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# Irrational Primes

Prove that the square root of a prime number is irrational.

Solution

Let $$p$$ be any prime number. For $$\sqrt{p}$$ to be rational, it must be expressible as the quotient of two coprime integers $$(Condition 1)$$.

$\sqrt{p}=\frac{m}{n}$

$p=\frac{{m}^{2}}{{n}^{2}}$

$p{n}^{2}={m}^{2}$

Since $$n,m,{n}^{2},{m}^{2}$$ are integers, this implies that $$m$$ has a factor of $$p$$. Therefore, if the expression $$m=pm'$$ is substituted into the third equation, then $${n}^{2}=p{(m' )}^{2}$$. By a similar argument, the integer $$n$$ must possess a factor of $$p$$ as well. This demonstrates the fact both $$m$$ and $$n$$ are not coprime, which contradicts $$Condition 1$$. Hence, the square root of a prime number is irrational.

Note by Steven Zheng
2 years, 1 month ago