \neq 0$ will be a counterexample (e.g., $a=5, b=-5$ or $a=-3, b=3$).

Given Statement (p): If x is an integer and $x^2$ is even, then x is also even.

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To Show:

The statement is true by the method of contrapositive.

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Solution:

The given statement is a conditional statement of the form "If $P$, then $Q$".

Hypothesis ($P$): x is an integer and $x^2$ is even.

Conclusion ($Q$): x is even.

The contrapositive of "If $P$, then $Q$" is "If not $Q$, then not $P$".

Not $Q$: It is not the case that x is even, which means x is odd.

Not $P$: It is not the case that (x is an integer and $x^2$ is even). Assuming x is an integer (as per the statement's domain), this becomes "It is not the case that $x^2$ is even", which means $x^2$ is odd.

So, the contrapositive statement is: If x is an integer and x is odd, then x² is odd.

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Proving the Contrapositive:

Assume the hypothesis of the contrapositive is true: x is an integer and x is odd.

By the definition of an odd integer, if x is odd, it can be written in the form $2k + 1$ for some integer $k$.

Now, consider the square of x, $x^2$:

$x^2 = (2k + 1)^2$

Expand the expression using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$x^2 = (2k)^2 + 2(2k)(1) + 1^2$

$x^2 = 4k^2 + 4k + 1$

Factor out 2 from the first two terms:

$x^2 = 2(2k^2 + 2k) + 1$

Let $m = 2k^2 + 2k$. Since $k$ is an integer, $2k^2$ is an integer and $2k$ is an integer, so their sum $m$ is also an integer.

So, $x^2 = 2m + 1$, where $m$ is an integer.

This means that $x^2$ has the form $2m + 1$, which is the definition of an odd integer.

Thus, if x is an integer and x is odd, then $x^2$ is odd. The conclusion of the contrapositive ($x^2$ is odd) is logically deduced from the hypothesis of the contrapositive (x is odd).

Hence, the contrapositive statement "If x is an integer and x is odd, then $x^2$ is odd" is true.

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Conclusion:

Since the contrapositive statement is true, the original statement "If x is an integer and $x^2$ is even, then x is also even" is also true (because a conditional statement and its contrapositive are logically equivalent).

Therefore, the statement is True.

To show that a universal statement or a negative existential statement is false, we provide a counterexample. A counterexample is a specific case that contradicts the statement.

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(i) Statement (p): If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.

This is a conditional statement "If $P$, then $Q$".

Hypothesis ($P$): All the angles of a triangle are equal.

Conclusion ($Q$): The triangle is an obtuse angled triangle.

A counterexample requires the hypothesis $P$ to be true and the conclusion $Q$ to be false.

Consider a triangle where all angles are equal. The sum of the interior angles of a triangle is 180 degrees. If all three angles are equal, each angle must be $\frac{180^\circ}{3} = 60^\circ$.

A triangle with all angles equal to $60^\circ$ is an equilateral triangle. So, the hypothesis "$$All the angles of a triangle are equal$$" is true for an equilateral triangle.

Now, check the conclusion for this triangle: Is an equilateral triangle (with angles $60^\circ, 60^\circ, 60^\circ$) an obtuse angled triangle?

An obtuse angled triangle is a triangle with one angle greater than $90^\circ$. In an equilateral triangle, all angles are $60^\circ$, which is not greater than $90^\circ$. Therefore, an equilateral triangle is not an obtuse angled triangle; it is an acute angled triangle (all angles are less than $90^\circ$). The conclusion "$$the triangle is an obtuse angled triangle$$" is false for an equilateral triangle.

We have found a triangle (an equilateral triangle) for which the hypothesis is true but the conclusion is false.

Counterexample:

Consider an equilateral triangle.

Check hypothesis: All angles of an equilateral triangle are $60^\circ$, so they are all equal. Hypothesis is true.

Check conclusion: An obtuse angled triangle has an angle greater than $90^\circ$. An equilateral triangle has no such angle. Conclusion is false.

Since we found a case where the hypothesis is true and the conclusion is false, the statement is false.

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(ii) Statement (q): The equation $x^2 – 1 = 0$ does not have a root lying between 0 and 2.

This statement is of the form "It is not the case that the equation $x^2 - 1 = 0$ has a root lying between 0 and 2".

To show this negative statement is false, we need to show that its negation is true. The negation is "The equation $x^2 - 1 = 0$ has a root lying between 0 and 2".

So, we need to find a root of the equation $x^2 - 1 = 0$ and check if it lies between 0 and 2. If we find such a root, it will be a counterexample to the original statement (q).

Consider the equation $x^2 - 1 = 0$.

$x^2 = 1$

Taking the square root of both sides, we get two real roots:

$x = \pm \sqrt{1}$

$x = 1$ or $x = -1$

Now, check if either of these roots lies between 0 and 2.

• For $x=1$: Is 1 between 0 and 2? Yes, $0 < 1 < 2$.
• For $x=-1$: Is -1 between 0 and 2? No, $-1$ is not greater than 0.

We have found a root of the equation, $x=1$, which lies between 0 and 2.

The existence of this root contradicts the statement (q) that "the equation does not have a root lying between 0 and 2".

Counterexample:

Consider the root $x = 1$ of the equation $x^2 - 1 = 0$.

Check if it is a root: $1^2 - 1 = 1 - 1 = 0$. Yes, it is a root.

Check if it lies between 0 and 2: $0 < 1 < 2$. Yes, it lies between 0 and 2.

The existence of this root ($x=1$) shows that the statement (q) is false.

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Conclusion:

Statement (p) is False. A counterexample is an equilateral triangle.

Statement (q) is False. A counterexample (a root between 0 and 2) is $x=1$.

We will determine the truth value of each statement and provide a reason.

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(i) Statement (p): Each radius of a circle is a chord of the circle.

A radius of a circle is a line segment from the center to a point on the circle. A chord of a circle is a line segment connecting two points on the circle.

For a line segment to be a chord, its endpoints must lie on the circle. A radius has one endpoint at the center (which is not on the circle) and the other endpoint on the circle.

Since a radius connects the center (not on the circle) to a point on the circle, it does not connect two points on the circle (unless the "two points" can be considered as the same point, but the definition implies distinct endpoints). Therefore, a radius does not fit the definition of a chord.

However, a diameter is a chord that passes through the center. A diameter consists of two radii joined end-to-end. While a diameter is a chord, a single radius is not.

Truth Value: False.

Reason: A chord connects two points on the circle. A radius connects the center (not on the circle) to a point on the circle. Therefore, a radius is not a chord.

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(ii) Statement (q): The centre of a circle bisects each chord of the circle.

To bisect a chord means to divide it into two equal segments. The center of a circle only lies on chords that are diameters.

Consider a chord that does not pass through the center of the circle. The center is not on this chord at all, let alone bisecting it.

For example, draw a circle and a chord that is not a diameter. The center will not lie on the chord, so it cannot bisect it.

Truth Value: False.

Reason: The center of a circle only bisects chords that pass through the center (i.e., diameters). For any other chord, the center does not lie on the chord, and thus does not bisect it.

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(iii) Statement (r): Circle is a particular case of an ellipse.

An ellipse is defined as the set of all points in a plane such that the sum of the distances from two fixed points (foci) is constant. The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes.

A circle is defined as the set of all points in a plane that are at a fixed distance (radius) from a fixed point (center). The equation of a circle centered at the origin with radius $r$ is $x^2 + y^2 = r^2$, or $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$.

Comparing the equations, if we set $a = b = r$ in the equation of an ellipse, we get the equation of a circle. In this case, the two foci of the ellipse coincide at the center of the circle.

Thus, a circle is a special type of ellipse where the two foci are at the same location and the semi-major and semi-minor axes are equal (and equal to the radius).

Truth Value: True.

Reason: A circle is an ellipse where the two foci coincide and the semi-major and semi-minor axes are equal (equal to the radius). The equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ becomes the equation of a circle $x^2 + y^2 = r^2$ when $a=b=r$.

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(iv) Statement (s): If x and y are integers such that x > y, then –x < – y.

This is a conditional statement. We need to check if, whenever the hypothesis is true, the conclusion is also true.

Hypothesis: x and y are integers such that $x > y$.

Conclusion: $-x < -y$.

Consider the property of inequalities: If you multiply both sides of an inequality by a negative number, you must reverse the direction of the inequality sign.

Start with the hypothesis: $x > y$.

Multiply both sides by -1 (which is a negative number): $(-1)x < (-1)y$.

This gives $-x < -y$, which is the conclusion.

This property holds for all real numbers, and thus specifically for all integers.

For example, if $x=5$ and $y=3$, then $x > y$ (since $5 > 3$). The statement says $-x < -y$, which is $-5 < -3$. This is true.

If $x=-2$ and $y=-5$, then $x > y$ (since $-2 > -5$). The statement says $-x < -y$, which is $-(-2) < -(-5)$, so $2 < 5$. This is true.

Truth Value: True.

Reason: If $x > y$, multiplying both sides by -1 reverses the inequality, resulting in $-x < -y$. This property holds for all real numbers, including integers.

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(v) Statement (t): $\sqrt{11}$ is a rational number.

A rational number is a number that can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

The square root of an integer is rational if and only if the integer is a perfect square (the square of an integer).

The perfect squares are $1^2=1$, $2^2=4$, $3^2=9$, $4^2=16$, and so on.

The number 11 is not a perfect square, as it lies between $3^2=9$ and $4^2=16$.

Therefore, $\sqrt{11}$ is not an integer, and it is not possible to express $\sqrt{11}$ as a fraction of two integers. $\sqrt{11}$ is an irrational number.

Truth Value: False.

Reason: $\sqrt{11}$ is not a perfect square, so $\sqrt{11}$ is an irrational number, not a rational number.

Given Compound Statement (t): you are wet when it rains or you are in a river.

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Identifying Component Statements:

The connecting word is "or". The component statements are:

$p$: You are wet when it rains. (More precisely: If it rains, then you are wet).

$q$: You are in a river. (More precisely: If you are in a river, then you are wet).

The statement $t$ is of the form "$p$ or $q$". Let's rephrase $p$ and $q$ to be simple conditions about being wet:

$P$: It is raining and you are exposed to the rain.

$Q$: You are in a river.

Then the statement "you are wet when it rains or you are in a river" can be interpreted as "(If it is raining and you are exposed, then you are wet) or (If you are in a river, then you are wet)". However, the typical interpretation in logic questions is that the compound statement asserts the conclusion ("you are wet") based on either of the conditions ("it rains" OR "you are in a river"). Let's follow the most likely intent for a logic problem at this level.

Let's interpret the statement $t$ as "You are wet if (it rains or you are in a river)". This is equivalent to "If (it rains or you are in a river), then you are wet".

Let $A$: It rains.

Let $B$: You are in a river.

Let $C$: You are wet.

The statement can be interpreted as "If ($A$ or $B$), then $C$".

In this case, the "or" connects two conditions ($A$ and $B$) that lead to a single outcome ($C$). Can it be raining while you are in a river? Yes, it's possible to be in a river while it is raining. In this scenario, both conditions for being wet (rain exposure and being in a river) are met, and you are wet. The "or" here does not exclude the possibility of both conditions occurring simultaneously.

Type of "Or": Since it is possible for both conditions (it rains, you are in a river) to happen together, the "or" connecting them is used in the inclusive sense.

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Check Truth Value of Compound Statement:

The compound statement is "If ($A$ or $B$), then $C$". This statement is true unless ($A$ or $B$) is true and $C$ is false.

Let's consider the conditions under which ($A$ or $B$) is true:

• $A$ is true and $B$ is false (It is raining, and you are not in a river). If you are exposed to the rain, you will be wet. If you are sheltered from the rain, you might not be wet. However, the statement "you are wet when it rains" implies exposure unless specified otherwise. Let's assume exposure. Then if it rains, you are wet.
• $A$ is false and $B$ is true (It is not raining, and you are in a river). If you are in a river, you will be wet.
• $A$ is true and $B$ is true (It is raining, and you are in a river). If you are in a river while it is raining, you will be wet.

In all cases where ($A$ or $B$) is true (at least one condition is met), the consequence $C$ ("you are wet") is also true. It is impossible for ($A$ or $B$) to be true while $C$ is false (you are not wet).

Since the scenario where the hypothesis ("it rains or you are in a river") is true and the conclusion ("you are wet") is false cannot occur, the conditional statement is true.

Truth Value: True.

Justification: The statement claims that either rain (implying exposure) or being in a river is sufficient to make you wet, and also allows for both conditions to be true. In any physically realistic scenario, if it is raining (and you are not sheltered) or you are in a river, you will indeed be wet. Thus, the implication holds.

The negation of a universal statement "For every A, P(A)" is "There exists an A such that P(A) is false". The negation of an existential statement "There exists an A such that P(A)" is "For every A, P(A) is false". The negation of "P(x) is true" is "P(x) is false".

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(i) Statement (p): For every real number x, x² > x.

This is a universal statement. The quantifier is "For every real number x". The assertion is "$x^2 > x$".

Negation: It is not the case that for every real number x, $x^2 > x$.

Equivalent phrasing: There exists a real number x such that it is not the case that $x^2 > x$.

"Not ($x^2 > x$)" means "$x^2 \le x$".

Negation of (p): There exists a real number x such that x² $\le$ x.

Note: Statement (p) is false (e.g., if $x = \frac{1}{2}$, $x^2 = \frac{1}{4}$, and $\frac{1}{4} \not> \frac{1}{2}$; if $x=0$, $x^2=0$, and $0 \not> 0$). So its negation is true.

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(ii) Statement (q): There exists a rational number x such that x² = 2.

This is an existential statement. The quantifier is "There exists a rational number x such that". The assertion is "$x^2 = 2$".

Negation: It is not the case that there exists a rational number x such that $x^2 = 2$.

Equivalent phrasing: For every rational number x, it is not the case that $x^2 = 2$.

"Not ($x^2 = 2$)" means "$x^2 \neq 2$".

Negation of (q): For every rational number x, x² $\neq$ 2.

Note: Statement (q) is false (as $\sqrt{2}$ is irrational, there is no rational number whose square is 2). So its negation is true.

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(iii) Statement (r): All birds have wings.

This is a universal statement. The quantifier is "All birds" (equivalent to "For every bird"). The assertion is "have wings".

Negation: It is not the case that all birds have wings.

Equivalent phrasing: There exists a bird such that it does not have wings.

Negation of (r): There exists a bird which does not have wings.

Note: Statement (r) is generally considered false (e.g., kiwis, penguins that don't fly, though they have rudimentary wings). So its negation is true.

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(iv) Statement (s): All students study mathematics at the elementary level.

This is a universal statement. The quantifier is "All students". The assertion is "study mathematics at the elementary level".

Negation: It is not the case that all students study mathematics at the elementary level.

Equivalent phrasing: There exists a student such that it is not the case that they study mathematics at the elementary level.

Negation of (s): There exists a student who does not study mathematics at the elementary level.

Note: Statement (s) is likely false in a global context (e.g., there might be educational systems or students who don't). So its negation is true.

Given Statement:

The integer n is odd if and only if n² is odd.

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To Rewrite:

Using "necessary and sufficient".

To Check:

Whether the statement is true.

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Rewriting the Statement:

The statement "$P$ if and only if $Q$" is equivalent to saying "$P$ is a necessary and sufficient condition for $Q$" and also "$Q$ is a necessary and sufficient condition for $P$".

In this case, let $P$: The integer n is odd.

Let $Q$: n² is odd.

The statement is "$P$ if and only if $Q$".

Using "necessary and sufficient":

1. P is a necessary and sufficient condition for Q:

For an integer n, n being odd is a necessary and sufficient condition for n² to be odd.

2. Q is a necessary and sufficient condition for P:

For an integer n, n² being odd is a necessary and sufficient condition for n to be odd.

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Checking the Truth Value of the Statement:

The statement "$P$ if and only if $Q$" is true if and only if both the conditional statement "If $P$, then $Q$" and its converse "If $Q$, then $P$" are true.

Conditional Statement ("If $P$, then $Q$"): If the integer n is odd, then n² is odd.

Assume n is an odd integer. Then $n = 2k + 1$ for some integer $k$.

$n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$.

Since $2k^2 + 2k$ is an integer, $n^2$ is of the form $2m + 1$, which means $n^2$ is odd. So, "If P, then Q" is true.

Converse ("If $Q$, then $P$"): If the integer n² is odd, then n is odd.

We can prove this using the contrapositive of the converse. The contrapositive of "If $Q$, then $P$" is "If not $P$, then not $Q$".

Not $P$: The integer n is not odd, which means n is even.

Not $Q$: n² is not odd, which means n² is even.

Contrapositive of the Converse: If the integer n is even, then n² is even.

Assume n is an even integer. Then $n = 2k$ for some integer $k$.

$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$.

Since $2k^2$ is an integer, $n^2$ is of the form $2m$, which means $n^2$ is even. So, the contrapositive of the converse is true.

Since the contrapositive of the converse is true, the converse ("If $Q$, then $P$") is also true.

Since both "If $P$, then $Q$" and "If $Q$, then $P$" are true, the biconditional statement "$P$ if and only if $Q$" is true.

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Conclusion:

The statement "The integer n is odd if and only if n² is odd" is True.

Given Statement (t): If you drive over 80 km per hour, then you will get a fine.

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This is a conditional statement of the form "If $P$, then $Q$".

Hypothesis ($P$): You drive over 80 km per hour.

Conclusion ($Q$): You will get a fine.

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In a conditional statement "If $P$, then $Q$":

• $P$ is a sufficient condition for $Q$. This means that $P$ is enough to guarantee $Q$. If $P$ happens, $Q$ must happen.
• $Q$ is a necessary condition for $P$. This means that $Q$ must happen if $P$ happens. If $Q$ does not happen, $P$ cannot have happened.

Based on the statement "If you drive over 80 km per hour, then you will get a fine":

• Driving over 80 km per hour is enough to guarantee that you will get a fine. So, "You drive over 80 km per hour" is a sufficient condition for "You will get a fine".
• If you got a fine (under this specific rule), it must be because you drove over 80 km per hour (assuming this is the *only* reason for getting this specific type of fine mentioned). So, "You will get a fine" is a necessary consequence of driving over 80 km per hour. Thus, "You will get a fine" is a necessary condition for "You drive over 80 km per hour".

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Therefore, for the statement "If you drive over 80 km per hour, then you will get a fine":

• Sufficient Condition: You drive over 80 km per hour.
• Necessary Condition: You will get a fine.

The negation of a universal statement "For every A, P(A)" is "There exists an A such that P(A) is false". The negation of an existential statement "There exists an A such that P(A)" is "For every A, P(A) is false". The negation of a statement $S$ is "It is not the case that $S$".

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(i) Statement (p): For every positive real number x, the number x – 1 is also positive.

This is a universal statement about positive real numbers. The assertion is "$x - 1$ is positive", which means $x - 1 > 0$.

Negation: It is not the case that for every positive real number x, the number x – 1 is positive.

Equivalent phrasing: There exists a positive real number x such that it is not the case that x – 1 is positive.

"Not ($x - 1$ is positive)" means "$x - 1$ is not positive", i.e., "$x - 1 \le 0$".

Negation of (p): There exists a positive real number x such that x – 1 $\le$ 0.

Note: Statement (p) is false (e.g., if $x = 0.5$, which is positive, $x-1 = -0.5$, which is not positive). So its negation is true (e.g., $x=0.5$ is a positive real number and $0.5-1 = -0.5 \le 0$).

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(ii) Statement (q): All cats scratch.

This is a universal statement about cats. The quantifier is "All cats" (equivalent to "For every cat"). The assertion is "scratch".

Negation: It is not the case that all cats scratch.

Equivalent phrasing: There exists a cat such that it does not scratch.

Negation of (q): There exists a cat which does not scratch.

Note: Statement (q) is likely false (e.g., declawed cats, or some cats that simply don't scratch furniture/people). So its negation is true.

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(iii) Statement (r): For every real number x, either x > 1 or x < 1.

This is a universal statement. The quantifier is "For every real number x". The assertion is "either $x > 1$ or $x < 1$". This assertion means $x \neq 1$.

So, the statement (r) is equivalent to "For every real number x, $x \neq 1$".

Negation of the equivalent statement: It is not the case that for every real number x, $x \neq 1$.

Equivalent phrasing: There exists a real number x such that $x = 1$.

Using the original phrasing of the assertion "$x > 1$ or $x < 1$":

Negation of "either $x > 1$ or $x < 1$" is "neither ($x > 1$) nor ($x < 1$)", which means "not ($x > 1$) AND not ($x < 1$)".

"Not ($x > 1$)" means "$x \le 1$".

"Not ($x < 1$)" means "$x \ge 1$".

So the negation of the assertion is "$x \le 1$ AND $x \ge 1$", which means $x = 1$.

Negation of the entire statement: There exists a real number x such that $x = 1$.

Negation of (r): There exists a real number x such that x = 1.

Note: Statement (r) is false (the real number 1 is neither > 1 nor < 1). So its negation is true (e.g., $x=1$ is a real number such that $x=1$).

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(iv) Statement (s): There exists a number x such that 0 < x < 1.

This is an existential statement. The quantifier is "There exists a number x such that". The assertion is "$0 < x < 1$".

Negation: It is not the case that there exists a number x such that $0 < x < 1$.

Equivalent phrasing: For every number x, it is not the case that $0 < x < 1$.

"Not ($0 < x < 1$)" means "Not ($x > 0$ AND $x < 1$)", which is "not ($x > 0$) OR not ($x < 1$)" by De Morgan's Law.

"Not ($x > 0$)" means "$x \le 0$".

"Not ($x < 1$)" means "$x \ge 1$".

So the negation of the assertion is "$x \le 0$ OR $x \ge 1$".

Negation of (s): For every number x, x $\le$ 0 or x $\ge$ 1.

Note: Statement (s) is true (e.g., $x=0.5$ is a number such that $0 < 0.5 < 1$). So its negation is false.

For statement (i) p: A positive integer is prime only if it has no divisors other than 1 and itself.

This can be rephrased in the standard 'If-then' form as: If a positive integer is prime, then it has no divisors other than 1 and itself.

The original statement is of the form "If P, then Q", where P is "A positive integer is prime" and Q is "It has no divisors other than 1 and itself".

Converse: The converse is "If Q, then P".

So, the converse is: If a positive integer has no divisors other than 1 and itself, then it is prime.

Contrapositive: The contrapositive is "If not Q, then not P".

So, the contrapositive is: If a positive integer has divisors other than 1 and itself, then it is not prime.

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For statement (ii) q: I go to a beach whenever it is a sunny day.

This can be rephrased in the standard 'If-then' form as: If it is a sunny day, then I go to a beach.

The original statement is of the form "If P, then Q", where P is "It is a sunny day" and Q is "I go to a beach".

Converse: The converse is "If Q, then P".

So, the converse is: If I go to a beach, then it is a sunny day.

Contrapositive: The contrapositive is "If not Q, then not P".

So, the contrapositive is: If I do not go to a beach, then it is not a sunny day.

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For statement (iii) r: If it is hot outside, then you feel thirsty.

This statement is already in the standard 'If-then' form.

The original statement is of the form "If P, then Q", where P is "It is hot outside" and Q is "You feel thirsty".

Converse: The converse is "If Q, then P".

So, the converse is: If you feel thirsty, then it is hot outside.

Contrapositive: The contrapositive is "If not Q, then not P".

So, the contrapositive is: If you do not feel thirsty, then it is not hot outside.

For statement (i) p: It is necessary to have a password to log on to the server.

The phrase "It is necessary to have Q to do P" means that Q is a necessary condition for P. In logical terms, this is equivalent to "If P, then Q".

Let P be the statement "You log on to the server".

Let Q be the statement "You have a password".

The statement "It is necessary to have a password to log on to the server" is equivalent to "If you log on to the server, then you have a password".

So, the statement in the form "if p, then q" is:

If you log on to the server, then you have a password.

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For statement (ii) q: There is traffic jam whenever it rains.

The phrase "Q whenever P" means that Q happens every time P happens. This is equivalent to "If P, then Q".

Let P be the statement "It rains".

Let Q be the statement "There is traffic jam".

The statement "There is traffic jam whenever it rains" is equivalent to "If it rains, then there is traffic jam".

So, the statement in the form "if p, then q" is:

If it rains, then there is traffic jam.

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For statement (iii) r: You can access the website only if you pay a subscription fee.

The phrase "P only if Q" means that P can only happen if Q happens. This implies that if P happens, then Q must have happened. This is equivalent to "If P, then Q".

Let P be the statement "You can access the website".

Let Q be the statement "You pay a subscription fee".

The statement "You can access the website only if you pay a subscription fee" is equivalent to "If you can access the website, then you pay a subscription fee".

So, the statement in the form "if p, then q" is:

If you can access the website, then you pay a subscription fee.

For statement (i) p: If you watch television, then your mind is free and if your mind is free, then you watch television.

This statement is of the form "If P, then Q and if Q, then P". This is the definition of "P if and only if Q".

Let P be the statement "You watch television".

Let Q be the statement "Your mind is free".

The given statement combines the implication $P \implies Q$ and the implication $Q \implies P$. The combination of these two implications is logically equivalent to the biconditional statement $P \iff Q$, which is read as "P if and only if Q".

So, the statement can be rewritten as:

You watch television if and only if your mind is free.

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For statement (ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework regularly.

The phrase "Q is necessary and sufficient for P" is a direct way of stating that P happens if and only if Q happens.

Let P be the statement "You get an A grade".

Let Q be the statement "You do all the homework regularly".

"Q is necessary for P" means if P happens, then Q must happen ($P \implies Q$).

"Q is sufficient for P" means if Q happens, then P will happen ($Q \implies P$).

When Q is both necessary and sufficient for P, it means $P \implies Q$ and $Q \implies P$, which is the definition of $P \iff Q$ or "P if and only if Q".

So, the statement can be rewritten as:

You get an A grade if and only if you do all the homework regularly.

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For statement (iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle, then it is equiangular.

This statement is also of the form "If P, then Q and if Q, then P". This is the definition of "P if and only if Q".

Let P be the statement "A quadrilateral is equiangular".

Let Q be the statement "It is a rectangle".

The given statement combines the implication $P \implies Q$ and the implication $Q \implies P$. The combination of these two implications is logically equivalent to the biconditional statement $P \iff Q$, which is read as "P if and only if Q".

So, the statement can be rewritten as:

A quadrilateral is equiangular if and only if it is a rectangle.

Given statements:

p: 25 is a multiple of 5.

q: 25 is a multiple of 8.

First, let's determine the truth value of each individual statement.

Statement p: "25 is a multiple of 5". Since $25 = 5 \times 5$, 25 is indeed a multiple of 5. Thus, statement p is True.

Statement q: "25 is a multiple of 8". To check if 25 is a multiple of 8, we can perform division. $25 \div 8$ gives a quotient of 3 and a remainder of 1. Since the remainder is not 0, 25 is not a multiple of 8. Thus, statement q is False.

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Compound statement connecting p and q with "And":

The compound statement is "p and q". In words, this is: 25 is a multiple of 5 and 25 is a multiple of 8.

Validity: For a compound statement connected by "And" to be true, both individual statements must be true. In this case, statement p is True, but statement q is False. Therefore, the compound statement "p and q" is False.

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Compound statement connecting p and q with "Or":

The compound statement is "p or q". In words, this is: 25 is a multiple of 5 or 25 is a multiple of 8.

Validity: For a compound statement connected by "Or" to be true, at least one of the individual statements must be true. In this case, statement p is True. Therefore, the compound statement "p or q" is True.

For statement (i) p: The sum of an irrational number and a rational number is irrational (by contradiction method).

Method of Proof: Contradiction.

Statement: The sum of an irrational number and a rational number is irrational.

Assumption for Contradiction: Assume the statement is false. That is, assume there exists an irrational number and a rational number whose sum is rational.

Let $x$ be an irrational number and $y$ be a rational number.

Assume that the sum $x+y$ is a rational number. Let $x+y = z$, where $z$ is rational.

By definition, a rational number can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$.

Since $y$ is rational, we can write $y = \frac{a}{b}$ for some integers $a, b$ with $b \neq 0$.

Since we assumed $z$ is rational, we can write $z = \frac{c}{d}$ for some integers $c, d$ with $d \neq 0$.

From our assumption $x+y = z$, we can write $x = z - y$.

Substituting the fractional forms of $z$ and $y$, we get:

$x = \frac{c}{d} - \frac{a}{b}$

To subtract these fractions, we find a common denominator:

$x = \frac{cb}{db} - \frac{ad}{db} = \frac{cb - ad}{db}$

Since $a, b, c, d$ are integers, the numerator $cb - ad$ is an integer.

Since $b \neq 0$ and $d \neq 0$, the denominator $db$ is a non-zero integer.

Therefore, $x$ is expressed as the ratio of two integers with a non-zero denominator. This means $x$ is a rational number.

This contradicts our initial premise that $x$ is an irrational number.

Since our assumption leads to a contradiction, the assumption must be false.

Therefore, the original statement is true.

Conclusion: The statement "The sum of an irrational number and a rational number is irrational" is valid.

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For statement (ii) q: If n is a real number with n > 3, then n 2 > 9 (by contradiction method).

Method of Proof: Contradiction.

Statement: If $n$ is a real number with $n > 3$, then $n^2 > 9$.

Assumption for Contradiction: Assume the statement is false. That is, assume there exists a real number $n$ such that $n > 3$ and $n^2 \leq 9$.

We are given that $n$ is a real number.

From our assumption, we have two conditions that hold simultaneously:

$n > 3$

and

$n^2 \leq 9$

Consider the inequality $n^2 \leq 9$. Taking the square root of both sides (and remembering that $\sqrt{x^2} = |x|$ for real $x$) gives:

$\sqrt{n^2} \leq \sqrt{9}$

$|n| \leq 3$

For real numbers, $|n| \leq 3$ means $-3 \leq n \leq 3$.

So, our assumption implies that both $n > 3$ and $n \leq 3$ must be true for the same real number $n$.

However, the conditions $n > 3$ and $n \leq 3$ are mutually exclusive; a single number cannot be strictly greater than 3 and less than or equal to 3 at the same time.

This is a contradiction.

Since our assumption leads to a contradiction, the assumption must be false.

Therefore, the original statement is true.

Conclusion: The statement "If $n$ is a real number with $n > 3$, then $n^2 > 9$" is valid.

The given statement is: p: If a triangle is equiangular, then it is an obtuse angled triangle.

This statement is in the form "If P, then Q", where:

P: A triangle is equiangular.

Q: It is an obtuse angled triangle.

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Here are five different ways to write the statement, conveying the same meaning:

1. If a triangle is equiangular, then it is an obtuse angled triangle.

2. A triangle being equiangular implies that it is an obtuse angled triangle.

3. A triangle is an obtuse angled triangle if it is equiangular.

4. A triangle is equiangular only if it is an obtuse angled triangle.

5. If a triangle is not an obtuse angled triangle, then it is not equiangular. (Contrapositive)