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Let $n$ be an integer. Show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square (i.e., the square of an integer).

How would I do this proof. In a simple manner.

If n is perfect square then the square root n is a rational number.

So if n is a perfect square there is a integer m such that $m(m)=n^2$

So then I guess I can do the logik from contrapostive.

If square root of n is irrational then not n is perfect square.

But how could I prove this?

Bill Dubuque
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Fernando Martinez
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  • First, is $n$ an integer? If so, it is obvious that if $\sqrt{n}$ is rational, or equivalently if $\sqrt{n}$ is irrational, then $n$ is not a perfect square. Somewhat more difficult is to show that if $\sqrt{n}$ is rational, then $n$ is a perfect square. Note that if $n$ is a perfect square there is an integer such that $(m)(m)=n$. – André Nicolas Sep 11 '15 at 23:45

2 Answers2

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Let $n\in\mathbb{N}$.

$\implies$ If n is a perfect square, then $n=m^2$ for some $m\in\mathbb{N}$, so $\sqrt{n}=m$ is rational.

$\impliedby$ If $\sqrt{n}$ is rational, let $\sqrt{n}=\frac{a}{b}$ where $a,b\in\mathbb{N}$ and $\text{gcd} (a,b)=1$.

$\hspace{.4 in}$Then $n=\frac{a^2}{b^2}$ gives $a^2=nb^2$; so if $p$ is a prime and $p\big|b$, then $p\big|a^2$ and therefore $p\big|a$.

$\hspace{.4 in}$This gives a contradiction, since $\text{gcd}(a,b)=1$;

$\hspace{.4 in}$so $b=1$ (since it has no prime divisors) and therefore $n=a^2$ is a perfect square.

user84413
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Use the rational root theorem, and note that if $x^m - n = 0$, then if $x = a / b$ it has to be that $b \mid 1$ (i.e., $b = 1$) and $a \mid n$, so that if $x$ is rational, it is an integer.

vonbrand
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  • Whenever I see this kind of proof, where an advanced theorem is used to prove a more elementary result, I wonder if the proof of the advanced theorem depends on the elementary one. – marty cohen Sep 12 '15 at 01:13
  • @marty.cohen, The rational root theorem is quite simple to prove: Substitute your root, multiply out the denominator so everything in sight is an integer. Use divisibility of all terms save one by numerator and denominator respectively, and the result follows. – vonbrand Sep 12 '15 at 13:09