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Let $n$ be an integer. Prove that if $3 \vert n^2$ then $3 \vert n$.

Use the result from above to prove $\sqrt{3}$ is also irrational.

So for the first part I did the contrapositive

$\neg (3 \vert n$) then $\neg (3 \vert n^2)$

But how do i do cases for this and how to do the second part?

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    $\begingroup$ Are you sure you did the contrapostive correctly? If you are claiming that $\lnot 9|12$ then $\lnot 9|12^2$ you are mistaken. $\endgroup$ – fleablood Feb 8 '18 at 7:33
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It is easier to use Euclid's lemma: if $3\mid n^2$, then $3\mid n$ or $3\mid n$. In other words. $3\mid n$.

If $\sqrt3$ was rational, then it would be a rational root of $x^2-3$. But, by the rational root theorem, the only possible rational roots of that polynomial are $\pm1$ and $\pm3$. But none of them is.

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If $\frac {a^2}{b^2} = 3$ then $a^2 = 3b^2$ so $3|a^2$ so $3|a$ and .... well somehow that has to lead to a contradiction. Can you figure it out?

For one thing if $a = 3c$ than $9c^2 = 3b^2$.....

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  • $\begingroup$ Is it because a and b share a common multiple? $\endgroup$ – Rick Feb 8 '18 at 8:11
  • $\begingroup$ Is what because they share a common multiple... The unspoken assumption is that any rational can be written "in lowest terms" and if $\frac {a^2}{b^2} = 3$ then $a, b$ can't be in lowest terms. This requires a teensy bit more sophistication why an infinite regress of fractions with common terms is a contradiction, but we can accept that it is (any strictly decreasing sequence of positive must end and have a last element so you can't have an infinite a = 3b; b = 3c; c = 3d.... you must eventually have a last value) $\endgroup$ – fleablood Feb 8 '18 at 16:58
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Hint: use the decompostion of $n$ in prime factors.

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