Irrational Numbers Math Example 3

Solution

1. 1
   Assume for contradiction that $\sqrt{3}$ is rational, so $\sqrt{3} = \frac{a}{b}$ where $a, b$ are integers with no common factors ($\gcd(a,b) = 1$).
2. 2
   Square both sides: $3 = \frac{a^2}{b^2}$, which gives $a^2 = 3b^2$. This means $a^2$ is divisible by 3, so $a$ must also be divisible by 3 (since 3 is prime). Write $a = 3k$ for some integer $k$.
3. 3
   Substitute $a = 3k$: $(3k)^2 = 3b^2 \Rightarrow 9k^2 = 3b^2 \Rightarrow b^2 = 3k^2$. Now $b^2$ is divisible by 3, so $b$ is also divisible by 3. But this contradicts $\gcd(a,b) = 1$, since both $a$ and $b$ share the factor 3. Therefore $\sqrt{3}$ is irrational.