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Given a+b=ab=a^b, probe a=b=2

  1. Apr 24, 2003 #1
    Please Help! given a+b=ab=a^b, probe a=b=2

    Please help me on this:

    Given a+b=ab=a^b
    prove a=b=2

    it seems like a very simple task but I have lost my sleep for the past
    couple of nights over it.

    please help or ill be
  2. jcsd
  3. Apr 24, 2003 #2
    Sorry, you realised it before I did.
  4. Apr 24, 2003 #3
    Okey doke, the (hopefully) non-erroneous method.

    a+b = ab => b = ab-a = a(b-1)

    ab = ab => b = ab-1

    so, b = a(b-1) = ab-1

    dividing by a;

    b-1 = 1b-2 = 1 => b = 2

    a+b = ab => a+2 = 2a => a=b=2
  5. Apr 24, 2003 #4


    User Avatar

    Good until here....
    Not good the first passage... it would be
    b-1 = ab-2
  6. May 24, 2003 #5
    This may well be a bit late, but I was bored and browsing through some old posts.
    Using a+b = ab, we get a = ab-b = b(a-1)
    So, ab = b(a-1)b
    But, b*b(a-1)b = ab, so, a = b*(a-1)b
    Now, b*(a-1)b = b(a-1), so (a-1)b = (a-1)
    This is true if a=2, or b=1.
    Substituting b=1 into ab = a+b, a*1 = a+1, which leads to 0=1, a contradiction, thus b is not 1.
    Substituting a=2 into ab = a+b, 2b = b+2, so b=2.
    Substituting these two values into ab= ab = a+b
    22 = 2*2 = 2+2, as required. Thus, a=b=2.
  7. May 24, 2003 #6

    Not quite sure I follow here.
    If a = b(a-1), then it follows that ab = ( b(a-1) )b = bb(a-1)b
  8. May 24, 2003 #7
    Quite right. Well spotted.
  9. May 24, 2003 #8
    I figured where I went wrong. It was my dodgy handwriting...

    So, (a-1) = bb-2(a-1)b
    Then, 1 = bb-2(a-1)b-1
    This is true if bb-2 is 1, and (a-1)b-1 is 1.
    bb-2 = 1 if b-2=0, so b=2. (a-1)=1, so a=2.
  10. May 24, 2003 #9
    I don't disagree, but that doesn't really prove that there are no other numbers a and b for which bb-2 and (a-1)b-1 are inverses.


    I think the problem essentially boils down to prove no real solutions other than a=2 for:

    Last edited by a moderator: May 24, 2003
  11. May 25, 2003 #10
    How about this?

    From (1):ab=ab we get (b)*ln(a)=ln(a)+ln(b) and
    From (2):a+b=ab we get a=b/(b-1) so ln(a)=ln(b/(b-1))

    subsitute (2) into (1) to get:

    (b)*ln[(b/b-1)]=ln(b/(b-1))+ln(b) so

    ln(b/(b-1))b=ln(b2/(b-1)) which means that

    bb=b2 therefore,

    b=2 and from a=b/(b-1) we get a=2/(2-1)=2
  12. May 25, 2003 #11
    I think it's good up until here. The b should be in the brackets to make


    This problem is harder than it first seems

    There are other inverses to my equation after all...take b=1, then a could be any real number, except 1...oh well. I think Thoth's onto something though.
    Last edited: May 25, 2003
  13. May 25, 2003 #12
    Re: PLEASE HELP!! given a+b=ab=a^b, probe a=b=2

    I. a+b=ab
    I. b=ab-a
    I. b=a(b-1)
    I. a=b/(b-1)

    II. ab=a^b
    I into II. b^2/(b-1)=(b/(b-1))^b

    2ln b - ln(b-1) = b ln b - b ln(b-1)
    (2-b)ln b = (1-b)ln(b-1)
    (2-b)/(1-b) = ln(b-1)/ln(b)

    Use definition of ln...

    (2-b)/(1-b) = lim(n->[oo]) n((b-1)^(1/n)-1) / n(b^(1/n)-1)
    n cancels ...
    (2-b)/(1-b) = lim(n->[oo]) ((b-1)^(1/n)-1) / (b^(1/n)-1)
    Using Bernoulli...
    (2-b)/(1-b) = lim(n->[oo]) (1 + b/n -1/n -1) / (1 + b/n -1)
    (2-b)/(1-b) = lim(n->[oo]) (b-1)/b
    (2-b)/(1-b) = (b-1)/b
    2b-b^2 = 2b-b^2-1
    No solution.
    Wow. Where's the flaw? I know the RHS must be zero.
    Hopital, maybe?
  14. May 25, 2003 #13
    I think your flaw may be in making the transition above, I don't know Bernoulli's rule or how you changed the ln's into a limit, but the limits don't match up:

    lim(n->inf) [ ((b-1)^(1/n)-1) / (b^(1/n)-1) ] = ln(b-1)/ln(b)


    lim(n->inf) [ (1 + b/n -1/n -1) / (1 + b/n -1) ] = (b-1)/b

    I would be a little hesitant to use ln's anyway because you don't know whether the quantities are negative. But it looks like a new avenue.
  15. May 26, 2003 #14
    Lonewolf, you are probably right and I might need to change my classes . But as far as I can tell , the result of both ln(b/(b-1))b and what you wrote comes out the same, which is: ln(bb/(b-1)b))=ln(b2/(b-1). However I give your opinion the benefit of the doubt since typing mathematics does not come easy for me .

    A major factor is to look at a=b/(b-1) and ln(a)=ln(b/(b-1)) and ask what values for b make sense for a?

    Here a=b/(b-1) first we notice that b cannot be =1 because of the singularity at b=1.
    After noticing this we then have to ask how about if b>1 and b<1 what would happen then. As b increases in value how a is being effected. We also need to know what values for both a and b satisfies the given conditions in the problem.

    In ln(bb/(b-1)b))=ln(b2/(b-1) means this: bb/(b-1)b=b2/(b-1). (By using eln definition). If bb=b2 then b=2 and if (b-1)b=(b-1)1 then b=1. however, we knew already that b cannot be 1.
    Last edited by a moderator: May 26, 2003
  16. May 26, 2003 #15
    Unfortunantly, for that last equation to be true, bb does not need to equal b2 and (b-1)b need not equal (b-1). As long as the ratio between the pairs is satisfied, the equation can be satisfied and so we haven't yet proved no other real numbers can satisfy the last equation.
  17. May 28, 2003 #16
    My teacher gave me an elegant solution

    This question has troubled me for some time, finally I gave it to my math teacher and he gave me an elegant solution today.

    b-1 = b/a
    a log b = b log a
    log a / a = log b / b

    a=b (as the function f(x)=logx/x is strictly increasing)
    =>2a = a2
    a=b= 2 or 0

    I think the proof would be more elegant if we can do it using pure algebra.
  18. May 28, 2003 #17


    User Avatar
    Staff Emeritus
    Gold Member

    Re: My teacher gave me an elegant solution

    It is not!

    f(1/e) = log(1/e)/e = -log(e)/e = -1/e
    f(e) = log e / e = 1/e
    f(e^2) = log(e^2)/(e^2) = 2/e^2 =(2/e)*(1/e) < 1/e

    i.e., it goes up and then back down.
  19. May 28, 2003 #18
    hey guys, i'm quite sure i solved the problem, but since i'm busy i'll try to post a formal proof by the end of the day.
  20. May 28, 2003 #19
    First note that neither a nor b can be equal to zero or one:

    from "a+b=ab"
    0 + b = 0*b -> b=0 -> ab = 00 (an undefined quantity)
    a + 0 = a*0 -> a=0 -> ab = 00
    1 + b = 1*b -> contradiction
    a + 1 = a*1 -> contradiction

    By establishing that a,b != {0,1}, we are now free to divide by {a, b, a-1, b-1}

    a*b = a + b
    a*b - a = b
    a*(b-1) = b
    a = b/(b-1)

    Let's pause for a moment and further realize that b cannot be less than zero:

    if (b<0) -> [a = -|b|/(-|b|-1)] -> (a>0)
    but this means a*b is negative and ab is postive, which is a contradiction since they must equal each other.

    ab = a*b
    ab-1 = b
    ( b/(b-1) )b-1 = b (substitution from above)
    bb-2 = (b-1)b-1
    |bb-2| = |(b-1)b-1|
    |b|b-2 = |b-1|b-1
    ln|b|b-2 = ln|b-1|b-1
    (b-2)ln|b| - (b-1)ln|b-1| = 0

    We now set f(x) = (x-2)ln|x| - (x-2)ln|x-1|, x!={0,1}. The zeros of f(x) are the only possible solutions for b as f(b) must equal zero.

    Finding these zeros is a difficult task and requires analyzing the derivatives.

    f'(x) = ln|x/(x-1)| - 2/x, x!={0,1}
    f''(x) = (x-2)/( x2(x-1) ), x!={0,1}

    Since f''(x) is the only function we really recognize, we us it to find out what f'(x) looks like and then we use that to find out what f(x) looks like.

    Code (Text):

                +        +          -        +
    f''(x) <---------|---------|---------|--------->
                     0(und)    1(und)    2
    Lim[ x-> +/-infinity, f'(x) ] = 0
    f'(0.1) < 0 and f'(0.95) > 0 (just plug it in)
    f'(1.1) > 0 and f'(1.90) < 0
    This means that f'(x) must have only two zero which exist on (0,1) and (1,2) (because f' is continous within the intervals inbetween) which we will label c1 and c2 respectively. This in turn implies that the sign chart of f'(x) looks as follows:

    Code (Text):

               +         -         +         +         -    
    f'(x) <---------|---------|---------|---------|--------->
                    0(und)    c1        1(und)    c2(c2<2)
    Limit[ x->1, f(x) ] = 0 (though not defined at 1)
    f(-4) < 0 and f(-3) > 0
    f(0.1) > 0 and f(0.9) < 0
    f(1.1) > 0 and f(2.1) < 0
    Finally, we use the last set of information to note that there are precisely three zeros of f(x). Since limit of f(x) as x approaches 1 is zero and f(x) is increasing on c1 to c2 there can't be a zero within this interval. However the other three intervals do have zeros because f(x) is continous on them and changes sign: (-inf, 0), (0, c1), (c2, +inf).
    We can label these zeros as b1, b2, and b3 respectively.

    b1, b2, and b3 are the only possible solutions for b because f(b) must equal zero as I stated above. But also remember that b > 0 so that b1, which is in (-inf, 0), is ruled out.

    If we assume b2 is correct than a is negative:

    0<b2<c1<1 -> a = |b|/(|b|-1) = pos/neg = neg.

    But this would mean that ab has a negative base. This leads to a complex number for ab which certainly cannot equal a+b. (Although if you simply take the neg base as meaning -|a|b, then this actually would look like a solution as i tried plugging in numbers and it comes extremely close).

    Anyways, now we have eliminated all but one possible solution, b3, which exists in the interval [c2, +inf) where c2 is less than 2. By inspection we know f(2) = 0 and therefore b3 must equal 2.

    We now try setting b=2 and seeing if this really is a solution:
    a = 2/(2-1) = 2
    2 +2 = 22 = 2*2 = 4

    I know this has to be the ugliest solution you would ever want to see, and i've omitted some of the stuff to keep this from being any longer... but i think this really proves that (a=2,b=2) is the only solution.

    Actually, like i said, b2 might work. Switch your calculator out of complex mode and into real mode and try it out:
    (a=-3.14104155643, b=.75851486)
  21. May 30, 2003 #20
    a, b integers ?

    If both a and b are integers, then a must equal to b.

    by the fundamental theorem of arithmetic, a = b

    On one hand, y=ab where y is an integer and it can be uniquely factorized. (a is a prime number)

    On the other hand, y = ba where b is a prime number.

    We can prove my contradition that a and b are even numbers.

    Therefore a=b=2

    I'm sorry to cause confusion here. Yes, f(x) is strictly increasing when e>x>0 and it is strictly decreasing when x>e. It seems that we have infinitely many solutions if a and b aren't integers.
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