Wade Bullock

2022-07-15

Prove that if $p$ is a prime number, then $\sqrt{p}$is an irrational number.

Elijah Benjamin

Beginner2022-07-16Added 10 answers

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 $\text{gcd}(a,b)\ne 1$

We can make this assumption, because we still lose no generality.

Now using $\text{gcd}(a,b)=d\ne 1$. 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.

We can make this assumption, because we still lose no generality.

Now using $\text{gcd}(a,b)=d\ne 1$. 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.

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