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Class 10th Chapters
1. Real Numbers 2. Polynomials 3. Pair of Linear Equations in Two Variables
4. Quadratic Equations 5. Arithmetic Progressions 6. Triangles
7. Coordinate Geometry 8. Introduction to Trigonometry 9. Some Applications of Trigonometry
10. Circles 11. Constructions 12. Areas Related to Circles
13. Surface Areas And Volumes 14. Statistics 15. Probability

Content On This Page
Euclid’s Division Lemma and Euclid’s Division ALgorithm Fundamental Theorem of Arithmetic HCF and LCM of Two Natural Numbers
Irrational Numbers and Proof of Irrationality Rational Numbers and Their Decimal Expansions

Chapter 1 Real Numbers (Concepts)

Welcome to this foundational chapter that delves deeper into the intricate properties of Real Numbers, building significantly upon the concepts introduced in Class 9. Our journey will take us through fundamental theorems governing integers, the rigorous proof of irrationality for certain numbers, and a precise understanding of the relationship between rational numbers and their decimal representations. This exploration solidifies our understanding of the number system that underpins virtually all of higher mathematics.

A cornerstone of this chapter is the formal introduction of Euclid's Division Lemma. This seemingly simple statement holds profound implications within number theory. It asserts that for any two given positive integers, let's call them $a$ (dividend) and $b$ (divisor), there exist unique integers $q$ (quotient) and $r$ (remainder) that satisfy the equation: $$ a = bq + r, \quad \text{where } 0 \le r < b $$ This lemma guarantees that when we divide $a$ by $b$, we get a unique quotient and a unique remainder $r$ which is non-negative and strictly less than the divisor $b$. This fundamental lemma provides the theoretical basis for Euclid's Division Algorithm. This algorithm is a highly efficient, step-by-step procedure used to determine the Highest Common Factor (HCF) of two positive integers. It involves repeatedly applying the division lemma, using the previous divisor as the new dividend and the previous remainder as the new divisor, until the remainder becomes zero. The divisor at this final stage (i.e., the last non-zero remainder) is precisely the HCF of the original two integers.

Another pillar of number theory presented here is the Fundamental Theorem of Arithmetic. This theorem makes a powerful statement about the building blocks of composite numbers. It asserts that every composite number can be factorized into a product of prime numbers, and critically, this prime factorization is unique for each composite number, irrespective of the order in which the prime factors are listed. For example, $12 = 2 \times 2 \times 3 = 2^2 \times 3$. No other combination of primes will multiply to 12. This theorem provides a robust method for calculating both the HCF and the Least Common Multiple (LCM) of two or more positive integers. Using their unique prime factorizations:

We revisit the concept of irrational numbers – numbers like $\sqrt{2}, \sqrt{3}, \sqrt{5}$ that cannot be expressed in the simple fraction form $\frac{p}{q}$. A key focus is on rigorously proving the irrationality of such numbers. The standard technique employed is the elegant method of proof by contradiction. This typically involves assuming the number *is* rational, leading logically (often using Euclid's Division Lemma or the Fundamental Theorem of Arithmetic) to a contradiction of a known fact, thereby proving the initial assumption false and establishing the number's irrationality.

Finally, the chapter illuminates the precise connection between the nature of a rational number's decimal expansion and the prime factors of its denominator. It is established that a rational number $\frac{p}{q}$ (expressed in its simplest form, where $p$ and $q$ have no common factors other than 1) will have a terminating decimal expansion if and only if the prime factorization of its denominator $q$ consists solely of powers of 2 and/or 5. That is, $q$ must be of the form $\mathbf{2^n 5^m}$, where $n$ and $m$ are non-negative integers. If the prime factorization of $q$ contains any prime factor other than 2 or 5, then the decimal expansion of the rational number will be non-terminating but repeating (recurring). This provides a powerful criterion to predict the type of decimal expansion without performing the actual long division.

Euclid’s Division Lemma and Euclid’s Division Algorithm

In mathematics, the set of Real Numbers encompasses all rational and irrational numbers. This chapter begins by exploring properties of integers, specifically focusing on concepts introduced by the ancient Greek mathematician Euclid. These concepts, known as Euclid's Division Lemma and Euclid's Division Algorithm, are fundamental in number theory and have practical applications, such as finding the Highest Common Factor (HCF) of two integers.

Difference Between a Lemma and an Algorithm

Before discussing Euclid's concepts, it's helpful to understand the distinction between a lemma and an algorithm:

Euclid's Division Lemma is a proven statement, and Euclid's Division Algorithm is a procedure based on that statement.

Euclid's Division Lemma

Euclid's Division Lemma is a precise statement about the outcome of the division of one positive integer by another. It formalises the familiar process of 'long division'.

Euclid’s Division Lemma. Given two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \le r < b$.

In this mathematical statement:

The lemma asserts that for any pair of positive integers $a$ and $b$, you can always find a unique quotient $q$ and a unique remainder $r$. The remainder $r$ must satisfy the condition that it is greater than or equal to zero and strictly less than the divisor $b$. This condition, $0 \le r < b$, is crucial and ensures the uniqueness of $q$ and $r$.

Examples Illustrating Euclid's Division Lemma:

The lemma is stated for positive integers $a$ and $b$ in the theorem, but it can be extended to include any integer $a$ and a positive integer $b$.

Example 1. Apply Euclid's Division Lemma for the pair of integers 13 and 3.

Answer:

Given:

The pair of integers 13 and 3.

To Apply:

Euclid's Division Lemma.

Solution:

We are given the positive integers $a = 13$ (dividend) and $b = 3$ (divisor).

According to Euclid's Division Lemma, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \le r < b$.

We perform the division of 13 by 3:

$$\begin{array}{r} 4\phantom{)} \\ 3{\overline{\smash{\big)}\,13\phantom{)}}} \\ \underline{-~12\phantom{)}} \\ 1\phantom{)} \end{array}$$

The quotient is 4, and the remainder is 1.

So, we can write the equation $a = bq + r$ as:

$13 = 3 \times 4 + 1$

... (1)

Here, we have $a = 13$, $b = 3$, $q = 4$, and $r = 1$.

We check the condition for the remainder: $0 \le r < b$.

$0 \le 1 < 3$

The condition is satisfied. The unique integers $q$ and $r$ found are indeed 4 and 1, respectively.

Answer: Applying Euclid's Division Lemma to 13 and 3, we get $13 = 3 \times 4 + 1$.

Euclid's Division Algorithm

Euclid's Division Algorithm is a systematic procedure based on Euclid's Division Lemma, used specifically to find the Highest Common Factor (HCF) of two positive integers. The HCF (also known as the Greatest Common Divisor or GCD) of two positive integers is the largest positive integer that divides both integers without leaving a remainder.

The algorithm relies on the property that the HCF of two positive integers $a$ and $b$ (with $a > b$) is the same as the HCF of the divisor $b$ and the remainder $r$ obtained when $a$ is divided by $b$. That is, $\text{HCF}(a, b) = \text{HCF}(b, r)$.

Steps of Euclid’s Division Algorithm to find the HCF of two positive integers, say $c$ and $d$, with $c > d$:

  1. Step 1: Apply Euclid's Division Lemma to $c$ and $d$. Find unique whole numbers $q$ and $r$ such that $c = dq + r$, where $0 \le r < d$.
  2. Step 2: If the remainder $r$ is 0, then the divisor $d$ is the HCF of $c$ and $d$. The algorithm stops here.
  3. Step 3: If the remainder $r$ is not 0, then replace the dividend with the previous divisor ($d$) and the divisor with the remainder ($r$). Now, apply Euclid's Division Lemma to $d$ and $r$. Find unique whole numbers $q_1$ and $r_1$ such that $d = r q_1 + r_1$, where $0 \le r_1 < r$.
  4. Step 4: Continue this process of replacing the dividend and divisor with the previous divisor and remainder, and applying the lemma, until the remainder at some step becomes 0. The divisor at the step where the remainder is 0 is the HCF of the original two numbers ($c$ and $d$).

The algorithm terminates because the sequence of remainders ($d > r > r_1 > r_2 > ... \ge 0$) is a strictly decreasing sequence of non-negative integers, and eventually, a remainder of 0 must be obtained.

Example 2. Use Euclid’s Division Algorithm to find the HCF of 135 and 225.

Answer:

Given:

Two positive integers, 135 and 225.

To Find:

The HCF of 135 and 225 using Euclid's algorithm.

Solution:

We need to find the HCF of 135 and 225. Since $225 > 135$, we take $a = 225$ and $b = 135$. We apply Euclid's Division Lemma to $a$ and $b$.

Step 1: Divide 225 by 135:

$225 = 135 \times 1 + 90$

($q=1, r=90$). ... (1)

The remainder is 90, which is not equal to 0. Now, we apply the lemma to the new divisor (135) and the new remainder (90).

Step 2: Divide 135 by 90:

$135 = 90 \times 1 + 45$

($q_1=1, r_1=45$). ... (2)

The remainder is 45, which is not equal to 0. Now, we apply the lemma to the new divisor (90) and the new remainder (45).

Step 3: Divide 90 by 45:

$90 = 45 \times 2 + 0$

($q_2=2, r_2=0$). ... (3)

The remainder is 0. The algorithm stops at this stage.

According to the algorithm, the divisor at the step where the remainder becomes 0 is the HCF of the original two numbers.

The divisor in Step 3 is 45.

Therefore, HCF$(135, 225) = 45$.

Answer: The HCF of 135 and 225 is 45.

Example 3. Use Euclid’s division algorithm to find the HCF of 4052 and 12576.

Answer:

Given:

Two positive integers, 4052 and 12576.

To Find:

The HCF of 4052 and 12576 using Euclid's algorithm.

Solution:

We need to find the HCF of 4052 and 12576. Since $12576 > 4052$, we apply Euclid's Division Lemma repeatedly, starting with $a = 12576$ and $b = 4052$.

Step 1: $12576 = 4052 \times 3 + 420$ (Remainder $420 \neq 0$)

Step 2: $4052 = 420 \times 9 + 272$ (Remainder $272 \neq 0$)

Step 3: $420 = 272 \times 1 + 148$ (Remainder $148 \neq 0$)

Step 4: $272 = 148 \times 1 + 124$ (Remainder $124 \neq 0$)

Step 5: $148 = 124 \times 1 + 24$ (Remainder $24 \neq 0$)

Step 6: $124 = 24 \times 5 + 4$ (Remainder $4 \neq 0$)

Step 7: $24 = 4 \times 6 + 0$ (Remainder $0$)

The remainder has become 0. The divisor at this last step is 4.

Therefore, HCF$(4052, 12576) = 4$.

Answer: The HCF of 4052 and 12576 is 4.

Example 4. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Answer:

Given:

Number of members in the army contingent = 616.

Number of members in the army band = 32.

Both groups must march in the same number of columns.

To Find:

The maximum number of columns.

Solution:

The number of columns must be a number that can divide both 616 and 32 without leaving a remainder, so that both groups can march in those columns with the same number of members per column. We want the maximum such number.

This means we need to find the Highest Common Factor (HCF) of 616 and 32.

We use Euclid's division algorithm to find the HCF of 616 and 32. Since $616 > 32$, we apply the division lemma repeatedly, starting with $a = 616$ and $b = 32$.

Step 1: Divide 616 by 32:

$616 = 32 \times 19 + 8$

($q=19, r=8$). ... (1)

The remainder is 8, which is not equal to 0. Now, we apply the lemma to the new divisor (32) and the new remainder (8).

Step 2: Divide 32 by 8:

$32 = 8 \times 4 + 0$

($q_1=4, r_1=0$). ... (2)

The remainder is 0. The algorithm stops at this stage.

The divisor at this last step is 8.

Therefore, HCF$(616, 32) = 8$.

The maximum number of columns in which both groups can march is the HCF of 616 and 32.

Answer: The maximum number of columns is 8.


Fundamental Theorem of Arithmetic

Following our discussion on Euclid's Division Lemma and Algorithm, we move to another cornerstone concept in number theory: the Fundamental Theorem of Arithmetic. This theorem is crucial as it establishes the basis for factoring any composite number into a product of prime numbers in a unique way.

Prime and Composite Numbers

To understand the Fundamental Theorem of Arithmetic, we first need to clearly define prime and composite numbers:

Any natural number greater than 1 is either a prime number or a composite number.

The Fundamental Theorem of Arithmetic

This theorem, also known as the Unique Factorization Theorem, is a foundational concept in number theory.

Statement: Every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur.

In simpler terms, the theorem states that any composite number can be broken down into a product of only prime numbers. Furthermore, there is only one possible set of prime numbers that can be multiplied together to form that particular composite number, regardless of the sequence in which you write those prime factors.

For example, the composite number 12 can be written as $2 \times 6$, and $6 = 2 \times 3$, so $12 = 2 \times 2 \times 3$. The prime factors are 2, 2, and 3. No other set of prime numbers will multiply to give 12.

The factorisation is usually written with the prime factors in ascending order, often using exponents to show repeated factors. For 12, the unique prime factorisation is $2^2 \times 3^1$.

Example 1. Find the prime factorisation of 12 and 2904.

Answer:

To Find:

The prime factorisation of 12 and 2904.

Solution:

We can find the prime factorisation using the division method, repeatedly dividing the number by the smallest possible prime factor until the quotient is 1.

Prime Factorisation of 12:

$\begin{array}{c|cc} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

From the division, we get the prime factors of 12 as 2, 2, and 3.

$12 = 2 \times 2 \times 3$

Using exponents to represent repeated prime factors:

$12 = 2^2 \times 3^1$

... (1)

Prime Factorisation of 2904:

$\begin{array}{c|cc} 2 & 2904 \\ \hline 2 & 1452 \\ \hline 2 & 726 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

From the division, we get the prime factors of 2904 as 2, 2, 2, 3, 11, and 11.

$2904 = 2 \times 2 \times 2 \times 3 \times 11 \times 11$

Using exponents:

$2904 = 2^3 \times 3^1 \times 11^2$

... (2)

These factorisations ($2^2 \times 3^1$ for 12 and $2^3 \times 3^1 \times 11^2$ for 2904) are unique in terms of the prime factors involved and their powers, as stated by the Fundamental Theorem of Arithmetic.

Answer: The prime factorisation of 12 is $2^2 \times 3$. The prime factorisation of 2904 is $2^3 \times 3^1 \times 11^2$.


HCF and LCM of Two Natural Numbers

Building upon the Fundamental Theorem of Arithmetic, which assures us that every composite number has a unique prime factorisation, we can use this theorem as a powerful tool to find the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two or more positive integers. These are important concepts in number theory and have applications in various areas, including solving problems involving fractions, ratios, and time/work.

HCF (Highest Common Factor)

The Highest Common Factor (HCF) of two or more positive integers is the largest positive integer that divides each of the given integers without leaving a remainder. It is also known as the Greatest Common Divisor (GCD).

Using the prime factorisation method, the HCF of two or more numbers is found by taking the product of the smallest power of each common prime factor involved in the prime factorisation of the numbers.

LCM (Least Common Multiple)

The Least Common Multiple (LCM) of two or more positive integers is the smallest positive integer that is a multiple of all the given integers. In other words, it is the smallest number that is divisible by each of the given numbers.

Using the prime factorisation method, the LCM of two or more numbers is found by taking the product of the highest power of each prime factor involved (whether common or not) in the prime factorisation of the numbers.

Method to Find HCF and LCM using Prime Factorisation

The prime factorisation method provides a systematic approach to finding the HCF and LCM of positive integers:

  1. Step 1: Find the prime factorisation of each of the given numbers. Write each factorisation in exponential form (e.g., $12 = 2^2 \times 3^1$).
  2. Step 2 (For HCF): Identify all the prime factors that are common to the prime factorisation of all the given numbers. For each common prime factor, select the smallest power (the lowest exponent) that appears in any of the factorisations. Multiply these selected smallest powers together to get the HCF. If there are no common prime factors, the HCF is 1.
  3. Step 3 (For LCM): List all the prime factors that appear in the prime factorisation of at least one of the given numbers. For each such prime factor, select the highest power (the largest exponent) that appears in any of the factorisations. Multiply these selected highest powers together to get the LCM.

Example 1. Find the HCF and LCM of 6 and 20 using the prime factorisation method.

Answer:

To Find:

The HCF and LCM of 6 and 20 using the prime factorisation method.

Solution:

Step 1: Find the prime factorisation of 6 and 20.

Prime factorisation of 6:

$\begin{array}{c|cc} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$6 = 2 \times 3 = 2^1 \times 3^1$

... (1)

Prime factorisation of 20:

$\begin{array}{c|cc} 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

... (2)

Step 2 (For HCF): Identify common prime factors and take the smallest power.

Comparing the prime factorisations in (1) and (2), the only prime factor common to both 6 and 20 is 2.

The powers of 2 are $2^1$ in 6 and $2^2$ in 20. The smallest power of the common prime factor 2 is $2^1$.

HCF$(6, 20) = 2^1 = 2$

... (3)

Step 3 (For LCM): List all prime factors involved and take the highest power.

The prime factors involved in the factorisations of 6 ($2^1, 3^1$) and 20 ($2^2, 5^1$) are 2, 3, and 5.

  • Highest power of 2 is $2^2$ (from 20).
  • Highest power of 3 is $3^1$ (from 6).
  • Highest power of 5 is $5^1$ (from 20).

Multiply these highest powers:

LCM$(6, 20) = 2^2 \times 3^1 \times 5^1$

LCM$(6, 20) = 4 \times 3 \times 5 = 60$

... (4)

Answer: HCF(6, 20) = 2 and LCM(6, 20) = 60.

Relationship between HCF and LCM of Two Numbers

There is a significant relationship between the HCF and LCM of any two positive integers. This relationship is often useful for finding either the HCF or the LCM if the other measure and the two numbers are known.

Property: For any two positive integers $a$ and $b$, the product of their HCF and LCM is equal to the product of the numbers themselves.

$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

This property holds true only for a pair of positive integers. It is not generally true for the HCF and LCM of three or more numbers.

Example 2. Verify the relationship $HCF(a, b) \times LCM(a, b) = a \times b$ for $a = 96$ and $b = 404$.

Answer:

Given:

Two positive integers $a = 96$ and $b = 404$.

To Verify:

The relationship $HCF(96, 404) \times LCM(96, 404) = 96 \times 404$.

Solution:

First, we find the HCF and LCM of 96 and 404 using the prime factorisation method.

Prime factorisation of 96:

$\begin{array}{c|cc} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3^1$

... (1)

Prime factorisation of 404:

$\begin{array}{c|cc} 2 & 404 \\ \hline 2 & 202 \\ \hline 101 & 101 \\ \hline & 1 \end{array}$

$404 = 2 \times 2 \times 101 = 2^2 \times 101^1$

... (2)

HCF(96, 404):

The only prime factor common to both 96 and 404 is 2. The lowest power of 2 is $2^2$ (from the factorisation of 404).

HCF$(96, 404) = 2^2 = 4$

... (3)

LCM(96, 404):

The prime factors involved in the factorisations of 96 and 404 are 2, 3, and 101.

  • The highest power of 2 is $2^5$ (from 96).
  • The highest power of 3 is $3^1$ (from 96).
  • The highest power of 101 is $101^1$ (from 404).

Multiply these highest powers:

LCM$(96, 404) = 2^5 \times 3^1 \times 101^1$

LCM$(96, 404) = 32 \times 3 \times 101$

LCM$(96, 404) = 96 \times 101 = 9696$

... (4)

Now, we verify the relationship $HCF(a, b) \times LCM(a, b) = a \times b$ using the values we found and the given numbers $a=96, b=404$.

Left Hand Side (LHS): $HCF(96, 404) \times LCM(96, 404)$

LHS $= 4 \times 9696$

Perform the multiplication:

LHS $= 38784$

... (5)

Right Hand Side (RHS): $a \times b$

RHS $= 96 \times 404$

Perform the multiplication:

RHS $= 38784$

... (6)

Comparing the results from equation (5) and equation (6), we see that LHS = RHS.

$HCF(96, 404) \times LCM(96, 404) = 96 \times 404$

**(Verified)**

The relationship is verified for the given numbers.


Irrational Numbers and Proof of Irrationality

In previous classes, you have learned about rational numbers, which are numbers that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is not zero. Real numbers consist of both rational and irrational numbers. Numbers that are real but cannot be expressed in the form $\frac{p}{q}$ are called irrational numbers.

Irrational Numbers

An irrational number is a real number that cannot be written as a simple fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Another way to identify irrational numbers is by looking at their decimal expansions. The decimal expansion of a rational number is either terminating (like 0.5, 2.34) or non-terminating and repeating (like $0.333... = 0.\overline{3}$, $1.272727... = 1.\overline{27}$).

In contrast, irrational numbers have decimal expansions that are non-terminating and non-repeating. This means the decimal goes on forever without any repeating pattern of digits.

Examples of well-known irrational numbers include:

Proof of Irrationality

One of the standard methods used to prove that a number is irrational is called proof by contradiction. This method involves the following steps:

  1. Assume that the number you want to prove is irrational is actually rational.
  2. Based on this assumption, express the number as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers, $q \neq 0$, and the fraction is in its simplest form (meaning $p$ and $q$ have no common factors other than 1).
  3. Perform mathematical manipulations on the equation involving the fraction.
  4. Show that these manipulations lead to a result that contradicts the initial assumption that $p$ and $q$ have no common factors other than 1.
  5. Since the assumption that the number is rational leads to a contradiction, the assumption must be false.
  6. Therefore, the number must be irrational.

Example 1. Prove that $\sqrt{2}$ is irrational.

Answer:

To Prove:

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

Proof (by contradiction):

Assume, for the sake of contradiction, that $\sqrt{2}$ is a rational number.

If $\sqrt{2}$ is rational, then it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and we can assume that $p$ and $q$ have no common factors other than 1 (i.e., the fraction $\frac{p}{q}$ is in its simplest or reduced form). This is a crucial part of the assumption.

$\sqrt{2} = \frac{p}{q}$

(Assuming $\sqrt{2}$ is rational, in simplest form)

Squaring both sides of the equation:

$(\sqrt{2})^2 = \left(\frac{p}{q}\right)^2$

$2 = \frac{p^2}{q^2}$

Multiply both sides by $q^2$:

$p^2 = 2q^2$

... (1)

From equation (1), $p^2$ is equal to $2$ times $q^2$. This means that $p^2$ is an even number, or $p^2$ is divisible by 2.

If the square of an integer ($p^2$) is divisible by 2, then the integer itself ($p$) must also be divisible by 2. (This is a property that can be proven: If $p$ is odd, $p=2k+1$, then $p^2=4k^2+4k+1=2(2k^2+2k)+1$, which is odd. So if $p^2$ is even, $p$ must be even).

Since $p$ is divisible by 2, we can write $p$ as $2k$ for some integer $k$.

$p = 2k$

(Since $p$ is divisible by 2)

Substitute this expression for $p$ into equation (1):

$(2k)^2 = 2q^2$

$4k^2 = 2q^2$

Divide both sides of the equation by 2:

$2k^2 = q^2$

... (2)

From equation (2), $q^2$ is equal to $2$ times $k^2$. This means that $q^2$ is an even number, or $q^2$ is divisible by 2.

Similar to the case for $p$, if the square of an integer ($q^2$) is divisible by 2, then the integer itself ($q$) must also be divisible by 2.

So, from the equation $p^2 = 2q^2$, we concluded that $p$ is divisible by 2. And from the equation $q^2 = 2k^2$, we concluded that $q$ is divisible by 2.

This implies that both $p$ and $q$ have a common factor of 2.

However, this contradicts our initial assumption that the fraction $\frac{p}{q}$ was in its simplest form, meaning $p$ and $q$ have no common factors other than 1.

Since our assumption that $\sqrt{2}$ is rational leads to a contradiction, the assumption must be false.

Therefore, $\sqrt{2}$ is an irrational number.

Answer: $\sqrt{2}$ is irrational.

Proving Irrationality of Numbers of the form $a + b\sqrt{c}$

We can use the same proof by contradiction technique to prove the irrationality of numbers that are formed by combining rational numbers and irrational numbers (which are square roots of non-perfect squares). The key idea is to assume the entire expression is rational and then rearrange it to show that the irrational part (like $\sqrt{c}$) must be rational, which contradicts the known fact that $\sqrt{c}$ is irrational.

Example 2. Prove that $5 - \sqrt{3}$ is irrational, given that $\sqrt{3}$ is irrational.

Answer:

To Prove:

$5 - \sqrt{3}$ is irrational.

Given:

$\sqrt{3}$ is irrational.

Proof (by contradiction):

Assume, for the sake of contradiction, that the number $5 - \sqrt{3}$ is a rational number.

If $5 - \sqrt{3}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

$5 - \sqrt{3} = \frac{p}{q}$

(Assuming $5 - \sqrt{3}$ is rational)

Our goal is to isolate the known irrational part ($\sqrt{3}$) on one side of the equation. Rearrange the equation:

$5 - \frac{p}{q} = \sqrt{3}$

Combine the terms on the left side into a single fraction:

$\frac{5q - p}{q} = \sqrt{3}$

... (1)

Now, let's analyse the expression on the left side of equation (1). Since $p$ and $q$ are integers, and $q$ is a non-zero integer, the terms $5q$, $p$, and $q$ are all integers. The difference $5q - p$ is an integer. The division of the integer $(5q - p)$ by the non-zero integer $q$ results in a rational number.

The expression $\frac{5q - p}{q}$ is a rational number.

Equation (1) states that the irrational number $\sqrt{3}$ is equal to the expression $\frac{5q - p}{q}$, which we have just shown to be a rational number. This leads to the statement that $\sqrt{3}$ is a rational number.

$\sqrt{3}$ is rational.

However, this conclusion directly contradicts the given information that $\sqrt{3}$ is irrational.

Since our initial assumption that $5 - \sqrt{3}$ is rational leads to a contradiction, the assumption must be false.

Therefore, $5 - \sqrt{3}$ must be an irrational number.

Answer: $5 - \sqrt{3}$ is irrational.


Rational Numbers and Their Decimal Expansions

We know that rational numbers are a subset of real numbers that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q$ is a non-zero integer. Irrational numbers, on the other hand, cannot be expressed in this fractional form. One way to distinguish between rational and irrational numbers is by examining their decimal representations.

Decimal Expansions of Rational Numbers

When we convert a rational number from its fractional form $\frac{p}{q}$ into a decimal form by performing the division, the resulting decimal expansion will exhibit one of two characteristics:

Case 1: Terminating Decimal Expansion

A rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime, meaning their HCF is 1) has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ contains only powers of 2 and/or powers of 5. That is, the prime factorisation of $q$ must be of the form $2^m \times 5^n$, where $m$ and $n$ are non-negative integers (they can be zero, meaning the denominator has only factors of 5 or only factors of 2). The decimal expansion terminates because the fraction can be rewritten with a denominator that is a power of 10 ($10^k = (2 \times 5)^k = 2^k \times 5^k$).

Example 1. Determine if the decimal expansion of $\frac{13}{125}$ is terminating or non-terminating. If it is terminating, write its decimal expansion.

Answer:

Given:

The rational number $\frac{13}{125}$.

To Determine:

The nature of its decimal expansion (terminating or non-terminating) and the expansion if it terminates.

Solution:

The given rational number is $\frac{13}{125}$. Here, $p = 13$ and $q = 125$.

First, check if the fraction is in its simplest form. The prime factors of 13 are just 13. The prime factors of 125 are $5 \times 5 \times 5$. There are no common factors between 13 and 125 other than 1, so the fraction is in simplest form.

Now, examine the prime factorisation of the denominator $q = 125$:

$\begin{array}{c|cc} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$125 = 5 \times 5 \times 5 = 5^3$

... (1)

The prime factorisation of the denominator is $5^3$. This can be written in the form $2^m \times 5^n$ by considering $m=0$ (since there are no factors of 2). So, the form is $2^0 \times 5^3$. Here, $m=0$ and $n=3$ are non-negative integers.

Since the prime factorisation of the denominator contains only the prime factor 5 (which fits the form $2^m \times 5^n$), the decimal expansion of $\frac{13}{125}$ is terminating.

To find the decimal expansion, we can perform the long division of 13 by 125, or we can rewrite the fraction with a denominator that is a power of 10. To get a power of 10 in the denominator, we need the same number of factors of 2 as factors of 5. The denominator $125 = 5^3$ has three factors of 5. We need three factors of 2, i.e., $2^3 = 8$. Multiply both the numerator and the denominator by $2^3 = 8$:

$\frac{13}{125} = \frac{13}{5^3} = \frac{13 \times 2^3}{5^3 \times 2^3}$

$\frac{13 \times 8}{(5 \times 2)^3} = \frac{104}{10^3} = \frac{104}{1000}$

Convert the fraction with a denominator of 1000 to a decimal:

$\frac{104}{1000} = 0.104$

... (2)

The decimal expansion of $\frac{13}{125}$ is 0.104, which is indeed terminating.

Answer: The decimal expansion of $\frac{13}{125}$ is terminating. Its decimal expansion is 0.104.

Case 2: Non-Terminating Repeating Decimal Expansion

A rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime, $\text{HCF}(p, q) = 1$) has a non-terminating repeating decimal expansion if and only if the prime factorisation of the denominator $q$ contains at least one prime factor other than 2 or 5.

Example 2. Determine if the decimal expansion of $\frac{1}{7}$ is terminating or non-terminating.

Answer:

Given:

The rational number $\frac{1}{7}$.

To Determine:

The nature of its decimal expansion.

Solution:

The given rational number is $\frac{1}{7}$. Here, $p = 1$ and $q = 7$.

The fraction is in simplest form since $\text{HCF}(1, 7) = 1$.

Now, examine the prime factorisation of the denominator $q = 7$. The only prime factor of 7 is 7 itself.

$7 = 7^1$

... (1)

The prime factorisation of the denominator contains the prime factor 7, which is a prime factor other than 2 or 5.

Therefore, according to the rule for non-terminating repeating decimal expansions, the decimal expansion of $\frac{1}{7}$ is non-terminating and repeating.

(Performing the division $1 \div 7$ yields $0.142857142857...$, where the block of digits 142857 repeats indefinitely, written as $0.\overline{142857}$).

Answer: The decimal expansion of $\frac{1}{7}$ is non-terminating and repeating.

Example 3. Write the decimal expansion of $\frac{3}{8}$.

Answer:

Given:

The rational number $\frac{3}{8}$.

To Find:

The decimal expansion of $\frac{3}{8}$.

Solution:

The given fraction is $\frac{3}{8}$. It is in simplest form (HCF(3, 8) = 1). The denominator is $q = 8$. The prime factorisation of 8 is $2 \times 2 \times 2 = 2^3$. This can be written as $2^3 \times 5^0$. Since the denominator only has the prime factor 2, the decimal expansion will be terminating.

To find the decimal expansion, we can perform long division or rewrite the fraction with a denominator that is a power of 10.

Method 1: Long Division

$\begin{array}{r} 0.375\phantom{} \\ 8{\overline{\smash{\big)}\,3.000\phantom{}}} \\ \underline{-~\phantom{(}(2\ 4\downarrow)}\phantom{} \\ 60\phantom{0} \\ \underline{-~\phantom{(}(56\downarrow)}\phantom{} \\ 40\phantom{} \\ \underline{-~\phantom{(}(40)} \\ 0\phantom{} \end{array}$

The long division terminates with a remainder of 0. The decimal expansion is 0.375.

Method 2: Using Powers of 10

The denominator is $8 = 2^3$. To make the denominator a power of 10 ($10^k = 2^k \times 5^k$), we need the same number of factors of 5 as factors of 2. We have $2^3$, so we need $5^3 = 125$. Multiply the numerator and the denominator by $5^3 = 125$:

$\frac{3}{8} = \frac{3}{2^3} = \frac{3 \times 5^3}{2^3 \times 5^3}$

$= \frac{3 \times 125}{(2 \times 5)^3} = \frac{375}{10^3} = \frac{375}{1000}$

Convert the fraction with a denominator of 1000 to a decimal:

$\frac{375}{1000} = 0.375$

... (1)

Answer: The decimal expansion of $\frac{3}{8}$ is 0.375.

Converting Decimal Expansions to Rational Form ($\frac{p}{q}$)

The converse is also true: any number with a terminating or non-terminating repeating decimal expansion is a rational number. We can convert such decimal expansions back into the $\frac{p}{q}$ form.

Example 4. Express $0.104$ in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Answer:

Given:

The decimal number $0.104$.

To Express:

The number in the form $\frac{p}{q}$.

Solution:

The decimal number is $0.104$. This is a terminating decimal.

To convert a terminating decimal to a fraction, write the number without the decimal point as the numerator. The denominator will be 1 followed by as many zeros as there are digits after the decimal point.

The number $0.104$ has 3 digits after the decimal point (1, 0, and 4).

$0.104 = \frac{104}{1000}$

Now, simplify the fraction by dividing the numerator and denominator by their HCF. Both 104 and 1000 are divisible by common factors like 2, 4, and 8. The HCF is 8.

$\frac{104}{1000} = \frac{104 \div 8}{1000 \div 8} = \frac{13}{125}$

Using the cancellation format:

$\frac{\cancel{104}^{13}}{\cancel{1000}_{125}}$

$\frac{13}{125}$

... (1)

So, $0.104 = \frac{13}{125}$. This is in the form $\frac{p}{q}$, where $p=13$ and $q=125$ are integers and $q \neq 0$. The fraction is in simplest form.

Answer: $0.104 = \frac{13}{125}$.

Example 5. Express $0.\overline{3}$ in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Answer:

Given:

The decimal number $0.\overline{3}$.

To Express:

The number in the form $\frac{p}{q}$.

Solution:

The decimal number is $0.\overline{3}$, which is a non-terminating repeating decimal where the digit 3 repeats indefinitely ($0.3333...$).

Let $x$ be equal to the given decimal number:

$x = 0.\overline{3}$

Write out the repeating decimal:

$x = 0.3333...$

... (1)

Since only one digit is repeating, multiply equation (1) by $10^1 = 10$ to shift the repeating part one place to the left:

$10x = 10 \times (0.3333...)$

$10x = 3.3333...$

... (2)

Subtract equation (1) from equation (2). This eliminates the repeating part:

$10x - x = (3.3333...) - (0.3333...)$

$9x = 3$

Solve for $x$ by dividing both sides by 9:

$x = \frac{3}{9}$

Simplify the fraction by dividing the numerator and denominator by their HCF, which is 3:

$x = \frac{\cancel{3}^1}{\cancel{9}_3}$

$x = \frac{1}{3}$

... (3)

So, $0.\overline{3} = \frac{1}{3}$. This is in the form $\frac{p}{q}$, where $p=1$ and $q=3$ are integers and $q \neq 0$. The fraction is in simplest form.

Answer: $0.\overline{3} = \frac{1}{3}$.