(a) Prove that $\sqrt{3}$ is an irrational number. (b) Find the HCF and LCM of 96 and 404 by the prime factorization met

Question

(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 $HCF 	imes LCM = 	ext{Product of the two numbers}$. (5 Marks)

Accepted Answer

(a) Proof of Irrationality of $\sqrt{3}$: Let us assume, to the contrary, that $\sqrt{3}$ is rational. Then, there exist co-prime integers $a$ and $b$ ($b 
eq 0$) such that $\sqrt{3} = \frac{a}{b}$. Squaring both sides: $3 = \frac{a^2}{b^2} \implies a^2 = 3b^2$. Since $a^2$ is divisible by 3, then by theorem, $a$ is also divisible by 3. Let $a = 3c$ for some integer $c$. Substituting this in $a^2 = 3b^2$: $(3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies 3c^2 = b^2$. Since $b^2$ is divisible by 3, then $b$ is also divisible by 3. Therefore, $a$ and $b$ have at least 3 as a common factor. This contradicts the fact that $a$ and $b$ are co-prime. Hence, our assumption was wrong, and $\sqrt{3}$ is irrational. (b) HCF and LCM Calculation: Prime Factorization: $96 = 2 	imes 2 	imes 2 	imes 2 	imes 2 	imes 3 = 2^5 	imes 3^1$ $404 = 2 	imes 2 	imes 101 = 2^2 	imes 101^1$ Finding HCF: HCF = Product of the smallest power of each common prime factor. $HCF(96, 404) = 2^2 = 4$. Finding LCM: LCM = Product of the greatest power of each prime factor involved. $LCM(96, 404) = 2^5 	imes 3^1 	imes 101^1 = 32 	imes 3 	imes 101 = 9696$. Verification: $HCF 	imes LCM = 4 	imes 9696 = 38784$. Product of numbers = $96 	imes 404 = 38784$. Since $38784 = 38784$, the relation is verified.

Real Numbers

❓ Sibling Questions in this Chapter