# How to prove that <msqrt>

How to prove that $\sqrt{3}$ is an irrational number?
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Leland Ochoa
Let $\sqrt{3}=\frac{p}{q},p,q$ relatively prime. $3=\frac{{p}^{2}}{{q}^{2}}$, so $3$ divides $p$ and so on.

aligass2004yi
Let $\sqrt{3}=\frac{a}{b}$, where a and b have no common factors besides $1$

As $3{b}^{2}={a}^{2}$ so ${a}^{2}$ is a multiple of $3$, and hence a should be a multiple of $3$. Let $a=3k$, then ${b}^{2}=3{k}^{2}$, and b must also be a multiple of three. You will arrive at a contradiction to the earlier assumption that $a$ and $b$ have no common factors.