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

Question

Accepted Answer

Step 1: Let $\sqrt{2}$ be a rational number. Then $\sqrt{2} = \frac{p}{q}$, where $q 
eq 0$ and $p$ & $q$ are co-primes.
Step 2: $2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$. So $p = 2a$, where $a$ is some integer. ... (i)
Step 3: $4a^2 = 2q^2 \Rightarrow q^2 = 2a^2 \Rightarrow q$ is divisible by $2$. So $q = 2b$, where $b$ is some integer. ... (ii)
Step 4: (i) and (ii) lead to contradiction as $p$ and $q$ are co-primes. $	herefore \sqrt{2}$ is an irrational number.

Question

Solution