MP Board · Class 10 · Mathematics · Real Numbers(a) Prove that $\sqrt{3}$ is an irrational number. (b) Find the HCF and LCM of 96 and 404 by the prime factorization method and verify that $HCF \times LCM = \text{Product of the two numbers}$. (5 Marks)

(a) Proof of Irrationality of $\sqrt{3}$:

1. Let us assume, to the contrary, that $\sqrt{3}$ is rational.
2. Then, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \frac{a}{b}$.
3. Squaring both sides: $3 = \frac{a^2}{b^2} \implies a^2 = 3b^2$.
4. Since $a^2$ is divisible by 3, then by theorem, $a$ is also divisible by 3.
5. Let $a = 3c$ for some integer $c$. Substituting this in $a^2 = 3b^2$: $(3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies 3c^2 = b^2$.
6. Since $b^2$ is divisible by 3, then $b$ is also divisible by 3.
7. Therefore, $a$ and $b$ have at least 3 as a common factor. This contradicts the fact that $a$ and $b$ are co-prime.
8. Hence, our assumption was wrong, and $\sqrt{3}$ is irrational.

(b) HCF and LCM Calculation:

1. Prime Factorization:
   • $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3^1$
   • $404 = 2 \times 2 \times 101 = 2^2 \times 101^1$
2. Finding HCF:
   • HCF = Product of the smallest power of each common prime factor.
   • $HCF(96, 404) = 2^2 = 4$.
3. Finding LCM:
   • LCM = Product of the greatest power of each prime factor involved.
   • $LCM(96, 404) = 2^5 \times 3^1 \times 101^1 = 32 \times 3 \times 101 = 9696$.
4. Verification:
   • $HCF \times LCM = 4 \times 9696 = 38784$.
   • Product of numbers = $96 \times 404 = 38784$.
   • Since $38784 = 38784$, the relation is verified.