# Square Root of Primes

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 Note by Steven Zheng
6 years, 3 months ago

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FYI - To display text in Latex, use \text{XXX}. Otherwise, it will appear italicized

Staff - 6 years, 2 months ago

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

- 6 years, 2 months 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 $\alpha \text{ Condition } \beta$.

Staff - 6 years, 2 months ago

Okay. I will fix it.

- 6 years, 2 months ago