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# 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
3 years, 6 months ago

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

Staff - 3 years, 5 months ago

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

- 3 years, 5 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 - 3 years, 5 months ago

Okay. I will fix it.

- 3 years, 5 months ago