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.

#Algebra

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$