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I Prove if x + (1/x) = 1 then x^7 + (1/x^7) = 1.

  1. Mar 9, 2016 #1
    I've just bought "Inside Interesting Integrals" by Paul Nahin and I've come to an impasse after the first paragraph in the preface. He posed this problem as a test to see whether or not the reader should continue on or put down the book. Basically, if the reader was interested and already trying to figure it out, then the reader should continue. I fit that category, and tried a proof. However I feel like I'm cheating. So would you please tell me if this is a legal proof? Any insights are welcomed. If there is an entirely different way to do this please show, but I'm partial to something similar to this.


    Without actually solving for x, prove the following: if x + 1/x = 1, then x7 + 1/x7 = 1.

    Assume that x7 + 1/x7 = 0.
    Then:
    x7 = -1/x7
    x7x7 = -1
    x14 = -1.

    Therefore x is an imaginary number.

    Now by hypothesis we know that x + x-1 = 1.

    Since x is imaginary, x-1 is also imaginary.

    By the closure property of the set of imaginary numbers, x + x-1 must also be imaginary.

    However, the hypothesis states that x + x-1 = 1, which is not imaginary. Thus by contradiction, if x + 1/x = 1, then x7 + 1/x7 = 1.​


    Does this work at all? Not only am I not sure, I feel like I cheated by just stating that x7 + 1/x7 = 0. What I was attempting to do was assume ~Q and then derive a contradiction using P, using the basic proof by contradiction technique. Saying it is equal to 0 is more specific than saying it is not equal to 1, but either way it's still stating it isn't equal to 1. Is this even a viable way to approach this proof?


    Thanks!



    EDIT- I should point out that the book actually does a direct proof using just a bit of recursive algebraic manipulation. I just want to see if this way is legal.
     
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  3. Mar 9, 2016 #2

    pwsnafu

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    Why is this true?
     
  4. Mar 9, 2016 #3

    blue_leaf77

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    Note that
    $$
    (x+1/x)^7 = (x^7+1/x^7)+7(x^5+1/x^5)+21(x^3+1/x^3)+35(x+1/x)
    $$
    In order to calculate ##(x^7+1/x^7)## in the right hand side, you need to know ##(x^3+1/x^3)## and ##(x^5+1/x^5)##. Start from ##(x^3+1/x^3)## first, how can you write it in terms of power(s) of ##(x+1/x)##?
     
  5. Mar 9, 2016 #4

    Samy_A

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    Two remarks about this part:
    x14 = -1 doesn't impy that x is imaginary: see on Wolfram Alpha.
    x is a complex number with non zero imaginary part, but the set of complex numbers with imaginary part is not closed under addition.
     
  6. Mar 9, 2016 #5

    Ssnow

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    If you want to start a ''contradiction proof'' you must assume that ## x^{7}+\frac{1}{x^{7}}\not=1## and not that ## x^{7}+\frac{1}{x^{7}}=0 ##, I don't understand this ... (and also other things...)
     
  7. Mar 10, 2016 #6
    It's not. In making this assumption I am assuming the conclusion of my original conditional statement is false so that I can show an contradiction arises.

    I was wondering if that is allowed. Normally with an "if P then Q" logical statement you can prove it is true by assuming Q is false and then deriving a contradiction, which is what I was trying to do. By saying it is equal to 0, that automatically means that I am contradicting that it equals 1. That is, setting it equal to zero was my "not Q" starting point.



    Well if x^14=-1 has any roots that have real parts then this entire attempt is worthless (and that picture makes it look like each point has both real and imaginary parts).

    But just for giggles, am I wrong in thinking that the SUBSET of the complex numbers that includes only the imaginary numbers is closed under addition? That is, if a and b are real numbers, then the sum ai + bi is imaginary? It would be (a+b)i, or if a+b=c just ci, which is also imaginary.

    Clearly I should invest in learning about complex numbers beyond the little bit I went over in elementary physics classes, though. I just kind of assumed it was imaginary since it's an even exponent and a negative result.
     
  8. Mar 10, 2016 #7
    Yes that's what I was wondering about. I figured since setting it equal to zero is also simultaneously claiming it is not equal to 1 it might somehow be allowed, but I had a suspicion that was cheating.
     
  9. Mar 10, 2016 #8

    Samy_A

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    Yes, the subset of imaginary numbers is closed under addition.
    One caveat: you have to consider 0 as an imaginary number. That's just a matter of definition, and most people would accept that. (There was a thread about this question awhile ago: here.)
     
  10. Mar 10, 2016 #9
    The way Paul Nahin did it in the book was he first multiplied by x to get x^2+1=x and went from there, doing some recursive substitution. I'll look at it from the perspective you have tomorrow morning. Thanks to all.
     
  11. Mar 10, 2016 #10
    Good. I thought I was crazy at first then I realized I misread what you said. But as I said if the root in question has real parts then the entire approach is useless. I am disappointed though. I thought it would be cool to use imaginary numbers (of course I know pretty much nothing about them).
     
  12. Mar 10, 2016 #11
    Okay what about this?
    As Samy_A pointed out, I was mistaken about x14=-1 implying that x is imaginary, however I still want to do a proof by contradiction. I think this works a lot better. What do you think?

    Once again, assume that the consequent is false in hopes of arriving at a contradiction:

    1. Assume that x7 + x-7≠1 (instead of assuming it's equal to 0 as that is an invalid approach).

    2. By our hypothesis, x + x-1 = 1.
    Multiplying by x gives
    x2 + 1 = x​
    and rearranging gives
    x2 = x -1.​
    Multiply by x again to get
    x3 = x2 -x​
    And now substituting x2 = x -1 into the above gives:
    x3 = -1.​
    3. Note that x7 + x-7≠1 is the same as x3x3x + 1/(x3x3x) ≠1.
    Substitute x3 = -1 to get:
    (-1)(-1)x + 1/((-1)(-1)x) ≠1,​
    which simplifies to
    x + x-1≠1.​
    4. But the original hypothesis was that x + x-1 = 1, and so we have arrived at a contradiction. Therefore if x + x-1 = 1, x7 + x-7=1.​


    I feel like that one is valid. Step two was taken in part from the suggestions above and from the direct proof given by Nahin. I think doing it by contradiction saves a little bit of work. Does this method seem sound to you all? Thanks again for your time.
     
  13. Mar 10, 2016 #12

    Samy_A

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    Very nice proof!

    One comment: you don't have to prove this by contradiction.
    You proved that x + x-1 = 1 implies x3 = -1.
    And that implies, as you noted, that x7 + x-7=x3x3x + 1/(x3x3x) = x + x-1 = 1.
     
  14. Mar 31, 2016 #13

    Zafa Pi

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    Notice that x = e^ipi/3. Now everything is easy.
     
  15. Mar 31, 2016 #14

    mfb

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    That violates the problem statement.
     
  16. Mar 31, 2016 #15
    Let me try on this one.
    Let P1be a predicate of your false assumption that
    P1: x7 + 1/x7 = 0 true
    ...your proof...
    and C1 is your conclusion which is true from your proof
    C1: x14 = -1. Therefore x is an imaginary number.

    Now let P0 be the "hypothesis" which is true by default.
    P0: x + x-1 = 1 true
    and you arrive at the another conclusion:
    C0: x-1 is also imaginary

    All in all, P1 -> C1 -> P0 -> C0 is a logical process but
    C0 becomes contradictory to P0 because your P1 is a false assumption made from the start.
     
  17. Oct 5, 2016 #16
    just found myself doing this yesterday and stumbled on this post.
    i did it like this
    start with x + 1/x = 1
    subtract 2/x from both sides & get: x - 1/x = 1 - 2/x
    multiply the equations together: x^2 - 1/x^2 = 1 - 2/x
    mult thru by x: x^3 - 1/x = x - 2
    rearrange: x^3 = x + 1/x -2 = 1 - 2 = -1
    so x^6 = 1
    so can shove a factor of x^6 anywhere, viz: x^7 + 1/x^7 = 1
    (BTW it's the complex cube roots of -1)
    no proof by contradiction & not sure if i breached the condition of not actually solving for x.
     
  18. Oct 5, 2016 #17

    Zafa Pi

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    Don't worry you didn't solve for x, since not all the cube roots of -1 are solutions for x.
     
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