# I Reason for the irrationality of (most) square roots?

1. Sep 9, 2017

### thebosonbreaker

The square root of any integer that is not a square number is always an irrational number.
I find this fact rather spectacular, but my question is why is this true? I have seen the formal proof for the irrationality of root 2 so I could vaguely see how one could prove that all (apart from sq numbers) integer square roots are irrational, but the real question I'm asking here is WHY is it true - is there a deeper mathematical reason? Or is it simply down to a coincidence?

2. Sep 9, 2017

### Orodruin

Staff Emeritus
What do you mean by "deeper mathematical reason"? It is possible to show this in general by using factorisation into primes. Meaning in mathematics never goes further than "follows from the axioms".

3. Sep 9, 2017

### Staff: Mentor

The rational numbers are those numbers, which extend the mulitiplicative halfgroup of the integers, that is $\{\ldots ,-2,-1,1,2,\ldots\}$ to a group, that is a set that allows inverses. In other words, we add all solutions of the equation $a\cdot x \stackrel{(1)}{=} 1$.

To get square roots, we would need to solve $x^2\stackrel{(2)}{=}a$ which is a different equation, which already says, that it's only solvable for squares. That a solution to (1) cannot lead to a solution of (2) except for squares is then the proof you know.

4. Sep 10, 2017

### Staff: Mentor

Most real numbers are irrational. As soon as you do something that is not guaranteed to give a rational number (like adding two rational numbers), you typically get irrational numbers.

5. Sep 10, 2017

### hilbert2

There are some funny "rules" about what kind of irrational numbers are obtained by certain calculations, for example taking the sine or cosine of rational multiples of $\pi$ will give square roots of rational numbers as a result, while taking the gamma function of half-integers $\frac{2n+1}{2}$ will give multiples of $\sqrt{\pi}$...

6. Sep 10, 2017

### Staff: Mentor

Wait... you mean this continued fraction terminates somewhere, after hundreds of terms that are not zero? Or do you mean just a few specific rational multiples?

Edit: Found a clear counterexample. The square is an algebraic number, but not a rational number.

7. Sep 10, 2017

### hilbert2

Oh, then I remembered it wrong. I just had an image of some tables of trigonometric functions in mind and didn't think more about it.

8. Sep 10, 2017

### thebosonbreaker

I meant that whilst it is possible to show the irrationality (there are a lot of different ways - factorisation into primes as you mention being one of them).. is there more of a "reason" for it? It just seemed to me like a heavy coincidence that these non-perfect square roots generate an infinite number of irrational numbers and all seem to follow this rule of irrationality.

Thank you for your insight, but what exactly do you mean when you say that "mathematics never goes further than follows from the axioms" in this context?

9. Sep 10, 2017

### thebosonbreaker

Yes thanks for your help, it seems easily proven but at the same time a large coincidence that these non-perfect square roots follow a certain rule of irrationality, so to speak. I was wondering if there was more of a logical / deeper reasoning behind it.

10. Sep 10, 2017

### thebosonbreaker

Hi, thanks for your help. Yes... irrational numbers continue to astonish me! It's incredible in many of the areas of Maths in which they appear. Usually irrational numbers aren't easy to find by calculation (yes ok - there are of course infinitely many of them, but...) it seems strange to me that using these square roots is such a seamless way of finding them. It's one thing to prove that they do this, but surely this property must have some sort of deeper reasoning.
Once again thanks for your insight into this.

11. Sep 10, 2017

### Staff: Mentor

Yes. $\mathbb{Q}= \operatorname{span}_\mathbb{Z} \left( \mathbb{Z} \cup \{x \,\vert \, \exists \, a \in \mathbb{Z}\, : \,x\cdot a = 1\} \right)$ and this set doesn't contain all solutions to $x^2\cdot a =1\,$. Or short: We made $\mathbb{Q}$ in order to divide, but division isn't enough to take all square roots. It simply isn't strong enough.

12. Sep 13, 2017

### FactChecker

Any number whose factorization has a prime to an odd power will have an irrational square root. As numbers get larger and have a greater number of distinct prime factors, the odds that one will be to an odd power increases. That would imply many more irrational square roots than rational.

13. Sep 17, 2017

### Erland

It can be proved that if $r$ is a rational zero of a polynomial $\sum_{k=0}^n a_kx^k$ with integer coefficients, then $a_nr$ is an integer. In particular, if this polynomial is monic, that is, if $a_n=1$, then $r$ is an integer.
A consequence of this is that if the $n$-th root of an integer is rational, it is an integer.

Last edited: Sep 18, 2017