# Can it be proven that a product of factors where two or more factors have non-integer exponentiation is a irrational number?

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?

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

$$\sqrt 4$$ 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.