Prove that under root 2 is a rational number sorry e-rational number

If $\sqrt{2}$ is rational, it can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ have no common factors other than 1 (i.e., the fraction is in its simplest form).

This equation implies that $p^2$ is an even number, because it is equal to 2 times an integer ($q^2$).

If $p^2$ is even, then $p$ itself must be even. (If $p$ were odd, $p^2$ would also be odd, e.g., $(2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$).
Since $p$ is even, we can write $p$ as $2k$ for some integer $k$.

Divide both sides by 2:

This equation implies that $q^2$ is an even number, because it is equal to 2 times an integer ($k^2$).

If $q^2$ is even, then $q$ itself must be even.

We have shown that $p$ is even and $q$ is even. This means that both $p$ and $q$ have a common factor of 2. However, in Step 1, we assumed that $p$ and $q$ have no common factors other than 1. This is a contradiction.

Since our initial assumption that $\sqrt{2}$ is rational leads to a contradiction, the assumption must be false. Therefore, $\sqrt{2}$ must be an irrational number.

Thus, $\text{under root 2 is an irrational number}$.