Prove that $\sqrt{45}$ is irrational (using Euclid’s Lemma)
Assume $\sqrt{45}$ is rational.
By definition of rational : $\sqrt{p}=\frac{a}{b}$= $\sqrt{45}$=$\frac{a}{b}$ for some $a,b$ integers and $b≠0$
Let $a,b$ have no common divisor $>1$
By algebra, $45=\frac{a^2}{b^2}$
$45b^2=a^2$
Since $45b^2$ is divisible by 45 it follows that $a^2$ is divisible by 45. Then a is divisble by 45 by corollary, if p is a prime and p divides $a^2$, then p divides a… and this proof continues
My question is that when using Euclid's Lemma doesn't 45 have to be prime? since it isn't prime how would i get around this to make the last part of my proof complete to be able to finish and show a contradition
Best Answer
$\sqrt{45}=3\sqrt{5}$. Thus it is enough to show that $\sqrt 5$ is irrati0nal. let $\sqrt{5}=\frac{a}{b}$ where $(a,b)=1$. then $5b^2=a^2\Rightarrow 5\vert a^2$ since $5$ is prime $5\vert a$ thus there is $k$ such that $a=5k\Rightarrow 5b^2=25k^2\Rightarrow b^2=5k^2 $ so $5\vert b$ contradiction with $(a,b)=1$