Proof Problem - 2

Let $\large \frac {p}{q}$ be a rational number whose square is $\large 2$

Assume that $\large p$ and $\large q$ have no common integer factors other than $\large \pm 1$

Then $\large (\frac {p}{q})^{2} = 2 \rightarrow \frac {p^2}{q^2} = 2 \rightarrow p^2 = {2q}^2$

So, $\large p^2$ is even and $\large p$ is even because if $\large p$ was odd, $\large p^2$ would be odd

We can substitute $\large p = 2m$ into $\large p^2 = {2q}^2$ , where $\large m$ is any positive integer

When we solve $\large ({2m})^2 = {2q}^2$ we get $\large {4m}^2 = {2q}^2 \rightarrow {2m}^2 = q^2$

So, $\large q^2$ is even and $\large q$ is even because if $\large q$ was odd, $\large q^2$ would be odd

From this, we can say that $\large p$ and $\large q$ have a common factor of $\large 2$ , as they are both even

However this contradicts our original assumption that $\large p$ and $\large q$ have no common factors other than $\large \pm 1$

Due to this contradiction, it is proven that there is no rational number whose square is $\large 2$