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
$\endgroup$ 83 Answers
$\begingroup$$\sqrt{45}=3\sqrt{5}$. Thus it is enough to show that $\sqrt 5$ is irrati0nal. let $\sqrt{5}=\frac{a}{b}$ where $(a,b)=1$. then $5b^2=a^2\Rightarrow 5\vert a^2$ since $5$ is prime $5\vert a$ thus there is $k$ such that $a=5k\Rightarrow 5b^2=25k^2\Rightarrow b^2=5k^2 $ so $5\vert b$ contradiction with $(a,b)=1$
$\endgroup$ 5 $\begingroup$Note that $5|a^2$ but $5$ is prime, then $5|a $. If $5|a $ then $5|9b^2$, but $(5,9)=1$ implies $5|b^2$ and for the precedent reason $5|b $,in other words $(a,b)\neq 1$.
$\endgroup$ $\begingroup$$$ 45b^2 = a^2 $$ Since $5$ divides $45$, we conclude that $5$ divides $45b^2$ and then $5$ divides $a^2$. This means that $5$ divides $a$, and that $25$ divides $a^2$. Hence $25$ divides $45b^2$ or that $5$ divides $b^2$ which means $5$ divides $b$. Contradiction.
$\endgroup$
