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Simple, yet annoying, proof

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]a+b = ab = a^b[/tex]
    Prove that [tex]a=b=2[/tex]


    2. Relevant equations



    3. The attempt at a solution
    Ok, I've tried this for more than an hour now, but can't figure out how to do it. I guess it could be done with Lambert's W-function (?) but this task is given on a much lower level. They haven't even learned how to prove that when x is a odd number, then x squared must be odd aswell. So this must be really simple, but still I can't figure anything out.

    I've tried to take the equations and isolate a or b, and then insert it into one of the other functions, but I've gotten nowhere. I really hope someone could help me se the obvious here.
     
  2. jcsd
  3. Jan 7, 2010 #2
    This is part of the proof:
    I'm in a hurry continue later(=

    [tex]a+b\equiv0(moda)[/tex]
    [tex]b\equiv0(moda)[/tex]
    Therefore,a divides b

    [tex]a^{b}\equivab(modb)[/tex]
    [tex]a^{b}\equiv0(modb)[/tex]
    Therefore,b divides a

    Concluding that [tex]a=\pmb[/tex]
     
  4. Jan 7, 2010 #3

    Hurkyl

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    Don't give complete answers. :grumpy: But your approach can't work anyways, because it doesn't say anything about non-integers.


    I presume that you're supposed to have an idea inspired by the qualitative behavior of addition vs. multiplication vs. exponentiation....
     
  5. Jan 8, 2010 #4

    ehild

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    At low level I would mean to substitute the values given for a and b and see if the relations are true :)

    ehild
     
  6. Jan 8, 2010 #5

    tiny-tim

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    Hi Norway! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Here's a start …

    the bit on the left gives you an ab - a - b … does that remind you of anything? :wink:
     
  7. Jan 8, 2010 #6

    HallsofIvy

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    ?? There were no "values given". The problem is to show that his is true for all numbers a and b.
     
  8. Jan 8, 2010 #7
    To be hones - no, it doesn't. :(
    First thing that comes to mind are those (a+b)2 rules, but I guess neither of them fits in here.
    Then I think; "So a and b multiplied minus a and b is zero. That must mean that a+b=ab", and we're back to the start...
    Sorry.
     
  9. Jan 8, 2010 #8

    tiny-tim

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    Try adding a constant to it, and then factoring it. :wink:
     
  10. Jan 8, 2010 #9
    Geez, I still don't get it.
    I'm trying to look at this task like I have no idea what a and b could be, I'm not doing anything wrong there, am I?

    Ok, [tex]ab-a-b=0[/tex]. If we knew that a=b, then OK, but we don't. That's what we're going to prove. Right?

    Or wait;
    [tex]ab - a - b + 1 = 1 \ \Rightarrow \ (a-1)(b-1) = 1[/tex]

    Was this what you had in mind?
    Then what? If they had to be integers, then sure, but they don't. Any more hints would be greatly appreciated. :D
     
  11. Jan 8, 2010 #10

    tiny-tim

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    :biggrin: Woohoo! :biggrin:

    ok, now have a go at the bit on the right … ab = ab

    how can you get an (a - 1) or a (b - 1) out of that? :wink:
     
  12. Jan 8, 2010 #11
    Hi everyone,
    tiny-tim, I tried: ab=a^b
    a-1=a^(b-1)-1 and since b-1=1/(a-1), then (a-1)(a^(1/(a-1))-1)=1...

    but it's not a likeable equation,is it?
    is there any other way?
     
  13. Jan 8, 2010 #12

    tiny-tim

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    Hi penguin007! :smile:

    (please use the X2 tag just above the Reply box :wink:)

    I managed to get as far as ab-2ba-2 = 1 …

    not sure what to do then. :redface:
     
  14. Jan 8, 2010 #13
    Hi Hurkyl!
    You are true i've neglected the non-integers. But if it is non-integers, i do not know how should i deal with my proof.

    Hi!Norway can you specify if a and b are real numbers or solely integers?
     
    Last edited: Jan 8, 2010
  15. Jan 8, 2010 #14
    Hi! The task was given exactly as I gave it to you. I would guess this means all numbers, but maybe it was integers. As I said, this was given on a relatively low level, to students who has never ever done any proofs before.

    Thank you guys.
     
  16. Jan 8, 2010 #15

    ehild

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    Norway, was it said that "Prove that a=b=2" with the same words? Or was it meant like that: prove that a=b=2 satisfies the equations above? A simple substitution of a=2 and b=2 into the equations is a complete proof then, the derivation of the solution is not necessary. If the question means "prove that a=b=2 is the unique solution", that is a nice problem for high level, as this thread illustrates it, but not for beginners in maths.

    ehild
     
  17. Jan 9, 2010 #16
    The expression you got seems to be interesting tiny-tim but I didn't manage to exploit it.

    We can try to introduce the function f:=x->(x-1)*(x(1/(x-1))-1) and then by studying the variations of this function, we would show that the equation (x-1)*(x(1/(x-1))-1) =1 has a single solution (x=2, which would give us a=2); BUT:
    1-I do not guarantee the study of f is easy;
    2-this problem seems to be a more arithmetic one and therefore should be solved differently;

    ...
     
  18. Jan 9, 2010 #17
    Well, it said "Given [tex]a+b=ab=a^b[/tex], prove: [tex]a=b=2[/tex]"

    But yes, I do start to wonder. They can express themselves quite sloppy sometimes, especially on lower levels. I just saw a good try on this on a Norwegian forum of maths (that's why I posted this here, I saw this question, and was unable to help, even though I'm on a much higher course :blushing:).

    [tex]ab = a^b[/tex]

    [tex](ab)^b = \left(a^b\right)^b[/tex]

    [tex]a^b b^b = a^{b \cdot b}[/tex]

    [tex]b^b = a^b[/tex]

    [tex]a = b[/tex]

    [tex]a+b = ab[/tex]

    [tex]2a = a \cdot a[/tex]

    [tex]a = b = 2[/tex]

    Obviously this turned out to be wrong, but I guess it was still a nice try.
    They said they found a solution to this now, so I'm going to check out the link they gave and then check back here.
     
  19. Jan 9, 2010 #18
    Okay, I just reviewed the other solution, and I can't see anything wrong with it. Maybe you guys can?

    [tex]ab = a^b \ \Rightarrow \ b = a^{b-1}[/tex]

    [tex]a+b = ab \ \Rightarrow \ b = 1 + \frac{b}{a} \ \Rightarrow \ b - 1 = \frac{b}{a}[/tex]

    Therefore:

    [tex]b = a^{b-1} = a^{\frac{b}{a}}[/tex]

    [tex]b = a^{\frac{b}{a}} \ \Rightarrow \ b^a = a^b[/tex]

    [tex]a \cdot \log{b} = b \cdot \log{a}[/tex]

    [tex]\frac{\log{a}}{a} = \frac{\log{b}}{b}[/tex]

    [tex]a = b[/tex]
    Is this correct here? Why not?

    [tex]a+b = a^2 \ \Rightarrow \ 2a = a \cdot a \ \Rightarrow \ a = b = 2 \; \; \; Q.E.D.[/tex]
     
  20. Jan 9, 2010 #19
    It's not. Because you believed too much in the guy who said it was strictly increasing. If you actually take a look at the graph, you'll see that a=b could mean a ton (infinite?) of possible (a,b) pairs.
     
  21. Jan 9, 2010 #20

    tiny-tim

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    Hi Norway! :smile:

    That requires that logx/x be single-valued.

    Unfortunately, it has a maximum value of 1/e at x = e (2.718…),

    so loga/a = logb/b does not imply a = b. :redface:

    EDIT: uhh? are you two different people (or countries)? :confused:
     
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