# $\sqrt {3}$ is irrational (proof verification) [duplicate]

$\sqrt {3} \in Q$. Then, $\sqrt{3} = \frac ab$ with the lowest term for $a,b \in Z$.

Then, $3b^2=a^2$, which implies that $a^2$ is divisible by 3.

That is, $a$ is also divisible by 3 (by fundamental theorem of arithmetic).

I don't understand here $a^2$ divisible by 3 implies $a$ divisible by 3.

Could you explain it?

## marked as duplicate by user296602, user223391, Mohammad Riazi-Kermani, Claude Leibovici, Paramanand Singh real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 6 '18 at 8:45

• – K B Dave Mar 6 '18 at 0:13
• Can $a^2$ be a multiple of three without $a$ being a multiple of three? Consider that $$(3n\pm 1)^2 = 3(3n^2\pm 2n)\color{red}{+1}.$$ – Jack D'Aurizio Mar 6 '18 at 0:17
• The linked question has 2 instead of 3, but the proof is virtually identical. – user296602 Mar 6 '18 at 0:17

Working with integers, recall that $cd$ divisible by a prime $k$ implies $c$ is divisible by $k$ or $d$ is divisible by $k$. $a^2$ divisible by 3 implies $a \cdot a$ divisible by 3. So, $a$ is divisible by $3$ or $a$ is divisible by $3$ (This statement occurs by taking the left $a$ and the right $a$; if you don't understand this, in the example mentioned let $c = a$ and $d=a$).
• cd divisible by k implies c is divisible by k or d is divisible by k Only if $k$ is a prime.Otherwise take $c=d, k=c^2$ for example. – dxiv Mar 6 '18 at 0:27
Consider the prime factorization of $a$.
Suppose on the contrary that $3$ is not a factor of $a$. Then squaring $a$ will also not make $3$ appears in the prime factorization of $a^2$. Hence $3$ will not be a factor of $a^2$.