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

**Solution**

\(\text{Condition 1}\)

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

\[\sqrt{p}=\frac{m}{n}\]

\[p=\frac{{m}^{2}}{{n}^{2}}\]

\[p{n}^{2}={m}^{2}\]

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 \(\text{Condition 1}\). Hence, the square root of a prime number is irrational.

Check out my other notes at Proof, Disproof, and Derivation

## Comments

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TopNewestFYI - To display text in Latex, use \text{XXX}. Otherwise, it will appear italicized – Calvin Lin Staff · 2 years, 1 month ago

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– Steven Zheng · 2 years, 1 month ago

Do you mean the equations in paragraphs? I actually like my equations italicized. They stand out.Log in to reply

– Calvin Lin Staff · 2 years, 1 month ago

Oh, I was refering to "\( (Condition 1) \)". I decided to remove the Latex brackets, so it displays as "Condition 1" instead. If you wanted to use latex, you could do " \alpha \text{ Condition } \beta" to get \( \alpha \text{ Condition } \beta\).Log in to reply

– Steven Zheng · 2 years, 1 month ago

Okay. I will fix it.Log in to reply