Real Numbers - Euclid's division lemma, fundamental theorem of arithmetic, and irrationality proofs

Welcome to Class 10 Mathematics! If you’ve ever wondered how numbers build up the mathematical universe around us, you are in the right place. Real Numbers is the very first chapter of your NCERT textbook, and it sets the foundation for algebra, geometry, and beyond.

In this tutorial, we will break down the three pillar concepts of this chapter into simple, bite-sized, and logical steps:

Euclid’s Division Lemma & Algorithm

The Fundamental Theorem of Arithmetic (FTA)

Proofs of Irrationality

Grab your notebook, a pen, and let's make Real Numbers easy and fun!

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1. Euclid’s Division Lemma & Algorithm

What is a "Lemma"?

In mathematics, a lemma is a proven statement used as a stepping stone to prove another statement. Think of it as a trusted tool in your math toolbox!

The Concept: Division with a Twist

Do you remember how you checked your division problems in primary school?

$$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$$

Euclid’s Division Lemma is nothing more than this exact concept written in formal mathematical language!

Euclid's Division Lemma:

Given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying:

$$a = bq + r \quad \text{where } 0 \le r < b$$

$a$: Dividend (the number being divided)

$b$: Divisor (the number dividing)

$q$: Quotient (how many times $b$ goes into $a$)

$r$: Remainder (what is left over)

Analogy Time! 🍫

Imagine you have 17 chocolates ($a$) and you want to pack them into boxes that hold 5 chocolates each ($b$).

How many full boxes can you fill? 3 full boxes ($q$).

How many chocolates remain unpacked? 2 chocolates ($r$).

Writing this mathematically:

$$17 = (5 \times 3) + 2$$

Notice that the remaining chocolates ($2$) must always be less than the box size ($5$). That’s why $0 \le r < b$!

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Euclid's Division Algorithm: Finding the HCF

An algorithm is a step-by-step procedure for solving a problem. Euclid's Division Algorithm is used to calculate the Highest Common Factor (HCF) of two large numbers.

Step-by-step process to find HCF of two numbers $a$ and $b$ ($a > b$):

Apply Euclid's Division Lemma: $a = bq + r$.

If $r = 0$, then $b$ is the HCF!

If $r \neq 0$, apply the lemma again to $b$ (new dividend) and $r$ (new divisor).

Repeat the process until the remainder becomes $0$. The divisor at this final stage is the HCF.

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2. The Fundamental Theorem of Arithmetic (FTA)

The "Lego Bricks" of Mathematics 🧱

In chemistry, every complex substance is made up of fundamental elements (atoms). In mathematics, prime numbers are the atoms! Every composite number can be built by multiplying prime numbers together.

The Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

For example:

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

No matter how you factorize $12$, you will always get two $2$'s and one $3$.

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Finding HCF and LCM using Prime Factorization

Using prime factorization, you can easily find the HCF and LCM of any set of positive integers:

HCF (Highest Common Factor): Product of the smallest power of each common prime factor involved in the numbers.

LCM (Lowest Common Multiple): Product of the greatest power of each prime factor involved in the numbers.

Important Formula:

For any two positive integers $a$ and $b$:

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

*(Note: This handy formula works only for two numbers, not three!)*

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3. Proofs of Irrationality

What is an Irrational Number?

A number is called irrational if it cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Examples include $\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi$, etc.

The Key Theorem Needed for Proofs

Before proving irrationality, keep this critical theorem in mind:

Let $p$ be a prime number. If $p$ divides $a^2$ (where $a$ is a positive integer), then $p$ also divides $a$.

*Example:* If $3$ divides $6^2$ ($36$), then $3$ also divides $6$.

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Proof by Contradiction: The "Math Detective" Method 🕵️‍♂️

To prove that a number like $\sqrt{2}$ is irrational, we use a clever technique called Proof by Contradiction:

We assume the opposite of what we want to prove (i.e., we assume $\sqrt{2}$ is *rational*).

We follow strict mathematical rules based on this assumption.

We arrive at a logical absurdity (a contradiction).

We conclude that our initial assumption was wrong, which means the number MUST be irrational!

Standard Proof: Prove that $\sqrt{2}$ is irrational

Assumption: Assume to the contrary that $\sqrt{2}$ is rational.

Formulation: Therefore, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:

$$\sqrt{2} = \frac{a}{b}$$

*(Co-prime means $a$ and $b$ have no common factor other than $1$.)*

Squaring both sides:

$$2 = \frac{a^2}{b^2} \implies 2b^2 = a^2 \quad \text{--- (Equation 1)}$$

Deduction 1: Since $2$ divides $a^2$, by our theorem, $2$ must also divide $a$.

So, we can write $a = 2c$ for some integer $c$.

Substitution: Substitute $a = 2c$ into Equation 1:

$$2b^2 = (2c)^2 \implies 2b^2 = 4c^2 \implies b^2 = 2c^2$$

Deduction 2: Since $2$ divides $b^2$, $2$ must also divide $b$.

The Contradiction:

From steps 4 and 6, both $a$ and $b$ have at least $2$ as a common factor. But this contradicts our initial fact that $a$ and $b$ are co-prime (have no common factors other than 1)!

Conclusion: This contradiction has arisen because of our incorrect assumption that $\sqrt{2}$ is rational. Hence, $\sqrt{2}$ is irrational. $\blacksquare$

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💡 Practice Questions with Solutions

Let’s test your understanding! Try solving these on your own first before reading the solutions.

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Question 1

Find the HCF and LCM of $90$ and $144$ using the Prime Factorization method. Verify that $\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$.

Solution:

Step 1: Write down the prime factorization of both numbers.

$90 = 2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1$

$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2$

Step 2: Calculate HCF (smallest powers of common factors).

Common prime factors: $2$ and $3$.

Smallest power of $2$ is $2^1$.

Smallest power of $3$ is $3^2$.

$$\text{HCF}(90, 144) = 2^1 \times 3^2 = 2 \times 9 = 18$$

Step 3: Calculate LCM (greatest powers of all factors involved).

Factors involved: $2, 3, 5$.

Greatest power of $2$ is $2^4$.

Greatest power of $3$ is $3^2$.

Greatest power of $5$ is $5^1$.

$$\text{LCM}(90, 144) = 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 = 720$$

Step 4: Verification.

$\text{LHS} = \text{HCF} \times \text{LCM} = 18 \times 720 = 12,960$

$\text{RHS} = \text{Product of numbers} = 90 \times 144 = 12,960$

Since $\text{LHS} = \text{RHS}$, the relationship $\text{HCF} \times \text{LCM} = a \times b$ is verified!

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Question 2

Prove that $5 - \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.

Solution:

Step 1: Let us assume to the contrary that $5 - \sqrt{3}$ is rational.

Step 2: Therefore, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that:

$$5 - \sqrt{3} = \frac{a}{b}$$

Step 3: Rearrange the equation to isolate $\sqrt{3}$:

$$\sqrt{3} = 5 - \frac{a}{b}$$

$$\sqrt{3} = \frac{5b - a}{b}$$

Step 4: Analyze the result:

Since $a$ and $b$ are integers, $(5b - a)$ and $b$ are also integers. Therefore, $\frac{5b - a}{b}$ is a rational number.

This implies that $\sqrt{3}$ must also be a rational number!

Step 5: Conclusion:

This contradicts the given fact that $\sqrt{3}$ is irrational.

This contradiction arose because of our wrong assumption that $5 - \sqrt{3}$ is rational. Therefore, $5 - \sqrt{3}$ is irrational. $\blacksquare$

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Question 3

Explain why $7 \times 11 \times 13 + 13$ is a composite number.

Solution:

A number is called composite if it has factors other than $1$ and itself (i.e., it can be expressed as a product of prime factors).

Given expression:

$$E = 7 \times 11 \times 13 + 13$$

Take out $13$ as a common factor:

$$E = 13 \times (7 \times 11 + 1)$$

$$E = 13 \times (77 + 1)$$

$$E = 13 \times 78$$

Now factorize $78$:

$$78 = 2 \times 3 \times 13$$

Substitute back:

$$E = 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2$$

Since the given expression can be written as a product of prime factors ($2$, $3$, and $13$), according to the Fundamental Theorem of Arithmetic, it must be a composite number. $\blacksquare$

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🌟 Quick Recap & Revision Tips

Euclid's Division Lemma: $a = bq + r$ where $0 \le r < b$.

Fundamental Theorem of Arithmetic: Every composite number has a unique prime factorization.

HCF & LCM Relation: $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$.

Irrationality Proofs: Always start by assuming the number is rational, use $p/q$ in simplest form, and find a logical contradiction!

Keep practicing your textbook exercises regularly. Math isn't about memorization—it's about building logical confidence step by step. You've got this!