 # Prove that if p is a prime number, Wade Bullock 2022-07-15 Answered
Prove that if $p$ is a prime number, then $\sqrt{p}$is an irrational number.
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By way of contradiction, assume $\sqrt{p}$ is rational. Then there exist $a,b\in \mathbb{Z}$ with $b\ne 0$ such that $\sqrt{p}=\frac{a}{b}$. Without loss of generality, we may assume $\text{gcd}\left(a,b\right)\ne 1$
We can make this assumption, because we still lose no generality.
Now using $\text{gcd}\left(a,b\right)=d\ne 1$. Then we can write $a=d\cdot {a}^{\prime }$and $b=d\cdot {b}^{\prime }$, for some relatively prime integers ${a}^{\prime }$and ${b}^{\prime }$.
Hence
$\sqrt{p}=\frac{a}{b}=\frac{d{a}^{\prime }}{d{b}^{\prime }}=\frac{{a}^{\prime }}{{b}^{\prime }},$
So we have shown that $\sqrt{p}$ is a ratio of two relatively prime integers.