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Discrete math problems

  1. Feb 22, 2012 #1
    The book works out the case with x and y irrational and xy rational. They used the nonconstructive existence proof method with x = sqrt(2) and y = sqrt(2). If that's rational, then you're finished. If it's irrational, then you can simply raise it to the power of sqrt(2) to get 2. I'm not sure how to adapt this approach to this problem. If it is irrational, then you're finished. If it's rational, then I'm not sure how to show you manipulate it to be irrational.

    I know this is an easy problem, but I'm stumped for some reason. I've tried to find a general expression for c. The uniqueness method is appropriate here. I first need to show that an integer c with that property does exist. I then need to show that it's unique.
  2. jcsd
  3. Feb 22, 2012 #2


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    Do you know anything about transcendental numbers?

    Try to think geometrically, i.e. draw a and b on a line. Where does c lie?
  4. Feb 22, 2012 #3
    I'm rusty on transcendental numbers.

    I didn't think to consider the problem geometrically. Good idea. Of course, c is the value that gives the intersection of the two "functions."
  5. Feb 22, 2012 #4


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    Or more geometrically, c is the integer right smack in between the two odd integers a and b.
  6. Feb 22, 2012 #5
    Last night I came up with a few distance-related equations. One was [abs(a) + abs(b)]/2.
  7. Feb 22, 2012 #6


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    Okay, that's fine. If you were comfortable with transcendental numbers, then there's a conceptual way of seeing how x^y could be irrational. (Here is the sketch: if both x and y were rational, then x^y would be algebraic. So if x is rational and x^y is transcendental, then y would necessarily be irrational. With these observations you can produce tons of examples of rational x and irrational y such that x^y is irrational - indeed transcendental.)

    Anyway, here's another approach. For simplicity let's suppose that x is an integer, and let's look at x^y. Choose your favorite irrational number z. If x^y=z, what can you say about y?
  8. Feb 22, 2012 #7
    I just made myself familiar with the terms algebraic and transcendental numbers. All transcendental numbers are irrational, but not all irrational numbers are transcendental. The square root of 2 is an example of an irrational algebraic number. I'll give an example. It's probably a bit nasty. Haha.

    x^y = e
    2^y = e

    y = log2(e) which is clearly irrational.

    For the second part, if z is irrational and x is rational, then y could be rational or irrational.
    Last edited: Feb 22, 2012
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