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I'm reading a course in Discrete Mathematics and exploring different ways things can be proven. One classic question seems to be prove, for instance, that ${\sqrt{30}}$ is a irrational number.

By proof through contradiction and going through the motion we do following: ${\sqrt{30}} = \frac{a}{b}$ where $a$, $b$ are integers, $b\ne0$ and $gcd(a,b)=1$. We end up with $2*3*5*b^2=a^2$

Now one can look at substitution and see that we end up with a contradiction.

But, a different argument at this point could be: Since all factors of $a^2$ must be of power $2n$, where $n$ is integer $\ge 1$, $a=(q^n_1*q^n_2*..) \Rightarrow a^2=(q^{2n}_1*q^{2n}_2*..)$. Since we have $2*3*5*b^2=a^2$ that would mean that if $a^2$ have these factors then they would have exponential of $^{2n+1}$ after multiplication with $b^2$, if $a^2$ doesn't have these factors they would have exponential of $^1$ after multiplication.

Taking the square root of $a^2$ would result in these factors have a exponential that are non-integer. Can we prove that for any combination of factors where the exponentials are fractions that the product is irrational?

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  • $\begingroup$ Welcome to MSE. In order to get $\sqrt{30}$, you should type \sqrt{30} and not \sqrt30, which will give you $\sqrt30$. $\endgroup$ – José Carlos Santos Jan 2 at 8:19
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    $\begingroup$ Thanks @JoséCarlosSantos, fixed that! $\endgroup$ – bej Jan 2 at 8:30
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Welcome to MSE!

$\sqrt 4$ is an integer, not irrational. In your argument, $4b^2=a^2$, so $2^2b^2=a^2$ and there is no way to find a contradiction.

So if the number is a perfect square, like 4, 9, 16, the proof of contraction fails as it should.

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