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Prove that [itex]a[/itex] is not an integer

  1. Jul 11, 2012 #1
    Prove that for any non-zero positive integers [itex]b,s[/itex]
    [itex]a = \large \log_x \bigg(\dfrac{ ( -3 + x^{2b} ) \pm \sqrt{(3-x^{2b})^2-4x^{2b}(1-s^2)}}{2x^{2b}}\bigg)\notin \mathbb{Z}.[/itex] ​

    The above expression comes from the result
    [itex]x^a = \dfrac{ ( -3 + x^{2b} ) \pm \sqrt{(3-x^{2b})^2-4x^{2b}(1-s^2)}}{2x^{2b}}.[/itex] ​

    Edit : The [itex]x^a[/itex] present there comes from the root of quadratic equation in terms of [itex]x^a[/itex] which is

    [itex]x^{2b}.x^{2a} +(3-x^{2b}) x^{a} + (1-s^2)=0 .[/itex]​

    There [itex]a \ge 1, b \ge 1, s \equiv 0 \mod 2, x[/itex] is a prime [itex]\gt3[/itex].

    P.S. : I don't know whether [itex]a[/itex] turns to integer or not . Any counter-examples that make [itex]a \in \mathbb{Z}[/itex] are highly appreciated. I have substituted some random values and got the numerator always less than denominator, and there by creating an obstruction for [itex]a[/itex] to be an integer. So can we have some comparisons on the numerator and denominator ? .

    Thank you.
    Last edited: Jul 11, 2012
  2. jcsd
  3. Jul 11, 2012 #2


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    Hey S.Iyengar.

    The first thing I think you should do is to decompose the logarithm with respect with the x terms (since log_x(x^(2b)) = 2b and similar for other terms) and then figure out when you get a non-integer expression from there.

    I'm going to assume that your expression is correct for getting a, and if it is then proving it will involve showing that your final log expressions give a non-integer number and this only happens when you get log_x(f(x)) is an integer.

    Maybe you could do this first and then offer any insight you have had from this.
  4. Jul 11, 2012 #3
    Yes, but given a surd, we can't bring that expression into that form. I hope if you see my latest edit you will understand what I mean
  5. Jul 11, 2012 #4


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    But you can at least pull some terms out and then see what the minimal condition is for the expression to be an integer. I didn't mean to imply that you can just get a clean answer, but instead an implicit constraint that needs to hold for the integer condition to hold.

    Also I'm wondering, are there any simplifications you can do for the +- part as opposed to just having it in that form (i.e. expanding explicitly the + and the - to get simplifications)?
  6. Jul 11, 2012 #5
    Yes may be you are right .

    I did something else to make it appear as follows . Log (a/b) = log a - log b. It makes a term [itex]Log_x(2)[/itex] in the expression ( that arises when we expand [itex] \log_x(2x^{2b})[/itex]. So that [itex]Log_x(2)[/itex] may contribute to some decimal part . How about this ?.
  7. Jul 11, 2012 #6


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    Yes that is the kind of thing I am getting at, but you will have to consider all parts of the relationship.

    In the case of an integer, you should always be able to cancel these kinds of terms and be left with a variable that you know has an integer expression. If you end up in a situation where you can not simplify or cancel things like log_x(2) or log_x(s) or something like that, then you have proven the non-integer condition.

    Of course the modulus conditions will give you more ways to simplify the problem, but the idea still stands.
  8. Jul 11, 2012 #7
    Dear sir,

    But that may not seem to work perfectly. If the term also gives some decimals and make it cancel both the sides then it would become an integer.
  9. Jul 11, 2012 #8


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    If you want to use this kind of case, you will need to specify exactly when this happens and this will tell you when you get this case, and if this case is not for all values of a,b,s,etc then you still have proven that at least some cases exists where you get a non-integer value and you're done (remember you only have to show one example of it not being true).

    So for example if you end up with log_x(2) + log_x(4s) = a (just an example, not the real answer) then it means that log_x(8s) = a which means that 8s = x^a. If we let say s > 0 then there are many ways of getting this, but again we can have lots of values that contradict a being an integer.
  10. Jul 12, 2012 #9
    Did you try anything better ?
  11. Jul 12, 2012 #10


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    No I haven't since this is not my problem. I'm merely trying to throw any ideas at you that may or may not help you solve your problem. I don't want to solve it for you, but certainly if I can think of a way that aids you in doing so, I will definitely mention that idea.
  12. Jul 13, 2012 #11


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  13. Jul 13, 2012 #12
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