Prove that `sqrt(7)` is irrational.

Monday, August 2, 2010

Prove that $\displaystyle\sqrt{{{7}}}$ is irrational.

Hello!

We'll prove this by a contradiction. Let $\displaystyle\sqrt{{{7}}}$ is a rational number, that means $\displaystyle\sqrt{{{7}}}=\frac{{m}}{{n}}$ where m and n are natural numbers.

Also let m and n be coprime. If they aren't, divide both by their GCF.

Now we square the equality and obtain $\displaystyle{7}{{n}}^{{2}}={{m}}^{{2}}.$ Therefore 7 is a factor of $\displaystyle{{m}}^{{2}}.$ Because 7 is a prime number, 7 is also a factor of m itself. So $\displaystyle{m}={7}{k}$ for some natural k.

Thus $\displaystyle{7}{{n}}^{{2}}={{7}}^{{2}}\cdot{{k}}^{{2}},$ or $\displaystyle{{n}}^{{2}}={7}{{k}}^{{2}}.$ So 7 is a factor of $\displaystyle{{n}}^{{2}}$ and therefore of n. But this means m and n have a common factor 7, which is a contradiction. This contradiction proves that $\displaystyle\sqrt{{{7}}}$ is irrational.