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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

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  • $\begingroup$ What is this $p$ you suddenly introduced? $\endgroup$ – Thomas Andrews Oct 19 '15 at 19:43
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    $\begingroup$ Start with $\sqrt{45}=3\sqrt{5}$ and proceed with the prime $5$. Or note that if $45\mid a^2$ then $5\mid a^2$ (because $5\mid45$) and proceed. $\endgroup$ – Did Oct 19 '15 at 19:47
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    $\begingroup$ Note that if $45\mid a^2$, it is not always true that $45\mid a$. For example, $45\mid 225=15^2$, but it is not true that $45\mid 15$. $\endgroup$ – Thomas Andrews Oct 19 '15 at 19:48
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    $\begingroup$ Hmmm... $3\sqrt{5}$ is rational if and only if $\sqrt{5}$ is $____$. $\endgroup$ – Did Oct 19 '15 at 19:54
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    $\begingroup$ "My question is that when using Euclid's Lemma doesn't 45 have to be prime?" The radicand $x$ doesn't have to be prime; it's just that it shouldn't contain any perfect square, or the step from $x|a^2 \Rightarrow x|a$ fails to hold. Hence the answers below that start with getting the factor of 9 out to begin with. $\endgroup$ – Daniel R. Collins Oct 19 '15 at 20:04
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$\sqrt{45}=3\sqrt{5}$. 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 $(a,b)=1$

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  • $\begingroup$ so right from the beginning i should break it up and then say showing squareroot of 5 is enough? $\endgroup$ – Charlene Oct 19 '15 at 19:55
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    $\begingroup$ yes, you are right. $\endgroup$ – R.N Oct 19 '15 at 20:00
  • $\begingroup$ I think that for this question-asker, an explicit step for why "$\sqrt{5}$ is irrational" implies "$3 \sqrt{5}$ is irrational" is appropriate. $\endgroup$ – Daniel R. Collins Oct 19 '15 at 20:08
  • $\begingroup$ @DanielR.Collins, dear it is can be easily checked by using $\frac{a}{b}$ where $a,b\in\mathbb{Z}$ and $b\neq0$ $\endgroup$ – R.N Oct 19 '15 at 20:20
  • $\begingroup$ @Razieh Noori: Yes, you should expand and add to your answer. $\endgroup$ – Daniel R. Collins Oct 19 '15 at 21:54
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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 $(a,b)\neq 1$.

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$$ 45b^2 = a^2 $$ 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.

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