Irrational or rational

We can prove √2 is irrational with proof through contradiction:

Let's suppose √2 was a rational number, that would mean we could write the number √2 as a/b, where a and b are real numbers, and b does not equal 0. In the expression a/b, a and b would be in their lowest possible terms so that we could no longer simplify a/b any further. Therefore, a or b has to be an odd number.

Knowing √2 = a/b, that means 2 = $a^2$ / $b^2$ , and $a^2$ = 2 $b^2$ . This means that a is an even number.

If a is an even number, then a can be written as a = 2k, where k is some other number.

if we substitute a into the original equation:

• 2 = $(2k)^2$ / $b^2$
• 2 = $4k^2$ / $b^2$
• 2 $b^2$ = $4k^2$
• $b^2$ = $2k^2$

$b^2$ is an even number, which means b is even too. Because a and b can't both be even, we have a contradiction, which means we can't write √2 as a rational number.

Because $\sqrt{2}$ can not be written as a fraction of $a$ and $b$ , where they are integers like this form $\to \frac{a}{b}$ . Therefore $\sqrt{2}$ is irrational.