Seems like all square roots of (positive) integers are either integers or irrational numbers. Can this be proven or disproven? Or: is it possible to prove that there is/can be or is not/cannot be a positive integer whose square root is a non-integer number that is rational?
Here is a go at it.
Take your rational root. Being a rational it must be of the form n/m where n and m are natural numbers. Further, lets take n/m as an irreducible rational expression - that is n and m have no factors in common.
The square of n/m is n[sup]2[/sup]/m[sup]2[/sup]. The unique prime factorisation theorem tell us that the factors of n[sup]2[/sup] can only be the factors of n repeated. The same for m[sup]2[/sup].
If the square is an integer (call it k), then n[sup]2[/sup] = m[sup]2[/sup]*k, or in other words, n[sup]2[/sup] is factored by m[sup]2[/sup]. However n and m have no common factors (by definition), and n[sup]2[/sup] and m[sup]2[/sup] have the same factors as n and m, thus they two also have none in common, thus m[sup]2[/sup] cannot factor n[sup]2[/sup], except for the degenerate case where the only common factor is one (and thus m is one.) So the only case is where the square root is a rational (n/m) when m is one, which means the square root is of the form n/1 and is thus an integer.
Why can’t n be 1?
Then the square would be of the form 1/m, and thus the only integer that satisfies this would be 1. But indeed, that is an additional corner case.
It’s not really an additional case; that’s just part of the m = 1 case. Your proof’s reasoning was perfectly fine and exhaustive.
(Though I would reword it a little: I would replace talk of no common factors with talk of being coprime, so there’s none of this “Does 1 count as a a factor? Does it not?” back-and-forth ambiguity. The last bit is noting that m is coprime to n[sup]2[/sup], and thus to every factor of n[sup]2[/sup], including m itself, but the only positive integer coprime to itself is 1.
Put more plain-spokenly, the proof is this: when you square a fraction in lowest terms by squaring its numerator and denominator, the result is automatically in lowest terms as well. So, if the result is an integer, its denominator is 1, which means the original fraction’s denominator must have been 1 (or -1) as well. Thus, every rational square root of an integer is itself an integer.)
Yeah, I think I was trying to include the “automatic” bit in the coprime factors. I felt it needed an explicit point that nn is coprime with mm due to unique prime factorisation - that there isn’t some way that m*m might gain a factor in common with n. Then again, I have a soft spot for that theorem, a couple of lovely summer’s afternoons in a lecture theatre when life was good, doing the proof. (Which I could not reproduce to save my life.) It was a very long time ago.
Right, right. You were certainly right to do so; the plain-spoken wording I gave left that part of the reasoning (essentially, gcd(n[sup]2[/sup], m[sup]2[/sup]) = gcd(n, m)[sup]2[/sup]) only implicit.
Very nice, thank you.
Let me take this discussion as a take-off point for my mini-rant about “New Math” or anything related to that. A common complaint is that math teaching is too abstract, too early. (A while back I cited the definition of “additive inverse” as I learned in Algebra I. The problem was, if you didn’t already grok negative numbers, it would not have been obvious what this “additive inverse” was and what it was all about.) And there was no “plain-spoken” explanation given along with it. That’s where I see the problem, and the obvious solution.
That’s where it relates to this thread. There’s the technical formal proof (not too arcane in this case, but there are other cases that are worse, and especially for beginning math students). And then there’s the “plain-spoken” informal proof, that makes clear what the hell the formal proof is getting at. A good pedagogical presentation would include both. Too often (in math classes I’ve sat through, and in textbooks I’ve read) that isn’t done.
Example: In that .999=1 thread, I asked for an epsilon-delta proof. I got one that was an informal “plain-spoken” proof, which I liked, but I was actually looking for the formal proof that day. So I asked again and got a more formal proof. I liked even better seeing both.
When I want to derive a proof of something myself, I always start with a “plain-spoken” or “intuitive” approach. When I get that, then I try to find the formal way to put it. I’m sure that’s what just about any good mathematician does. But I’ve never seen that taught as an actual technique.
If you want a plain-spoken explanation, you could do a lot worse than to look at BetterExplained and see if it has a write-up.
For example, here’s the BetterExplained guide to linear algebra.
This is why I enjoy reading Indistinguishable’s posts so much. The formality is there and nuanced carefully (to the point that he usually edits his posts two or three times or appends them.) But he also steps back and gives a big picture overview that is usually straightforward to grasp and puts things in perspective.
So, thanks.
Maybe this will help. Every prime will divide a square an even number of times. E.g. 144 is divisible by 2 four times, by 3 twice and by all other primes zero times. Moreover a number with that property is a square. To find the square just halve the number of times each prime divides it. In the case of 144, half of four is two and half of two is one, so the square root is 2^2*3.
Now if (n/m)^2 = k, then n^2 = m^2k and any prime divides the left hand side an even number of times and therefore also divides the right hand side an even number of times. Since it also divides m^2 an even number of times, it also divides k an even number of times, so k has to be the square of an integer.
This argument does not differ materially from the ones above, but is perhaps just a bit easier to understand. All arguments for this depend on unique prime factorization and it can fail in non-UFD rings.
Sorry, Francis Vaughan provided a clear, concise demonstration of a mathematical concept in mathematical language, what you have done here is obscure the proof in superfluous verbiage. Reading it seemed to me a bit like traveling from Baltimore to Washington DC by way of Pittsburgh.
It’s just different strokes for different folks, For You. For you, Francis’s explanation was the clearest, but others might find Hari’s approach clearer, or at least additionally useful, to them (for example, the first time I came across that presentation of the argument, I felt it helped me understand things best; it’s still how I think of it in my own head). No harm in a multitude of perspectives.
Is the same true of Gaussian Integers? Is the square root of a Gaussian integer always either itself a Gaussian integer or an “irrational complex number”? (My terminology is failing me here. What is the technical term for complex numbers that cannot be expressed and m/n +j/k*i (m,n,j,k all integers)?)
My intuition suggests that this is true but I am expecting that any proof I attempt will come unstuck because of the non-unique factorisation of Gaussian integers.
I intend to give it a go. But does anyone know the actual answer before I do so?
What do you mean by “the non-unique factorisation of Gaussian integers”? Gaussian integers are a unique factorization domain (indeed, a principal ideal domain!); it’s just that the Gaussian primes aren’t quite the same as the ordinary primes.
But, yes, any Gaussian rational square root of a Gaussian integer is itself a Gaussian integer, for the exact same reasons.
Ok then. I was under the impression that Gaussian integers did not necessarily have unique factorisation.
I was probably misremembering something – Probably a demonstration that prime naturals are not necessarily Gaussian primes but do in some cases factorise.
I have started work on a proof of sorts but see a hole in it. I will have to come back to it later.
J.
You do have to be a little bit careful about defining “unique factorization” in any domain with multiple unit elements. For example, even just allowing negative real integers, you could argue that 10 can be factored as 25 or (-2)(-5), since you have two unit elements, 1 and -1. The Gaussian integers have the same issue, except now you also have i and -i as units.
Yeah, that’s probably it. For example, 2 = (1 + i)(1 - i)
Good luck!
Good point! What we really mean is “unique factorization of numbers into prime numbers, where numbers are considered equivalent if both are multiples of each other, and zero doesn’t count as a number”.