Chapter 3 Integers

Given temperatures on Sunday:

Temperature in Leh = $-4^\circ\text{C}$

Temperature in Tawang = $-2^\circ\text{C}$

To determine which place was cooler, we need to compare the two temperatures.

Comparing $-4$ and $-2$, we know that a smaller number represents a lower temperature.

On the number line, $-4$ is to the left of $-2$. This means:

$-4 < -2$

So, $-4^\circ\text{C}$ is a lower temperature than $-2^\circ\text{C}$.

A lower temperature means a place is cooler.

Since the temperature in Leh ($-4^\circ\text{C}$) is lower than the temperature in Tawang ($-2^\circ\text{C}$), Leh was cooler than Tawang on that Sunday.

Let's check the given options:

(A) Leh was cooler than Tawang. This is consistent with our conclusion ($ -4^\circ\text{C} < -2^\circ\text{C} $).

(B) Leh was hotter than Tawang. This would mean $-4^\circ\text{C} > -2^\circ\text{C}$, which is false.

(C) Leh was as cool as Tawang. This would mean $-4^\circ\text{C} = -2^\circ\text{C}$, which is false.

(D) Tawang was cooler than Leh. This would mean $-2^\circ\text{C} < -4^\circ\text{C}$, which is false.

The correct statement is that Leh was cooler than Tawang.

The correct option is (A) Leh was cooler than Tawang.

Let's analyse each statement.

(a) Every positive integer is greater than 0.

The set of positive integers is $\{1, 2, 3, 4, \ldots\}$.

By definition, a positive number is a number greater than zero.

Therefore, every positive integer is indeed greater than 0.

This statement is True.

(b) Every integer is either positive or negative.

The set of integers includes positive integers, negative integers, and zero.

Integers = $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$.

The number 0 is an integer, but it is neither considered positive nor negative.

Since there is an integer (zero) which is neither positive nor negative, the statement is not universally true for all integers.

This statement is False.

We need to compare the values of the expressions on both sides of the blank.

Consider the left side of the blank:

$3 + (-2)$

Adding a negative integer is equivalent to subtracting the corresponding positive integer.

$3 + (-2) = 3 - 2 = 1$

Consider the right side of the blank:

$3 + (-3)$

Adding a negative integer is equivalent to subtracting the corresponding positive integer.

$3 + (-3) = 3 - 3 = 0$

Now, we compare the results from the left and right sides.

Left side value = 1

Right side value = 0

Comparing 1 and 0, we know that 1 is greater than 0.

$1 > 0$

Therefore, the correct symbol to fill in the blank is '>'.

The completed statement is:

3 + (–2) > 3 + (–3)

Integers can be used to represent quantities that have a direction or are relative to a reference point, such as sea level, profit/loss, or temperatures above/below zero.

(a) 3 km above sea level

Sea level is often considered the zero reference point.

Positions above sea level are represented by positive numbers.

Positions below sea level are represented by negative numbers.

Therefore, 3 km above sea level is represented by the integer +3 or simply 3.

(b) A loss of Rs 500

Financial gains (profit or income) are typically represented by positive numbers.

Financial losses (expenses or debt) are typically represented by negative numbers.

Therefore, a loss of $\textsf{₹}500$ is represented by the integer -500.

We will find the sum of each pair of integers.

(a) Sum of – 6 and – 4:

To add two negative integers, we add their absolute values and place a negative sign before the sum.

$(-6) + (-4) = -(6 + 4) = -10$

(b) Sum of +3 and – 4:

To add integers with different signs, we find the difference between their absolute values and use the sign of the integer with the larger absolute value.

Absolute value of +3 is 3.

Absolute value of –4 is 4.

The difference is $4 - 3 = 1$.

The integer with the larger absolute value is –4, which is negative.

So, $(+3) + (-4) = -1$

(c) Sum of +4 and –2:

To add integers with different signs, we find the difference between their absolute values and use the sign of the integer with the larger absolute value.

Absolute value of +4 is 4.

Absolute value of –2 is 2.

The difference is $4 - 2 = 2$.

The integer with the larger absolute value is +4, which is positive.

So, $(+4) + (-2) = +2$ or simply $2$

To find the sum of –2 and –3 using the number line, follow these steps:

Step 1: Draw a number line and mark the integers.

Step 2: Start from 0. The first integer is –2. Since it is negative, move 2 units to the left from 0. You reach –2.

Step 3: From –2, add the second integer, which is –3. Since –3 is negative, move 3 units further to the left from –2.

Moving 3 units left from –2 means:

From –2 to –3 is 1 unit left.

From –3 to –4 is 1 unit left.

From –4 to –5 is 1 unit left.

So, moving 3 units left from –2, you reach –5.

The final position on the number line is –5.

Therefore, $(-2) + (-3) = -5$.

Subtracting an integer is the same as adding its additive inverse (opposite).

(i) Subtract 3 from –4:

This means we need to calculate $(-4) - (3)$.

The additive inverse of 3 is –3.

$(-4) - (3) = (-4) + (-3)$

To add two negative integers, we add their absolute values and put a negative sign.

$(-4) + (-3) = -(4 + 3) = -7$

So, $3$ subtracted from $–4$ is $–7$.

(ii) Subtract –3 from –4:

This means we need to calculate $(-4) - (-3)$.

The additive inverse of –3 is +3 (or simply 3).

$(-4) - (-3) = (-4) + (3)$

To add integers with different signs, we find the difference between their absolute values and use the sign of the integer with the larger absolute value.

Absolute value of –4 is 4.

Absolute value of 3 is 3.

The difference is $4 - 3 = 1$.

The integer with the larger absolute value is –4, which is negative.

So, $(-4) + (3) = -1$

So, $–3$ subtracted from $–4$ is $–1$.

We will use a number line to perform the subtraction.

(a) Subtract 2 from –3:

This is equivalent to calculating $(-3) - 2$.

Step 1: Start at 0 on the number line.

Step 2: Move 3 units to the left from 0 to reach –3.

Step 3: To subtract a positive number (2), move further to the left from –3. Move 2 units to the left from –3.

Moving 2 units left from –3 takes us to –4, and then to –5.

The final position on the number line is –5.

So, $(-3) - 2 = -5$.

(b) Subtract –2 from –3:

This is equivalent to calculating $(-3) - (-2)$.

Subtracting a negative integer is the same as adding the corresponding positive integer. So, $(-3) - (-2) = (-3) + 2$.

Step 1: Start at 0 on the number line.

Step 2: Move 3 units to the left from 0 to reach –3.

Step 3: To add a positive number (2), move to the right from –3. Move 2 units to the right from –3.

Moving 2 units right from –3 takes us to –2, and then to –1.

The final position on the number line is –1.

So, $(-3) - (-2) = -1$.

We need to find the number of integers that are strictly greater than –9 and strictly less than –2.

The integers between –9 and –2 are the integers that come after –9 and before –2 on the number line.

Starting from –9 and moving towards the right, the integers are: –8, –7, –6, –5, –4, –3.

The first integer greater than –9 is –8.

The last integer less than –2 is –3.

So, the integers between –9 and –2 are: –8, –7, –6, –5, –4, –3.

Let's count these integers:

–8 (1st)

–7 (2nd)

–6 (3rd)

–5 (4th)

–4 (5th)

–3 (6th)

There are 6 integers between –9 and –2.

Alternatively, we can use the formula for the number of integers between two integers $a$ and $b$ (where $a < b$), which is $(b - a - 1)$.

Here, $a = -9$ and $b = -2$.

Number of integers between –9 and –2 = $(-2) - (-9) - 1$

$= -2 + 9 - 1$

$= 7 - 1 = 6$

This confirms the count.

We need to calculate the value of the given expression: $1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10$.

We can group the terms in pairs:

$(1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)$

Evaluate each pair:

$1 – 2 = -1$

$3 – 4 = -1$

$5 – 6 = -1$

$7 – 8 = -1$

$9 – 10 = -1$

Substitute these values back into the grouped expression:

$(-1) + (-1) + (-1) + (-1) + (-1)$

Adding five negative ones:

$5 \times (-1) = -5$

Alternatively, we can add all the positive numbers together and all the negative numbers together, and then find the sum:

Positive numbers: $1, 3, 5, 7, 9$

Sum of positive numbers: $1 + 3 + 5 + 7 + 9 = 25$

Negative numbers: $-2, -4, -6, -8, -10$

Sum of negative numbers: $(-2) + (-4) + (-6) + (-8) + (-10) = -(2 + 4 + 6 + 8 + 10) = -30$

Now, add the sum of positive numbers and the sum of negative numbers:

$25 + (-30)$

Adding a positive and a negative number: find the difference between their absolute values and use the sign of the number with the larger absolute value.

$|25| = 25$

$|-30| = 30$

Difference = $30 - 25 = 5$

The number with the larger absolute value is -30, which is negative.

So, $25 + (-30) = -5$

The value of the expression is $-5$.

Let the two integers be $a$ and $b$.

We are given that the sum of the two integers is 47. So,

$a + b = 47$

We are also given that one of the integers is –24. Let's assume $a = -24$.

Substituting this value into the equation:

$(-24) + b = 47$

To find the other integer $b$, we need to isolate $b$. We can do this by adding the additive inverse of –24 to both sides of the equation. The additive inverse of –24 is +24 (or simply 24).

$(-24) + b + 24 = 47 + 24$

$0 + b = 47 + 24$

$b = 47 + 24$

Now, we calculate the sum $47 + 24$:

So, $b = 71$.

The other integer is 71.

We can verify this by adding the two integers: $(-24) + 71$.

$(-24) + 71 = 71 - 24 = 47$.

This matches the given information that the sum is 47.

We need to insert '+' or '–' signs between the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, in that specific order, to achieve the given results.

(a) Result = 5

We are looking for signs such that $0 \text{ op}_1 1 \text{ op}_2 2 \text{ op}_3 3 \text{ op}_4 4 \text{ op}_5 5 \text{ op}_6 6 \text{ op}_7 7 \text{ op}_8 8 \text{ op}_9 9 = 5$, where $\text{op}_i$ is either '+' or '–'.

One possible combination of signs is + + – + + + + – –.

Let's insert these signs between the digits:

$0 + 1 + 2 - 3 + 4 + 5 + 6 + 7 - 8 - 9$

Let's calculate the value step-by-step:

$0 + 1 = 1$

$1 + 2 = 3$

$3 - 3 = 0$

$0 + 4 = 4$

$4 + 5 = 9$

$9 + 6 = 15$

$15 + 7 = 22$

$22 - 8 = 14$

$14 - 9 = 5$

This matches the required result.

So, one way to get 5 is: $0 + 1 + 2 - 3 + 4 + 5 + 6 + 7 - 8 - 9$.

(b) Result = –3

We are looking for signs such that $0 \text{ op}_1 1 \text{ op}_2 2 \text{ op}_3 3 \text{ op}_4 4 \text{ op}_5 5 \text{ op}_6 6 \text{ op}_7 7 \text{ op}_8 8 \text{ op}_9 9 = -3$, where $\text{op}_i$ is either '+' or '–'.

One possible combination of signs is + + + + + + – – –.

Let's insert these signs between the digits:

$0 + 1 + 2 + 3 + 4 + 5 + 6 - 7 - 8 - 9$

Let's calculate the value step-by-step:

$0 + 1 = 1$

$1 + 2 = 3$

$3 + 3 = 6$

$6 + 4 = 10$

$10 + 5 = 15$

$15 + 6 = 21$

$21 - 7 = 14$

$14 - 8 = 6$

$6 - 9 = -3$

This matches the required result.

So, one way to get –3 is: $0 + 1 + 2 + 3 + 4 + 5 + 6 - 7 - 8 - 9$.

We need to find five different integers whose sum is 5.

Let the five distinct integers be $a, b, c, d,$ and $e$. We require $a \neq b$, $a \neq c$, ..., $d \neq e$, and their sum must be 5:

$a + b + c + d + e = 5$

We can choose a set of distinct integers, including positive, negative, and zero, and check if their sum is 5. If not, we can adjust them.

Let's try the following set of five distinct integers: 0, 1, 2, 3, and -1.

Check if they are distinct:

0, 1, 2, 3, -1 are all different from each other. Yes, they are distinct.

Check their sum:

$0 + 1 + 2 + 3 + (-1)$

$= (0 + 1 + 2 + 3) + (-1)$

$= 6 + (-1)$

$= 6 - 1$

$= 5$

The sum is 5, which is the required result.

Therefore, five distinct integers whose sum is 5 are 0, 1, 2, 3, and –1.

Integers less than 0 are the negative integers.

Examples of integers less than 0 include -1, -5, -100, etc.

All negative numbers are preceded by a negative sign, which is '–'.

Thus, every integer less than 0 has the sign –.

Comparing this with the given options:

(A) +

(B) –

(C) ×

(D) ÷

The correct option is (B).

On a number line, positive integers are located to the right of 0, and negative integers are located to the left of 0.

The distance of an integer from 0 is its absolute value.

Moving 5 units to the right from 0 means we are moving in the positive direction.

Starting at 0 and moving 5 units to the right gives us the integer $0 + 5 = 5$.

The integer is 5, which is a positive number and can be represented as +5.

Comparing with the given options:

(A) +5

(B) –5

(C) +4

(D) – 4

The correct option is (A).

The predecessor of an integer is the integer that comes just before it.

To find the predecessor of an integer, we subtract 1 from the integer.

We need to find the predecessor of –1.

Predecessor of –1 = $-1 - 1$

Calculating the value: $-1 - 1 = -2$.

So, the predecessor of –1 is –2.

Comparing with the given options:

(A) 0

(B) 2

(C) –2

(D) 1

The correct option is (C).

We need to find the number of integers that are strictly greater than –1 and strictly less than 1.

The integers are ..., –3, –2, –1, 0, 1, 2, 3, ...

The integers between –1 and 1 are the integers $x$ such that $-1 < x < 1$.

The only integer that satisfies this condition is 0.

Therefore, there is only one integer lying between –1 and 1.

Comparing with the given options:

(A) 1

(B) 2

(C) 3

(D) 0

The correct option is (A).

Whole numbers are the set of non-negative integers, i.e., $\{0, 1, 2, 3, 4, ...\}$.

We need to find the number of whole numbers $w$ such that $w$ is between –5 and 5.

This means we are looking for whole numbers $w$ that satisfy the inequality $-5 < w < 5$.

Since whole numbers must be greater than or equal to 0, the condition $w > -5$ is automatically satisfied for any whole number.

So, we are looking for whole numbers $w$ such that $0 \leq w < 5$.

The whole numbers less than 5 are 0, 1, 2, 3, and 4.

Listing the whole numbers between –5 and 5: $\{0, 1, 2, 3, 4\}$.

Counting these numbers, we find there are 5 whole numbers.

Comparing with the given options:

(A) 10

(B) 3

(C) 4

(D) 5

The correct option is (D).

We are looking for integers that lie strictly between –10 and –15.

This means we are looking for integers $x$ such that $x$ is greater than the smaller value and less than the larger value.

Since –15 is less than –10, the condition is $-15 < x < -10$.

Let's list the integers that are greater than –15:

–14, –13, –12, –11, –10, –9, ...

Let's list the integers that are less than –10:

..., –16, –15, –14, –13, –12, –11

The integers that are both greater than –15 and less than –10 are the integers common to both lists within the specified range.

These integers are –14, –13, –12, and –11.

We need to find the greatest integer among these values: –14, –13, –12, –11.

On the number line, numbers increase as we move from left to right. Thus, the integer furthest to the right among these is the greatest.

Comparing the values, –11 is the greatest among –14, –13, –12, and –11.

Therefore, the greatest integer lying between –10 and –15 is –11.

Comparing with the given options:

(A) –10

(B) –11

(C) –15

(D) –14

The correct option is (B).

We are looking for integers that lie strictly between –10 and –15.

This means we are looking for integers $x$ such that $x$ is greater than the smaller value and less than the larger value.

Since –15 is less than –10, the condition is $-15 < x < -10$.

The integers that are greater than –15 and less than –10 are –14, –13, –12, and –11.

We need to find the least integer among these values: –14, –13, –12, –11.

On the number line, numbers decrease as we move from right to left. Thus, the integer furthest to the left among these is the least.

Comparing the values, –14 is the least among –14, –13, –12, and –11.

Therefore, the least integer lying between –10 and –15 is –14.

Comparing with the given options:

(A) –10

(B) –11

(C) –15

(D) –14

The correct option is (D).

On a standard horizontal number line, integers are arranged in increasing order from left to right.

Zero (0) is typically the origin or center point.

Positive integers (like 1, 2, 3, 4, 5, ...) are located to the right of 0.

Negative integers (like –1, –2, –3, ...) are located to the left of 0.

Since 5 is a positive integer, it is located to the right of 0 on the number line.

Let's check the other options:

(A) to the left of 0: This is incorrect as 5 is positive.

(C) to the left of 1: This is incorrect as 5 > 1, so 5 is to the right of 1.

(D) to the left of –2: This is incorrect as 5 > –2, so 5 is to the right of –2.

Comparing with the given options:

(A) to the left of 0

(B) to the right of 0

(C) to the left of 1

(D) to the left of –2

The correct option is (B).

On a number line, if an integer $a$ is to the left of an integer $b$, it means that $a$ is less than $b$ ($a < b$).

The question asks for the pair where the first integer is not on the left of the other integer.

This means we are looking for a pair $(a, b)$ where the first integer $a$ is not less than the second integer $b$. In other words, $a \geq b$.

Let's examine each option:

(A) (–1, 10)

Here, $a = -1$ and $b = 10$. We check if $-1 < 10$. Yes, $-1 < 10$. So, –1 is to the left of 10.

(B) (–3, –5)

Here, $a = -3$ and $b = -5$. We check if $-3 < -5$. No, $-3$ is not less than $-5$. In fact, $-3 > -5$. So, –3 is to the right of –5.

(C) (–5, –3)

Here, $a = -5$ and $b = -3$. We check if $-5 < -3$. Yes, $-5 < -3$. So, –5 is to the left of –3.

(D) (–6, 0)

Here, $a = -6$ and $b = 0$. We check if $-6 < 0$. Yes, $-6 < 0$. So, –6 is to the left of 0.

The pair where the first integer is not on the left of the other integer is (–3, –5), because –3 is greater than –5 and therefore to the right of –5 on the number line.

Comparing with the given options:

(A) (–1, 10)

(B) (–3, –5)

(C) (–5, –3)

(D) (–6, 0)

The correct option is (B).

An integer with a negative sign (–) is a negative integer.

The set of negative integers is $\{-1, -2, -3, -4, ...\}$.

We need to find a value from the options that every negative integer is always less than.

Let's examine each option:

(A) 0: By definition, all negative integers are less than 0. For any negative integer $x$, we have $x < 0$. This statement is always true.

(B) –3: Is every negative integer less than –3? Consider the negative integer –1. Is $-1 < -3$? No, $-1 > -3$. So, this statement is not always true.

(C) –1: Is every negative integer less than –1? Consider the negative integer –1 itself. Is $-1 < -1$? No, $-1 = -1$. So, this statement is not always true.

(D) –2: Is every negative integer less than –2? Consider the negative integer –1. Is $-1 < -2$? No, $-1 > -2$. Also, consider –2. Is $-2 < -2$? No, $-2 = -2$. So, this statement is not always true.

Only option (A) is always true for every integer with a negative sign.

The correct option is (A).

An integer with a positive sign (+) is a positive integer.

The set of positive integers is $\{1, 2, 3, 4, ...\}$.

We need to find a value from the options that every positive integer is always greater than.

Let's examine each option:

(A) 0: All positive integers are located to the right of 0 on the number line. This means every positive integer is greater than 0. For any positive integer $x$, we have $x > 0$. This statement is always true.

(B) 1: Is every positive integer greater than 1? No. The positive integer 1 is not greater than 1, since $1 = 1$. So, this statement is not always true.

(C) 2: Is every positive integer greater than 2? No. The positive integers 1 and 2 are not greater than 2 ($1 < 2$ and $2 = 2$). So, this statement is not always true.

(D) 3: Is every positive integer greater than 3? No. The positive integers 1, 2, and 3 are not greater than 3 ($1 < 3$, $2 < 3$, and $3 = 3$). So, this statement is not always true.

Only option (A) is always true for every integer with a positive sign.

The correct option is (A).

First, let's find the predecessor of –50.

The predecessor of an integer is found by subtracting 1 from it.

Predecessor of –50 = $-50 - 1 = -51$.

Now, we need to find the successor of the predecessor of –50, which is the successor of –51.

The successor of an integer is found by adding 1 to it.

Successor of –51 = $-51 + 1 = -50$.

Thus, the successor of the predecessor of –50 is –50.

Alternatively, for any integer $n$, the predecessor is $n-1$ and the successor is $n+1$.

The predecessor of –50 is $-50 - 1$.

The successor of the predecessor of –50 is the successor of $(-50 - 1)$.

Successor of $(-50 - 1) = (-50 - 1) + 1 = -50 - 1 + 1 = -50 + 0 = -50$.

The result is –50.

Comparing with the given options:

(A) –48

(B) –49

(C) –50

(D) –51

The correct option is (C).

The additive inverse of an integer $a$ is an integer $b$ such that $a + b = 0$. The additive inverse of $a$ is denoted by $-a$.

We are considering the additive inverse of a negative integer.

Let the negative integer be $n$. Since $n$ is a negative integer, $n < 0$.

We want to find the additive inverse of $n$, which is $-n$.

Let's take an example. Consider the negative integer –5.

The additive inverse of –5 is $-(-5)$.

We know that $-(-a) = a$. So, $-(-5) = 5$.

The number 5 is a positive integer.

Let's consider another example. Consider the negative integer –10.

The additive inverse of –10 is $-(-10) = 10$.

The number 10 is a positive integer.

In general, if $n$ is a negative integer, it can be written as $n = -k$ where $k$ is a positive integer ($k > 0$).

The additive inverse of $n = -k$ is $-n = -(-k) = k$.

Since $k$ is a positive integer, the additive inverse of a negative integer is always positive.

Comparing with the given options:

(A) is always negative

(B) is always positive

(C) is the same integer

(D) zero

The correct option is (B).

The minimum temperature at place A is $-4^\circ$C.

The minimum temperature at place B is $-1^\circ$C.

When comparing temperatures, a lower temperature means it is cooler.

We need to compare the integers –4 and –1.

On the number line, –4 is located to the left of –1.

Therefore, $-4 < -1$.

Since $-4^\circ$C is less than $-1^\circ$C, place A is cooler than place B.

Let's evaluate the given options:

(A) A is cooler than B: This statement is true because $-4^\circ$C is less than $-1^\circ$C.

(B) B is cooler than A: This statement is false because $-1^\circ$C is greater than $-4^\circ$C, meaning B is warmer than A.

(C) There is a difference of 2°C in the temperature: The difference in temperature is $|-4 - (-1)| = |-4 + 1| = |-3| = 3^\circ$C. The difference is 3°C, not 2°C. This statement is false.

(D) The temperature at A is 4°C higher than that at B: This would mean Temperature at A = Temperature at B + 4°C. So, $-4^\circ$C = $-1^\circ$C + $4^\circ$C = $3^\circ$C. This is false as $-4 \neq 3$. This statement is false.

The only true statement among the options is (A).

The correct option is (A).

Let the two negative integers be $a$ and $b$. We are considering the subtraction $a - b$, where both $a < 0$ and $b < 0$.

Subtracting a negative integer is equivalent to adding its positive additive inverse. So, $a - b = a + (-b)$. Since $b$ is negative, $-b$ is positive.

We are essentially adding a negative number ($a$) and a positive number ($-b$). The sign of the result depends on the absolute values of the numbers being added.

Let's consider some examples:

Example 1: Subtract –5 from –3.

Expression: $-3 - (-5)$

Calculation: $-3 - (-5) = -3 + 5 = 2$. The result is positive.

Example 2: Subtract –3 from –5.

Expression: $-5 - (-3)$

Calculation: $-5 - (-3) = -5 + 3 = -2$. The result is negative.

Example 3: Subtract –5 from –5.

Expression: $-5 - (-5)$

Calculation: $-5 - (-5) = -5 + 5 = 0$. The result is zero (which has no sign, but is neither positive nor negative).

From these examples, we can see that the sign of the result is not always negative, not always positive, and it can be negative (as in Example 2). The sign of the result depends on the specific numerical values (or their absolute values) of the two negative integers involved in the subtraction.

Specifically, if $|b| > |a|$, the result $a + (-b)$ will have the same sign as $-b$, which is positive. (e.g., $-3 - (-5) = -3 + 5 = 2$, here $|-5| > |-3|$).

If $|b| < |a|$, the result $a + (-b)$ will have the same sign as $a$, which is negative. (e.g., $-5 - (-3) = -5 + 3 = -2$, here $|-3| < |-5|$).

If $|b| = |a|$, the result is zero. (e.g., $-5 - (-5) = -5 + 5 = 0$, here $|-5| = |-5|$).

Therefore, the sign of the result depends on the numerical values of the integers.

Comparing with the given options:

(A) is always negative

(B) is always positive

(C) is never negative

(D) depends on the numerical value of the integers

The correct option is (D).

Let the integer be $n$.

When the integer is added to itself, the sum is $n + n = 2n$.

The statement is "The sum is greater than the integer", which can be written as the inequality: $2n > n$.

We need to determine for which integers $n$ this inequality holds true.

Consider the inequality $2n > n$. Subtracting $n$ from both sides, we get:

$2n - n > n - n$

$n > 0$

So, the statement "$2n > n$" is equivalent to the condition "$n > 0$".

Let's consider different types of integers:

Case 1: When the integer $n$ is positive.

If $n$ is a positive integer, then $n > 0$. Our condition $n > 0$ is satisfied.

For example, if $n = 3$, the sum is $3 + 3 = 6$. Is $6 > 3$? Yes, true.

Case 2: When the integer $n$ is negative.

If $n$ is a negative integer, then $n < 0$. Our condition $n > 0$ is not satisfied.

For example, if $n = -3$, the sum is $-3 + (-3) = -6$. Is $-6 > -3$? No, $-6$ is less than $-3$. False.

Case 3: When the integer $n$ is zero.

If $n = 0$, then $n$ is neither positive nor negative. Our condition $n > 0$ is not satisfied.

The sum is $0 + 0 = 0$. Is $0 > 0$? No, $0$ is equal to $0$. False.

From the analysis, the statement "$2n > n$" is true only when the integer $n$ is positive.

Comparing with the given options:

(A) always true - False (not true for negative integers or zero)

(B) never true - False (true for positive integers)

(C) true only when the integer is positive - True (matches our conclusion)

(D) true for non-negative integers - False (non-negative integers include 0, and it's false for 0)

The correct option is (C).

To find the rise in temperature, we subtract the initial temperature from the final temperature.

Let's calculate the rise for each option:

(A) From $0^\circ$C to $10^\circ$C

Rise = Final temperature - Initial temperature

Rise$_A = 10^\circ\text{C} - 0^\circ\text{C} = 10^\circ\text{C}$.

(B) From $-4^\circ$C to $8^\circ$C

Rise = Final temperature - Initial temperature

Rise$_B = 8^\circ\text{C} - (-4^\circ\text{C}) = 8^\circ\text{C} + 4^\circ\text{C} = 12^\circ\text{C}$.

(C) From $-15^\circ$C to $-8^\circ$C

Rise = Final temperature - Initial temperature

Rise$_C = -8^\circ\text{C} - (-15^\circ\text{C}) = -8^\circ\text{C} + 15^\circ\text{C} = 7^\circ\text{C}$.

(D) From $-7^\circ$C to $0^\circ$C

Rise = Final temperature - Initial temperature

Rise$_D = 0^\circ\text{C} - (-7^\circ\text{C}) = 0^\circ\text{C} + 7^\circ\text{C} = 7^\circ\text{C}$.

Comparing the rises in temperature for each option:

Rise$_A = 10^\circ$C

Rise$_B = 12^\circ$C

Rise$_C = 7^\circ$C

Rise$_D = 7^\circ$C

The maximum rise in temperature is the largest value among 10°C, 12°C, 7°C, and 7°C.

The maximum rise is $12^\circ$C, which occurs in option (B).

Comparing with the given options:

(A) 0°C to 10°C (Rise = 10°C)

(B) –4°C to 8°C (Rise = 12°C)

(C) –15°C to –8°C (Rise = 7°C)

(D) –7°C to 0°C (Rise = 7°C)

The correct option is (B).

Natural numbers are the counting numbers.

The set of natural numbers typically starts from 1: $\{1, 2, 3, 4, ...\}$.

According to this definition, the smallest natural number is 1.

The number zero (0) is not included in the set of natural numbers according to this common definition.

Therefore, the statement "The smallest natural number is zero" is incorrect based on the standard definition where natural numbers begin with 1.

The statement is False.

Integers are the set of whole numbers and their opposites (negative counting numbers).

The set of integers is $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

This set explicitly includes zero (0).

While it is true that zero is neither positive nor negative, this characteristic does not exclude it from being an integer.

Zero serves as the origin on the number line and separates the positive integers from the negative integers.

Therefore, zero is an integer.

The statement "Zero is not an integer as it is neither positive nor negative" is incorrect.

The statement is False.

We need to find the integers that lie strictly between –5 and –1.

These are the integers $x$ such that $-5 < x < -1$.

Listing the integers between –5 and –1: –4, –3, –2.

Now, we need to find the sum of these integers:

Sum = $(-4) + (-3) + (-2)$.

Adding negative numbers: Sum = $-(4 + 3 + 2) = -9$.

The sum of the integers between –5 and –1 is –9.

The given statement says the sum is –6.

Since $-9 \neq -6$, the statement is incorrect.

The statement is False.

The successor of an integer is the integer that comes just after it on the number line.

To find the successor of an integer, we add 1 to the integer.

We need to find the successor of the integer 1.

Successor of 1 = $1 + 1 = 2$.

The successor of the integer 1 is 2.

The given statement says the successor of the integer 1 is 0.

Since $2 \neq 0$, the statement is incorrect.

The statement is False.

Positive integers are $\{1, 2, 3, ...\}$.

Negative integers are $\{..., -3, -2, -1\}$.

On the number line, positive integers are located to the right of 0, and negative integers are located to the positive integer is to the right of 0, and every negative integer is to the left of 0. This means every positive integer is to the right of every negative integer.

Thus left of 0.

Any number to the right on the number line is greater than any number to its left.

Since all positive integers are, every positive integer is greater than every negative integer.

For example, $1 > -1$, $5 > -100$, $1000 > -0 to the right of 0, and all negative integers are to the left of 0, every positive integer is located to the right of every negative integer on the number line.

For.5$ (although -0.5 is not an integer, this illustrates the concept across 0). For integers, $1 > -1$, $1 > -10$, $5 > -2$, $5 > -5 any positive integer $p$ and any negative integer $n$, we have $p > 0$ and $n < 0$. This implies $p > 0 > n$, so $p > n$.

The statement "Every positive integer is larger than every negative integer" is correct.

The statement is True.

Let the two negative integers be $a$ and $b$.

Since $a$ is a negative integer, $a < 0$.

Since $b$ is a negative integer, $b < 0$.

The sum of the two integers is $a + b$.

Let's consider an example. Let $a = -2$ and $b = -3$. Both are negative integers.

The sum is $a + b = -2 + (-3) = -5$.

Now we compare the sum (–5) with the original integers (–2 and –3).

Is $-5 > -2$? On the number line, –5 is to the left of –2, so $-5 < -2$.

Is $-5 > -3$? On the number line, –5 is to the left of –3, so $-5 < -3$.

In this example, the sum (–5) is less than both of the original integers (–2 and –3).

In general, when you add a negative number to any number, the result is always less than the original number.

Since $b$ is negative, $a + b < a$.

Since $a$ is negative, $a + b < b$.

Thus, the sum of any two negative integers is always less than both the integers.

The statement "The sum of any two negative integers is always greater than both the integers" is incorrect.

The statement is False.

Let the two negative integers be $a$ and $b$.

Since $a$ is a negative integer, $a < 0$.

Since $b$ is a negative integer, $b < 0$.

The sum of the two integers is $a + b$.

Consider the inequality comparing the sum $a+b$ with $a$. We are adding $b$ to $a$. Since $b$ is negative ($b < 0$), adding $b$ to $a$ will result in a number smaller than $a$.

$a + b < a$

Similarly, consider the inequality comparing the sum $a+b$ with $b$. We are adding $a$ to $b$. Since $a$ is negative ($a < 0$), adding $a$ to $b$ will result in a number smaller than $b$.

$a + b < b$

Since the sum $a+b$ is smaller than $a$ and also smaller than $b$, the sum of any two negative integers is always smaller than both the integers.

Let's verify with an example: $a = -5$, $b = -2$.

Sum = $-5 + (-2) = -7$.

Is $-7 < -5$? Yes, true.

Is $-7 < -2$? Yes, true.

The statement "The sum of any two negative integers is always smaller than both the integers" is correct.

The statement is True.

Solution:

Let the two positive integers be $a$ and $b$.

According to the statement, the sum $a+b$ must be greater than $a$ and also greater than $b$.

For any positive integer $b > 0$, adding $b$ to $a$ will result in a number greater than $a$.

So, $a+b > a$.

Similarly, for any positive integer $a > 0$, adding $a$ to $b$ will result in a number greater than $b$.

So, $a+b > b$.

This property holds true for any pair of positive integers.

For example, let the integers be $3$ and $5$. Their sum is $3+5=8$.

$8 > 3$ and $8 > 5$.

Therefore, the statement "The sum of any two positive integers is greater than both the integers" is true.

True (T)

Solution:

The set of whole numbers is $\{0, 1, 2, 3, 4, \dots\}$.

The set of integers is $\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$.

By comparing the two sets, we can see that every element in the set of whole numbers is also present in the set of integers.

In other words, the set of whole numbers is a subset of the set of integers.

Therefore, the statement "All whole numbers are integers" is true.

True (T)

Solution:

The set of integers is $\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$.

The set of whole numbers is $\{0, 1, 2, 3, \dots\}$.

Integers include negative numbers such as $-1, -2, -3$, etc.

Whole numbers do not include negative numbers.

For example, $-5$ is an integer, but it is not a whole number.

Therefore, the statement "All integers are whole numbers" is false.

False (F)

Solution:

We are given that $5 > 3$. This is a true statement.

The statement claims that because $5 > 3$, it implies that $-5 > -3$.

When comparing negative numbers, the number with the smaller absolute value is greater.

The absolute value of $-5$ is $|-5| = 5$.

The absolute value of $-3$ is $|-3| = 3$.

Since $5 > 3$, the number with the smaller absolute value is $-3$.

Therefore, $-3$ is greater than $-5$.

In inequality form, this is written as $-5 < -3$.

The statement claims $-5 > -3$, which contradicts the correct comparison.

Therefore, the statement "Since 5 > 3, therefore –5 > –3" is false.

False (F)

Solution:

Positive integers are the set of natural numbers: $\{1, 2, 3, 4, \dots\}$.

Comparing zero ($0$) with any positive integer ($n$), we know that $0$ is to the left of any positive number on the number line.

Thus, for any positive integer $n$, the inequality $0 < n$ holds true.

For example, $0 < 1$, $0 < 100$, etc.

Therefore, the statement "Zero is less than every positive integer" is true.

True (T)

Solution:

Negative integers are the integers that are less than zero: $\{\dots, -3, -2, -1\}$.

On a number line, numbers increase in value as you move from left to right.

Zero is located to the right of all negative integers on the number line.

Therefore, any negative integer is to the left of zero, meaning zero is greater than any negative integer.

For example, $0 > -1$, $0 > -10$, $0 > -1000$.

Therefore, the statement "Zero is larger than every negative integer" is true.

True (T)

Solution:

Positive numbers are defined as numbers greater than zero ($x > 0$).

Negative numbers are defined as numbers less than zero ($x < 0$).

Zero is exactly $0$.

Zero is not greater than zero, so it is not a positive number.

Zero is not less than zero, so it is not a negative number.

Zero serves as the boundary between positive and negative numbers on the number line.

Therefore, the statement "Zero is neither positive nor negative" is true.

True (T)

Solution:

The number line is constructed such that numbers increase in value as you move from left to right.

If an integer $b$ is located to the right of an integer $a$ on the number line, it means that $b$ is greater than $a$.

This is a fundamental property of the number line and the ordering of integers.

For example, consider the integer $2$. The integer to its right, $3$, is larger ($3 > 2$).

Consider the integer $-5$. The integer to its right, $-4$, is larger ($-4 > -5$).

Therefore, the statement "On the number line, an integer on the right of a given integer is always larger than the integer" is true.

True (T)

Solution:

On the number line, numbers increase from left to right. This means a number further to the left is smaller than a number to its right.

We need to compare the values of $-2$ and $-5$.

We know that $-2$ is greater than $-5$ ($-2 > -5$).

Since $-2$ is greater than $-5$, $-2$ is located to the right of $-5$ on the number line.

Conversely, $-5$ is less than $-2$, so $-5$ is located to the left of $-2$ on the number line.

Therefore, the statement "–2 is to the left of –5 on the number line" is false.

False (F)

Solution:

The set of integers is $\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$.

The integers extend infinitely in both the positive and negative directions.

For any given integer, say $n$, there is always a smaller integer, such as $n-1$.

For example, $-1$ is smaller than $0$, $-100$ is smaller than $0$, and so on.

There is no integer that is smaller than all other integers.

Therefore, the statement "The smallest integer is 0" is false.

False (F)

Solution:

The distance of a number from zero on the number line is its absolute value.

The distance of $6$ from $0$ is $|6| = 6$.

The distance of $-6$ from $0$ is $|-6| = 6$.

Since $|6| = |-6| = 6$, both $6$ and $-6$ are $6$ units away from $0$ on the number line.

Therefore, the statement "6 and –6 are at the same distance from 0 on the number line" is true.

True (T)

Solution:

Let the integer be $a$.

The additive inverse of an integer $a$ is the number that, when added to $a$, gives zero. This number is $-a$.

The difference between an integer and its additive inverse can be expressed in two ways:

Difference $1 = a - (-a) = a + a = 2a$.

Difference $2 = (-a) - a = -a - a = -2a$.

We need to determine if $2a$ and $-2a$ are always even for any integer $a$.

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

In the expression $2a$, the value is $2$ multiplied by an integer $a$. By definition, any integer multiplied by $2$ is an even number.

So, $2a$ is always an even number for any integer $a$.

Similarly, $-2a = 2(-a)$. Since $-a$ is also an integer, $2(-a)$ is also always an even number.

For example:

If $a = 3$ (odd integer), the difference is $2(3) = 6$ (even) or $-2(3) = -6$ (even).

If $a = -4$ (even integer), the difference is $2(-4) = -8$ (even) or $-2(-4) = 8$ (even).

If $a = 0$, the difference is $2(0) = 0$ (even).

Since the difference between an integer and its additive inverse is always of the form $2a$ or $-2a$, and both are always even, the statement is true.

True (T)

Solution:

Let the integer be $a$.

The additive inverse of an integer $a$ is the number $-a$ such that $a + (-a) = 0$.

The sum of an integer and its additive inverse is $a + (-a)$.

By the definition of additive inverse, $a + (-a) = 0$.

This holds true for any integer $a$.

For example:

If the integer is $5$, its additive inverse is $-5$. The sum is $5 + (-5) = 0$.

If the integer is $-10$, its additive inverse is $-(-10) = 10$. The sum is $-10 + 10 = 0$.

Therefore, the statement "The sum of an integer and its additive inverse is always zero" is true.

True (T)

Solution:

Let the two negative integers be $a$ and $b$.

Since $a$ and $b$ are negative integers, $a < 0$ and $b < 0$.

When we add two negative numbers, the result is always a negative number.

The sum $a+b$ will be less than $0$.

For example, consider the two negative integers $-3$ and $-5$.

Their sum is $(-3) + (-5) = -8$.

$-8$ is a negative integer, not a positive integer.

Consider another example, $-1$ and $-2$.

Their sum is $(-1) + (-2) = -3$.

$-3$ is a negative integer.

Therefore, the statement "The sum of two negative integers is a positive integer" is false. The sum of two negative integers is always a negative integer.

False (F)

Solution:

The statement claims that it is impossible for the sum of three integers, which are all distinct from each other, to be equal to zero.

Let's test this claim by considering some examples of three different integers.

Consider the integers $1, 2$, and $3$. These are three different integers. Their sum is $1 + 2 + 3 = 6$, which is not zero. This example does not contradict the statement.

Consider the integers $-1, 0$, and $1$. These are three different integers ($(-1) \neq 0$, $0 \neq 1$, and $(-1) \neq 1$).

Let's find their sum:

Sum = $(-1) + 0 + 1$.

$(-1) + 0 = -1$.

$-1 + 1 = 0$.

So, the sum of the three different integers $-1, 0$, and $1$ is $0$.

Since we have found a case where the sum of three different integers IS zero, the statement "The sum of three different integers can never be zero" is false.

Therefore, the statement is false.

False (F)

Solution:

On the number line, numbers are arranged in increasing order from left to right.

Positive numbers are located to the right of zero.

Negative numbers are located to the left of zero.

The integer $-15$ is a negative number, as it is less than $0$.

Therefore, $-15$ is located to the left of zero on the number line.

The completed statement is: On the number line, –15 is to the left of zero.

Solution:

On the number line, numbers are arranged in increasing order from left to right.

Positive numbers are located to the right of zero.

Negative numbers are located to the left of zero.

The integer $10$ is a positive number, as it is greater than $0$.

Therefore, $10$ is located to the right of zero on the number line.

The completed statement is: On the number line, 10 is to the right of zero.

Solution:

The additive inverse of an integer is the number that, when added to the integer, results in a sum of zero.

If the integer is $a$, its additive inverse is $-a$.

We are asked for the additive inverse of $14$.

Let the additive inverse be $x$. Then, according to the definition:

$14 + x = 0$

Subtracting $14$ from both sides, we get:

$x = 0 - 14$

$x = -14$.

So, the additive inverse of $14$ is $-14$.

The completed statement is: The additive inverse of 14 is –14.

Solution:

The additive inverse of an integer is the number that, when added to the integer, results in a sum of zero.

If the integer is $a$, its additive inverse is $-a$.

We are asked for the additive inverse of $-1$.

Let the additive inverse be $x$. Then:

$(-1) + x = 0$

Adding $1$ to both sides, we get:

$x = 0 + 1$

$x = 1$.

So, the additive inverse of $-1$ is $1$. This is also found by $-(-1) = 1$.

The completed statement is: The additive inverse of –1 is 1.

Solution:

The additive inverse of an integer is the number that, when added to the integer, results in a sum of zero.

If the integer is $a$, its additive inverse is $-a$.

We are asked for the additive inverse of $0$.

Let the additive inverse be $x$. Then:

$0 + x = 0$

Subtracting $0$ from both sides, we get:

$x = 0 - 0$

$x = 0$.

So, the additive inverse of $0$ is $0$.

The completed statement is: The additive inverse of 0 is 0.

Solution:

We need to find the integers that are greater than $-5$ and less than $5$.

These integers are listed in increasing order: $-4, -3, -2, -1, 0, 1, 2, 3, 4$.

Now, let's count the number of integers in this list.

There are $9$ integers: $-4, -3, -2, -1, 0, 1, 2, 3, 4$.

Alternatively, the number of integers between $a$ and $b$ (where $a < b$ and $a, b$ are integers) is $(b - a - 1)$.

In this case, $a = -5$ and $b = 5$.

Number of integers = $5 - (-5) - 1 = 5 + 5 - 1 = 10 - 1 = 9$.

The completed statement is: The number of integers lying between –5 and 5 is 9.

Solution:

We need to find the sum of the three negative integers $-11$, $-2$, and $-1$.

When adding negative integers, we add their absolute values and place a negative sign before the result.

The absolute values are:

$|-11| = 11$

$|-2| = 2$

$|-1| = 1$

Sum of absolute values = $11 + 2 + 1 = 14$.

Since all integers are negative, the sum is $-14$.

Alternatively, we can group the terms:

$(-11) + (-2) + (-1) = (-11 - 2) + (-1)$

$= (-13) + (-1)$

$= -13 - 1$

$= -14$.

The completed statement is: (–11) + (–2) + (–1) = –14.

Solution:

Let the missing integer be $x$.

The given equation is:

$x + (-11) + 111 = 130$

Simplify the left side of the equation:

$x - 11 + 111 = 130$

Combine the constant terms $-11$ and $111$:

$-11 + 111 = 100$

So the equation becomes:

$x + 100 = 130$

To find $x$, subtract $100$ from both sides of the equation:

$x = 130 - 100$

$x = 30$

Thus, the missing integer is $30$.

The completed statement is: 30 + (–11) + 111 = 130.

Solution:

We need to find the sum of the integers $-80$, $0$, and $-90$.

The sum is $(-80) + 0 + (-90)$.

Adding $0$ to any number does not change the value. So, $(-80) + 0 = -80$.

The expression becomes:

$(-80) + (-90)$

To add two negative integers, we add their absolute values and place a negative sign before the result.

$|-80| = 80$ and $|-90| = 90$.

Sum of absolute values $= 80 + 90 = 170$.

The sum is therefore $-(170) = -170$.

$\begin{array}{cc} & 8 & 0 \\ + & 9 & 0 \\ \hline 1 & 7 & 0 \\ \hline \end{array}$

So, $(-80) + (-90) = -170$.

The completed statement is: (–80) + 0 + (–90) = –170.

Solution:

Let the missing number be $x$.

The given equation is:

$x - 3456 = -8910$

To find $x$, we need to isolate $x$ on one side of the equation. We can do this by adding $3456$ to both sides of the equation.

$x - 3456 + 3456 = -8910 + 3456$

$x = -8910 + 3456$

Now we need to calculate the sum of $-8910$ and $3456$. This is equivalent to subtracting the smaller absolute value from the larger absolute value and taking the sign of the number with the larger absolute value.

$|-8910| = 8910$

$|3456| = 3456$

Since $8910 > 3456$, the result will be negative.

We calculate $8910 - 3456$:

$\begin{array}{cccc} & 8 & 9 & 1 & 0 \\ - & 3 & 4 & 5 & 6 \\ \hline & 5 & 4 & 5 & 4 \\ \hline \end{array}$

So, $8910 - 3456 = 5454$.

Therefore, $-8910 + 3456 = -5454$.

$x = -5454$.

The completed statement is: –5454 –3456 = –8910.

Solution:

We need to compare the values of the expressions on both sides of the blank.

Left side: $(-11) + (-15)$

Adding two negative integers: add their absolute values and put a negative sign.

$(-11) + (-15) = -(11 + 15) = -26$.

Right side: $11 + 15$

Adding two positive integers:

$11 + 15 = 26$.

Now we need to compare $-26$ and $26$.

Negative numbers are always less than positive numbers.

So, $-26 < 26$.

The completed comparison is: (–11) + (–15) < 11 + 15.

Solution:

We need to evaluate the expressions on both sides of the blank and compare their values.

Left side: $(-71) + (+9)$

This is the sum of a negative and a positive integer. Subtract the smaller absolute value from the larger absolute value and take the sign of the number with the larger absolute value.

$|-71| = 71$ and $|+9| = 9$.

The difference is $71 - 9 = 62$.

Since $|-71| > |+9|$, the result is negative.

$(-71) + (+9) = -62$.

Right side: $(-81) + (–9)$

This is the sum of two negative integers. Add their absolute values and put a negative sign.

$|-81| = 81$ and $|-9| = 9$.

The sum of absolute values is $81 + 9 = 90$.

Since both integers are negative, the result is negative.

$(-81) + (-9) = -90$.

Now we need to compare the values of the left side and the right side: $-62$ and $-90$.

On the number line, $-62$ is to the right of $-90$.

Therefore, $-62$ is greater than $-90$.

$-62 > -90$.

The completed comparison is: (–71) + (+9) > (–81) + (–9).

Solution:

We need to compare the values of $0$ and $1$.

On the number line, $0$ is located to the left of $1$.

Numbers increase in value from left to right on the number line.

Therefore, $0$ is less than $1$.

This is written using the less than symbol ($<$).

$0 < 1$.

The completed comparison is: 0 < 1.

Solution:

We need to compare the values of $-60$ and $50$.

$-60$ is a negative integer.

$50$ is a positive integer.

On the number line, negative numbers are always to the left of positive numbers.

Therefore, any negative number is less than any positive number.

So, $-60 < 50$.

The completed comparison is: –60 < 50.

Solution:

We need to compare the values of the negative integers $-10$ and $-11$.

On the number line, numbers increase from left to right.

When comparing two negative integers, the one with the smaller absolute value is greater because it is closer to zero on the number line.

The absolute value of $-10$ is $|-10| = 10$.

The absolute value of $-11$ is $|-11| = 11$.

Since $10 < 11$, $-10$ is closer to zero than $-11$.

Therefore, $-10$ is greater than $-11$.

This is written using the greater than symbol ($>$).

$-10 > -11$.

The completed comparison is: –10 > –11.

Solution:

We need to compare the values of the negative integers $-101$ and $-102$.

On the number line, numbers increase from left to right.

When comparing two negative integers, the one with the smaller absolute value is greater because it is closer to zero on the number line.

The absolute value of $-101$ is $|-101| = 101$.

The absolute value of $-102$ is $|-102| = 102$.

Since $101 < 102$, $-101$ is closer to zero than $-102$.

Therefore, $-101$ is greater than $-102$.

This is written using the greater than symbol ($>$).

$-101 > -102$.

The completed comparison is: –101 > –102.

Solution:

We need to evaluate the expressions on both sides of the blank and compare their values.

Left side: $(-2) + (-5) + (-6)$

Sum of negative integers:

$(-2) + (-5) + (-6) = (-7) + (-6) = -13$.

Right side: $(-3) + (–4) + (–6)$

Sum of negative integers:

$(-3) + (-4) + (-6) = (-7) + (-6) = -13$.

Now we need to compare the values of the left side and the right side: $-13$ and $-13$.

The two values are equal.

$-13 = -13$.

The completed comparison is: (–2) + (–5) + (–6) = (–3) + (–4) + (–6).

Solution:

We need to compare the values of $0$ and $-2$.

$0$ is neither positive nor negative.

$-2$ is a negative integer.

On the number line, $0$ is located to the right of all negative numbers.

Therefore, $0$ is greater than any negative number.

So, $0 > -2$.

The completed comparison is: 0 > –2.

Solution:

We need to evaluate the expressions on both sides of the blank and compare their values.

Left side: $1 + 2 + 3$

Sum of positive integers:

$1 + 2 + 3 = 6$.

Right side: $(-1) + (–2) + (–3)$

Sum of negative integers: Add their absolute values and put a negative sign.

$(-1) + (-2) + (-3) = -(1 + 2 + 3) = -(6) = -6$.

Now we need to compare the values of the left side and the right side: $6$ and $-6$.

$6$ is a positive number, and $-6$ is a negative number.

On the number line, positive numbers are always to the right of negative numbers.

Therefore, $6$ is greater than $-6$.

$6 > -6$.

The completed comparison is: 1 + 2 + 3 > (–1) + (–2) + (–3).

Solution:

Let's evaluate each item in Column I:

(i) The additive inverse of $+2$: The additive inverse of a number is the number that sums to $0$ with the original number. The additive inverse of $+2$ is $-2$.

(ii) The greatest negative integer: Negative integers are $\dots, -3, -2, -1$. The greatest among these is the one closest to zero, which is $-1$.

(iii) The greatest negative even integer: Negative even integers are $\dots, -6, -4, -2$. The greatest among these is the one closest to zero, which is $-2$.

(iv) The smallest integer greater than every negative integer: The negative integers are $\{\dots, -3, -2, -1\}$. Any integer greater than all negative integers must be $\geq 0$. The smallest such integer is $0$.

(v) Sum of predecessor and successor of –1: The predecessor of $-1$ is $-1 - 1 = -2$. The successor of $-1$ is $-1 + 1 = 0$. The sum is $(-2) + 0 = -2$.

Matching these values with Column II:

(i) $-2$ matches with (B).

(ii) $-1$ matches with (E).

(iii) $-2$ matches with (B).

(iv) $0$ matches with (A).

(v) $-2$ matches with (B).

The matches are:

(i) - (B)

(ii) - (E)

(iii) - (B)

(iv) - (A)

(v) - (B)

Solution (a):

We need to compute $30 + (-25) + (-10)$.

$30 + (-25) = 30 - 25 = 5$.

Now, add this result to $(-10)$:

$5 + (-10) = 5 - 10 = -5$.

Solution (b):

We need to compute $(-20) + (-5)$.

When adding two negative integers, add their absolute values and put a negative sign before the sum.

$(-20) + (-5) = -(20 + 5) = -25$.

Solution (c):

We need to compute $70 + (-20) + (-30)$.

First, compute $70 + (-20)$:

$70 + (-20) = 70 - 20 = 50$.

Now, add this result to $(-30)$:

$50 + (-30) = 50 - 30 = 20$.

Solution (d):

We need to compute $-50 + (-60) + 50$.

We can rearrange the terms using the commutative property of addition:

$(-50) + (-60) + 50 = (-50) + 50 + (-60)$.

Since $-50$ and $50$ are additive inverses, their sum is $0$:

$(-50) + 50 = 0$.

So the expression becomes:

$0 + (-60) = -60$.

Solution (e):

We need to compute $1 + (-2) + (-3) + (-4)$.

We can group the negative integers:

$1 + ((-2) + (-3) + (-4))$.

Sum of negative integers: $(-2) + (-3) + (-4) = -(2 + 3 + 4) = -(9) = -9$.

So the expression becomes:

$1 + (-9) = 1 - 9 = -8$.

Solution (f):

We need to compute $0 + (-5) + (-2)$.

Adding $0$ to any number does not change its value:

$0 + (-5) = -5$.

Now, add this result to $(-2)$:

$(-5) + (-2)$.

Sum of two negative integers: add absolute values and put a negative sign.

$(-5) + (-2) = -(5 + 2) = -7$.

Solution (g):

We need to compute $0 - (-6) - (+6)$.

Subtracting a negative number is the same as adding its positive counterpart:

$0 - (-6) = 0 + 6 = 6$.

Subtracting a positive number is the same as adding its negative counterpart:

$-(+6) = -6$.

So the expression becomes:

$6 - 6$.

$6 - 6 = 0$.

Solution (h):

We need to compute $0 - 2 - (-2)$.

$0 - 2 = -2$.

Subtracting a negative number is the same as adding its positive counterpart:

$-(-2) = +2$.

So the expression becomes:

$-2 + 2$.

$-2 + 2 = 0$.

Solution:

We are given that height above sea level is represented by a positive integer, and depth below sea level is represented by a negative integer. Sea level is the reference point, which is represented by $0$.

(a) 200 m above sea level:

Since the height is above sea level, it is represented by a positive integer.

The integer is $+200$ or simply $200$.

Answer for (a) is 200.

(b) 100 m below sea level:

Since the depth is below sea level, it is represented by a negative integer.

The integer is $-100$.

Answer for (b) is –100.

(c) 10 m above sea level:

Since the height is above sea level, it is represented by a positive integer.

The integer is $+10$ or simply $10$.

Answer for (c) is 10.

(d) sea level:

Sea level is the point of reference, which is represented by $0$.

The integer is $0$.

Answer for (d) is 0.

Solution:

We need to provide the opposite statement for each given phrase.

(a) Decrease in size: The opposite of decrease is increase.

Opposite: Increase in size

(b) Failure: The opposite of failure is success.

Opposite: Success

(c) Profit of $\textsf{₹}10$: The opposite of profit is loss.

Opposite: Loss of $\textsf{₹}10$

(d) 1000 A.D.: The opposite direction in time from A.D. (Anno Domini) is B.C. (Before Christ).

Opposite: 1000 B.C.

(e) Rise in water level: The opposite of rise is fall or decrease.

Opposite: Fall in water level (or Decrease in water level)

(f) 60 km south: The opposite direction of south is north.

Opposite: 60 km north

(g) 10 m above the danger mark of river Ganga: The opposite of above is below.

Opposite: 10 m below the danger mark of river Ganga

(h) 20 m below the danger mark of the river Brahmaputra: The opposite of below is above.

Opposite: 20 m above the danger mark of the river Brahmaputra

(i) Winning by a margin of 2000 votes: The opposite of winning by a margin is losing by the same margin.

Opposite: Losing by a margin of 2000 votes

(j) Depositing $\textsf{₹}100$ in the Bank account: The opposite of depositing is withdrawing.

Opposite: Withdrawing $\textsf{₹}100$ from the Bank account

(k) 20°C rise in temperature: The opposite of a rise in temperature is a fall (or decrease) in temperature.

Opposite: 20°C fall in temperature (or 20°C decrease in temperature)

Solution:

The initial temperature at 12:00 noon was given as $+5^\circ\text{C}$.

In the first hour, from 12:00 noon to 1:00 pm, the temperature increased by $3^\circ\text{C}$.

Temperature at 1:00 pm = Initial temperature + Increase in temperature

Temperature at 1:00 pm = $5^\circ\text{C} + 3^\circ\text{C} = 8^\circ\text{C}$.

In the second hour, from 1:00 pm to 2:00 pm, the temperature decreased by $1^\circ\text{C}$.

Temperature at 2:00 pm = Temperature at 1:00 pm - Decrease in temperature

Temperature at 2:00 pm = $8^\circ\text{C} - 1^\circ\text{C} = 7^\circ\text{C}$.

Therefore, the temperature at 2:00 pm was $7^\circ\text{C}$.

The temperature at 2:00 pm was $7^\circ\text{C}$.

Solution:

We need to insert '+' or '–' signs between the digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ in order to obtain the sum $3$.

The expression will have the form:

$0 \text{ op}_1 1 \text{ op}_2 2 \text{ op}_3 3 \text{ op}_4 4 \text{ op}_5 5 \text{ op}_6 6 \text{ op}_7 7 \text{ op}_8 8 \text{ op}_9 9 = 3$

where each $\text{op}_i$ is either '+' or '–'.

Let's find a combination of signs that satisfies the condition. One such combination is:

$0 + 1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 - 9$

Let's compute the value of this expression:

$(0 + 1 + 2 + 3 + 5 + 6 + 7) + (-4 - 8 - 9)$

$(1 + 2 + 3 + 5 + 6 + 7) + (-4 - 8 - 9)$

Sum of positive numbers: $1 + 2 + 3 + 5 + 6 + 7 = 24$.

Sum of negative numbers: $-4 - 8 - 9 = -(4 + 8 + 9) = -(21) = -21$.

The total sum is $24 + (-21) = 24 - 21 = 3$.

This matches the required result.

One possible solution is:

$0 + 1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 - 9 = 3$

(Note: Other combinations of signs may also yield the result 3).

Solution:

The additive inverse of an integer $a$ is the number $-a$ such that $a + (-a) = 0$.

We are looking for an integer that is its own additive inverse. This means the integer must be equal to its additive inverse.

Let the integer be $x$. According to the condition, the integer must be equal to its additive inverse.

$x = -x$

To solve for $x$, we can add $x$ to both sides of the equation:

$x + x = -x + x$

$2x = 0$

Now, divide both sides by $2$:

$\frac{2x}{2} = \frac{0}{2}$

$x = 0$.

The only integer that is equal to its own additive inverse is $0$.

Let's check: The additive inverse of $0$ is $-0$, which is equal to $0$. So, $0$ is its own additive inverse.

The integer which is its own additive inverse is 0.

Solution:

We need to find a set of six integers such that all the integers are different from each other, and their sum is equal to $7$.

Let's choose some integers, including positive, negative, and possibly zero, and check if they are distinct and their sum is $7$.

Consider the following six integers: $1, 2, 3, -1, -2, 4$.

These integers are all distinct from each other.

Let's find their sum:

Sum = $1 + 2 + 3 + (-1) + (-2) + 4$

Sum = $(1 + 2 + 3 + 4) + ((-1) + (-2))$

Sum = $10 + (-3)$

Sum = $10 - 3$

Sum = $7$.

The sum of these six distinct integers is $7$.

Therefore, one set of six distinct integers whose sum is $7$ is $\{1, 2, 3, -1, -2, 4\}$.

Other sets are possible, for example, $\{0, 1, 2, -1, -2, 7\}$ also sums to $7$.

Six distinct integers whose sum is 7 are: $1, 2, 3, -1, -2, 4$.

Solution:

Let the integer be $x$.

The additive inverse of the integer $x$ is $-x$.

According to the problem statement, the integer $x$ is $4$ more than its additive inverse $-x$.

We can write this as an equation:

To solve for $x$, add $x$ to both sides of the equation (i):

$x + x = -x + 4 + x$

$2x = 4$

Now, divide both sides by $2$:

$\frac{2x}{2} = \frac{4}{2}$

$x = 2$

Thus, the integer is $2$.

Let's verify the answer. The integer is $2$. Its additive inverse is $-2$.

Is $2$ equal to $-2 + 4$?

$-2 + 4 = 2$.

Since $2 = 2$, the condition is satisfied.

The integer which is 4 more than its additive inverse is 2.

Solution:

Let the integer be $x$.

The additive inverse of the integer $x$ is $-x$.

According to the problem statement, the integer $x$ is $2$ less than its additive inverse $-x$.

This can be written as an equation:

To solve for $x$, add $x$ to both sides of the equation (i):

$x + x = -x - 2 + x$

$2x = -2$

Now, divide both sides by $2$:

$\frac{2x}{2} = \frac{-2}{2}$

$x = -1$

Thus, the integer is $-1$.

Let's verify the answer. The integer is $-1$. Its additive inverse is $-(-1) = 1$.

Is $-1$ equal to $1 - 2$?

$1 - 2 = -1$.

Since $-1 = -1$, the condition is satisfied.

The integer which is 2 less than its additive inverse is –1.

Solution:

Let the two integers be $a$ and $b$. We are looking for integers $a$ and $b$ such that their sum $(a+b)$ is less than $a$ and also less than $b$.

This can be written as two inequalities:

$a + b < a$

$a + b < b$

Consider the first inequality: $a + b < a$. If we subtract $a$ from both sides, we get $b < 0$. This means the integer $b$ must be negative.

Consider the second inequality: $a + b < b$. If we subtract $b$ from both sides, we get $a < 0$. This means the integer $a$ must be negative.

Thus, both integers must be negative for their sum to be less than both of them.

Let's choose two negative integers, for example, $a = -2$ and $b = -3$.

These are integers.

Their sum is $(-2) + (-3) = -5$.

Now we check if the sum is less than both integers:

Is $-5 < -2$? Yes, because $-5$ is to the left of $-2$ on the number line.

Is $-5 < -3$? Yes, because $-5$ is to the left of $-3$ on the number line.

Since both conditions are met, $-2$ and $-3$ are two such integers. Any pair of negative integers will satisfy the condition.

Two integers whose sum is less than both the integers are –2 and –3.

Solution:

Let the two distinct integers be $a$ and $b$. We are given that $a \neq b$.

The problem states that their sum ($a+b$) is equal to one of the integers. This means either:

Case 1: $a + b = a$

or

Case 2: $a + b = b$

Let's consider Case 1: $a + b = a$.

Subtracting $a$ from both sides of the equation, we get:

$a + b - a = a - a$

$b = 0$

So, if one integer is $0$, their sum is equal to the other integer ($a + 0 = a$). Since the integers must be distinct ($a \neq b$), the other integer ($a$) must be a non-zero integer.

Let's consider Case 2: $a + b = b$.

Subtracting $b$ from both sides of the equation, we get:

$a + b - b = b - b$

$a = 0$

So, if one integer is $0$, their sum is equal to the other integer ($0 + b = b$). Since the integers must be distinct ($a \neq b$), the other integer ($b$) must be a non-zero integer.

In both cases, one of the distinct integers must be $0$ and the other must be any non-zero integer.

Let's choose the integer $5$ and the integer $0$.

These are distinct integers ($5 \neq 0$).

Their sum is $5 + 0 = 5$.

The sum ($5$) is equal to one of the integers ($5$).

Two distinct integers whose sum is equal to one of the integers are 5 and 0 (or any non-zero integer and 0).

Solution:

On a number line, numbers are arranged in increasing order from left to right. To compare two integers using a number line, we locate their positions. The integer that is positioned further to the right on the number line is the greater integer, and the integer that is positioned further to the left is the smaller integer.

(a) Comparing two negative integers:

Negative integers are located to the left of zero on the number line. When comparing two negative integers, say $a$ and $b$, the integer that is closer to zero (i.e., has a smaller absolute value) will be located to the right of the other negative integer.

So, the negative integer with the smaller absolute value is greater.

Example: To compare $-5$ and $-2$, we locate them on the number line. $-5$ is to the left of $-2$. Therefore, $-5 < -2$. Alternatively, $|-5|=5$ and $|-2|=2$. Since $2 < 5$, $-2$ is closer to $0$ and thus $-2 > -5$.

(b) Comparing two positive integers:

Positive integers are located to the right of zero on the number line. When comparing two positive integers, say $c$ and $d$, the integer with the larger value will be located further to the right on the number line.

So, the positive integer located further to the right is greater.

Example: To compare $3$ and $7$, we locate them on the number line. $7$ is to the right of $3$. Therefore, $7 > 3$.

(c) Comparing one positive and one negative integer:

Positive integers are always located to the right of zero, and negative integers are always located to the left of zero on the number line. This means that any positive integer is always located to the right of any negative integer.

Therefore, a positive integer is always greater than a negative integer.

Example: To compare $4$ and $-6$, we locate them on the number line. $4$ is positive and $-6$ is negative. $4$ is to the right of $-6$. Therefore, $4 > -6$.

Solution:

The given expression is $1 + 2 - 3 + 4 + 5 - 6 - 7 + 8 - 9$.

Let's first calculate the value of the given expression to confirm the sum is $-5$.

Sum of positive terms: $1 + 2 + 4 + 5 + 8 = 20$.

Sum of negative terms: $-3 - 6 - 7 - 9$. We can write this as $-(3 + 6 + 7 + 9) = -25$.

The total sum is $20 + (-25) = 20 - 25 = -5$. This matches the given result.

We need to change one '–' sign to a '+' sign such that the new sum is $9$.

The '–' signs are before the numbers $3, 6, 7,$ and $9$.

Let's see the effect of changing a '–' sign before a number $k$ to a '+' sign. The term changes from $-k$ to $+k$. The overall change in the sum is $+k - (-k) = k + k = 2k$.

We want the new sum to be $9$. The original sum is $-5$.

Let the change in sum be $\Delta S$. Original sum $+ \Delta S =$ New sum.

$-5 + \Delta S = 9$.

Adding $5$ to both sides, $\Delta S = 9 + 5 = 14$.

We need the change in sum ($\Delta S$) to be $14$.

The change in sum caused by changing the '–' sign before a number $k$ to a '+' sign is $2k$.

So, we need $2k = 14$.

Dividing by $2$, we get $k = 7$.

This means we need to find the term where the '–' sign is before the number $7$.

In the given expression, the term with the '–' sign before $7$ is $–7$.

We should change the '–' sign before $7$ to a '+'. The term $–7$ becomes $+7$.

The modified expression is $1 + 2 - 3 + 4 + 5 - 6 \underline{\mathbf{+}} 7 + 8 - 9$.

Let's verify the sum of the modified expression:

$1 + 2 - 3 + 4 + 5 - 6 + 7 + 8 - 9$

Sum of positive terms: $1 + 2 + 4 + 5 + 7 + 8 = 27$.

Sum of negative terms: $-3 - 6 - 9 = -(3 + 6 + 9) = -18$.

The new total sum is $27 + (-18) = 27 - 18$.

$\begin{array}{cc} & 2 & 7 \\ - & 1 & 8 \\ \hline & & 9 \\ \hline \end{array}$

The new sum is $9$, which is the required result.

To get the sum 9, we need to change the ‘–’ sign before 7 to a ‘+’ sign.

Given:

The set of integers: –2, 1, 0, –3, +4, –5

To Arrange:

The given integers in ascending order.

Solution:

Ascending order means arranging numbers from the smallest to the largest.

Negative integers are smaller than zero and positive integers.

Among negative integers, the integer with the greater absolute value is smaller.

Among positive integers, the integer with the greater value is larger.

Let's list the given integers: -2, 1, 0, -3, 4, -5.

The negative integers are: -2, -3, -5.

Comparing these: -5 < -3 < -2.

The positive integers are: 1, 4.

Comparing these: 1 < 4.

Zero is greater than all negative integers and less than all positive integers.

Combining them in increasing order:

-5 is the smallest.

-3 is next.

-2 is next.

0 is next.

1 is next.

4 is the largest.

Therefore, the integers arranged in ascending order are:

-5, -3, -2, 0, 1, 4

Given:

The set of integers: –3, 0, –1, –4, –3, –6

To Arrange:

The given integers in descending order.

Solution:

Descending order means arranging numbers from the largest to the smallest.

Positive integers are larger than zero and negative integers.

Zero is larger than all negative integers.

Among negative integers, the integer with the smaller absolute value is larger.

Let's list the given integers: -3, 0, -1, -4, -3, -6.

The unique integers are: -3, 0, -1, -4, -6.

The positive integers are none in this set.

Zero is present: 0.

The negative integers are: -3, -1, -4, -6.

Comparing the negative integers: The largest negative integer is the one closest to zero, which is -1.

Next largest negative integer is -3 (which appears twice in the original list).

Next is -4.

The smallest negative integer is -6.

Combining all the numbers in decreasing order:

0 is the largest.

-1 is next.

-3 is next (appears twice).

-4 is next.

-6 is the smallest.

Therefore, the integers arranged in descending order are:

0, -1, -3, -3, -4, -6

Given:

Two integers whose sum is 6 and difference is 6.

To Find:

The two integers.

Solution:

Let the two integers be $x$ and $y$.

According to the problem statement, we can set up a system of two linear equations:

1. The sum of the two integers is 6:

2. The difference of the two integers is 6:

We can solve this system by adding the two equations.

Add equation (i) and equation (ii):

$(x + y) + (x - y) = 6 + 6$

$2x = 12$

Now, solve for $x$:

$x = \frac{12}{2}$

$x = 6$

Substitute the value of $x=6$ into equation (i):

$6 + y = 6$

Subtract 6 from both sides to solve for $y$:

$y = 6 - 6$

$y = 0$

So, the two integers are 6 and 0.

Verification:

Sum: $6 + 0 = 6$ (Correct)

Difference: $6 - 0 = 6$ (Correct)

The two integers are 6 and 0.

Given:

Condition for integers: less than –100 and greater than –150.

To Find:

Five integers satisfying the given conditions.

Solution:

We are looking for integers $x$ such that $–150 < x < –100$.

The integers greater than –150 start from –149, –148, –147, and so on.

The integers less than –100 end at –101, –102, –103, and so on.

The integers that are both less than –100 and greater than –150 are the integers between –150 and –100, excluding –150 and –100 themselves.

This range of integers is {-149, -148, -147, ..., -103, -102, -101}.

We need to write any five integers from this list.

For example, we can choose:

-140, -135, -120, -115, -105

These five integers are all greater than –150 and less than –100.

Given:

The reference point on the number line is the integer 2.

To Find:

Four pairs of integers that are at the same distance from 2 on the number line.

Solution:

Two integers are at the same distance from a point on the number line if they are located equally far away from that point, one on the left side and one on the right side.

Let the point be $p=2$. If two integers $x_1$ and $x_2$ are at the same distance $d$ from $p$, then the distance $|x_1 - p| = d$ and $|x_2 - p| = d$.

This means $x_1$ can be $p-d$ and $x_2$ can be $p+d$ (or vice versa).

So, for the point 2, pairs of integers at a distance $d$ will be of the form $(2-d, 2+d)$.

$d$ must be a positive integer for the pair to consist of distinct integers that are a positive distance away.

Let's choose four different positive integer values for the distance $d$:

1. Let $d = 1$.

The integers are $2 - 1 = 1$ and $2 + 1 = 3$.

The pair is (1, 3).

Distance from 2 to 1 is $|1 - 2| = |-1| = 1$.

Distance from 2 to 3 is $|3 - 2| = |1| = 1$.

2. Let $d = 2$.

The integers are $2 - 2 = 0$ and $2 + 2 = 4$.

The pair is (0, 4).

Distance from 2 to 0 is $|0 - 2| = |-2| = 2$.

Distance from 2 to 4 is $|4 - 2| = |2| = 2$.

3. Let $d = 3$.

The integers are $2 - 3 = -1$ and $2 + 3 = 5$.

The pair is (-1, 5).

Distance from 2 to -1 is $|-1 - 2| = |-3| = 3.

Distance from 2 to 5 is $|5 - 2| = |3| = 3$.

4. Let $d = 4$.

The integers are $2 - 4 = -2$ and $2 + 4 = 6$.

The pair is (-2, 6).

Distance from 2 to -2 is $|-2 - 2| = |-4| = 4$.

Distance from 2 to 6 is $|6 - 2| = |4| = 4$.

Four pairs of integers that are at the same distance from 2 on the number line are:

(1, 3), (0, 4), (-1, 5), (-2, 6).

(Other valid pairs exist, e.g., (-3, 7), (-4, 8), etc., by choosing different values for $d$).

Given:

The sum of two integers is 30.

One of the integers is –42.

To Find:

The other integer.

Solution:

Let the two integers be $x$ and $y$.

According to the problem statement, the sum of the two integers is 30.

We are given that one of the integers is –42. Let's assign this value to $x$.

$x = -42$

Substitute the value of $x$ into equation (i):

$-42 + y = 30$

To find the value of $y$, we need to isolate it. Add 42 to both sides of the equation:

$y = 30 + 42$

Perform the addition:

$y = 72$

Thus, the other integer is 72.

Verification:

Check the sum of the two integers: $-42 + 72 = 30$. This matches the given information.

The other integer is 72.

Given:

The sum of two integers is –80.

One of the integers is –90.

To Find:

The other integer.

Solution:

Let the two integers be $a$ and $b$.

According to the problem statement, the sum of the two integers is –80.

We are given that one of the integers is –90. Let's assign this value to $a$.

$a = -90$

Substitute the value of $a$ into equation (i):

$-90 + b = -80$

To find the value of $b$, we need to isolate it. Add 90 to both sides of the equation:

$b = -80 + 90$

Perform the addition:

$b = 10$

Thus, the other integer is 10.

Verification:

Check the sum of the two integers: $-90 + 10 = -80$. This matches the given information.

The other integer is 10.

Given:

Starting point on the number line: 8.

Target integers: (a) –5, (b) 11, (c) 0.

To Find:

The direction of movement from 8 to reach each target integer.

Solution:

On a number line, moving to the right corresponds to moving towards larger numbers, and moving to the left corresponds to moving towards smaller numbers.

Our starting point is 8.

(a) Target integer: –5

We compare the target integer (–5) with the starting point (8).

Since –5 is less than 8 ($-5 < 8$), we need to move towards smaller numbers from 8.

Moving towards smaller numbers on a number line is moving to the left.

(b) Target integer: 11

We compare the target integer (11) with the starting point (8).

Since 11 is greater than 8 ($11 > 8$), we need to move towards larger numbers from 8.

Moving towards larger numbers on a number line is moving to the right.

(c) Target integer: 0

We compare the target integer (0) with the starting point (8).

Since 0 is less than 8 ($0 < 8$), we need to move towards smaller numbers from 8.

Moving towards smaller numbers on a number line is moving to the left.

Directions of movement:

(a) To reach –5 from 8, we should move to the left.

(b) To reach 11 from 8, we should move to the right.

(c) To reach 0 from 8, we should move to the left.

Given:

Operations on the number line starting from a given integer:

(a) 4 more than –5

(b) 3 less than 2

(c) 2 less than –2

To Find:

The resulting integer after each operation using the number line concept.

Solution:

On a number line, moving to the right indicates adding (finding 'more than'), and moving to the left indicates subtracting (finding 'less than').

(a) 4 more than –5

Start at –5 on the number line.

Move 4 steps to the right (since it's 'more than').

Starting at –5, move right: –4, –3, –2, –1.

We land on –1.

Alternatively, this can be represented as an addition: $-5 + 4 = -1$.

The integer is –1.

(b) 3 less than 2

Start at 2 on the number line.

Move 3 steps to the left (since it's 'less than').

Starting at 2, move left: 1, 0, –1.

We land on –1.

Alternatively, this can be represented as a subtraction: $2 - 3 = -1$.

The integer is –1.

(c) 2 less than –2

Start at –2 on the number line.

Move 2 steps to the left (since it's 'less than').

Starting at –2, move left: –3, –4.

We land on –4.

Alternatively, this can be represented as a subtraction: $-2 - 2 = -4$.

The integer is –4.

Given:

The expression: $49 – (–40) – (–3) + 69$

To Find:

The value of the given expression.

Solution:

We need to evaluate the expression $49 – (–40) – (–3) + 69$.

Recall the rules for operations with integers, especially subtracting negative numbers: $-(-\text{a}) = +\text{a}$.

Apply this rule to the expression:

The term $– (–40)$ becomes $+40$.

The term $– (–3)$ becomes $+3$.

The expression can be rewritten as:

$49 + 40 + 3 + 69$

Now, perform the addition from left to right:

$49 + 40 = 89$

The expression becomes $89 + 3 + 69$.

Next, add 3 to 89:

$89 + 3 = 92$

The expression becomes $92 + 69$.

Finally, add 69 to 92:

$92 + 69 = 161$

We can perform this addition as:

$\begin{array}{cc} & 9 & 2 \\ + & 6 & 9 \\ \hline 1 & 6 & 1 \\ \hline \end{array}$

The value of the expression $49 – (–40) – (–3) + 69$ is 161.

Given:

Integers to consider: –2100, –2001, and –5308.

To Find:

The value obtained by subtracting –5308 from the sum of –2100 and –2001.

Solution:

First, find the sum of the two integers –2100 and –2001.

Sum $= (–2100) + (–2001)$

Adding two negative integers results in a negative integer. We add their absolute values and keep the negative sign.

$|-2100| = 2100$

$|-2001| = 2001$

$2100 + 2001 = 4101$

So, the sum is $-4101$.

We can show the addition vertically:

$\begin{array}{cc} & 2 & 1 & 0 & 0 \\ + & 2 & 0 & 0 & 1 \\ \hline & 4 & 1 & 0 & 1 \\ \hline \end{array}$

Therefore, $(–2100) + (–2001) = –4101$.

Next, we need to subtract –5308 from this sum (–4101).

Result $= (\text{Sum}) - (–5308)$

Result $= –4101 - (–5308)$

Subtracting a negative number is the same as adding its positive counterpart. $-(-\text{a}) = +\text{a}$.

So, $-4101 - (–5308)$ becomes $-4101 + 5308$.

Result $= 5308 - 4101$

Now, perform the subtraction:

$\begin{array}{cc} & 5 & 3 & 0 & 8 \\ - & 4 & 1 & 0 & 1 \\ \hline & 1 & 2 & 0 & 7 \\ \hline \end{array}$

The value of the expression is 1207.

Subtracting –5308 from the sum [(–2100) + (–2001)] gives 1207.