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Number Theory - Proving x is not a perfect square?

  1. Mar 4, 2014 #1
    Hey Guys,

    This is my first post to Physics Forums. If I posted this question in the wrong area or am violating some other etiquette, please let me know!

    I'm working on a proof and am currently stuck. I'm trying to prove that x (a weird number, an infinite product to be specific), is NOT a perfect square. At first I thought I could say that if x converges to a value between two perfect squares, it then can't be a perfect square. Then I realized that 1, 4, 9, 16, 25,... are not the only perfect squares. There are also 0.16 (0.4^2), 6.25 (2.5^2), etc. How could I go about showing that x is a perfect square? or that sqrt(x) is irrational?

    I'd really appreciate some guidance on this!

    Thanks so much!
  2. jcsd
  3. Mar 4, 2014 #2
    Usually perfect square refers to integers. That shouldn't be too hard (well, maybe it is... I don't really know what the problem is). If you want to show that sqrt(x) is irrational you might be able to follow the method used for sqrt(2), but, again, the specifics matter.
  4. Mar 4, 2014 #3
    I tried using the method for sqrt(2), but ran into problems because radicand is not an integer. Any ideas on proving irrationality for the square root of a non integer?
  5. Mar 5, 2014 #4
    Apparently this is the product of an infinite series of rational numbers.

    Is there a way of counting the number of any particular prime value in either the numerator or denominator?
    For example, can you show that no matter how many terms are evaluated, the total number of 2's in the numerator and denominator will always be odd? If so, then the square root will be irrational.

    Bear in mind that may not be a firm yes or no answer. Take this product series, for example:
    (2/1) (2/4) (2/9) (2/16) (2/25) (2/36) (2/49) ...

    The square root of that value is both rational and not - or perhaps always undefined.

    On the other hand:
    would always be rational - assuming it is even defined.
  6. Mar 5, 2014 #5

    Thanks for the reply, which will surely be useful somepoint in the future. For this specific problem, though, it is quite the opposite actually. Every number is *irrational*. The product contains a bunch of nth roots of 2 and a bunch of nested radicals. How could I go about proving irrationality for an infinite product of irrational numbers?

  7. Mar 5, 2014 #6
    Since every number is irrational, that says a lot.
    We have to assume that the series approaches 1, otherwise the product will not approach a set number and will be undefined.

    But even if it approaches 1, we still have a problem because, being irrational, it can never reach 1.
    Let's call the ratio of two perfect squares that is not a perfect square a "good square". A good square is what we are looking for and good squares are always rational.

    Let's say that up to term "n" we have a good square. Then we know that on term n+1 it will not be a good square. This suggests that the product of an infinite series of irrationals can never be rational - but I'm not actually convinced of this. I think that I could write such a series if I took the time.

    The key thing to do is to keep tally of the terms in the series to see if the irrationals combine to form rationals.
    For example, let's say every element of the series is a rational number to a power. Then perhaps you can sum all the exponents of any given prime in the numerator and subtract all the exponents of that same prime that appear in denominators. If the result is not an even number, then you aren't going to end up with a good square. Of course, the same can be done with any prime.
  8. Mar 6, 2014 #7
    The problem is that there are products that are always irrational, but the limit can still be rational

    Consider the function

    [tex] f(x) = \prod_{n=1}^\infty 1+e^{-nx}[/tex]

    is defined for x>0
    lim x->0 f(x) = infinite
    limit x->inf f(x) = 1,
    it is strictly decreasing
    it can take on all values >1
  9. Mar 6, 2014 #8
    That's incredibly helpful! Thank you so much everyone!
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