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Math Help - b^(x+y)=b^x*b^y

  1. #1
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    b^(x+y)=b^x*b^y

    let b>1. My textbook has asked me to do a prelimanary proof which I have answered

    Define B(x) to be the set with elements b^t where t E Q and t<_ x, where x is real.

    Show Sup(B(r))=b^r where r E Q

    r>_t so b^r >_ b^t

    Since Q is dense in R, for all y < r, there is a c E Q such that y<c<r, implying b^y<b^c<b^r so b^y is not an upperbound. Note all reals > 1 can be expressed b^y for some real y.

    Hence b^x=Sup(B(x)) for real x.

    So now to the proof in the title. b^x*b^y =Sup(B(x))Sup(B(y)) for real x,y

    B(x+y)={b^t:t E Q, t<x+y}. I see I have to show Sup(B(x))Sup(B(y))=Sup(B(x+y))
    but can't see how to proceed. Thanks
    Last edited by Duke; July 6th 2011 at 01:10 AM.
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  2. #2
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    Re: b^(x+y)=b^x*b^y

    Quote Originally Posted by Duke View Post
    let b>1. My textbook has asked me to do a prelimanary proof which I have answered

    Define B(x) to be the set with elements b^t where t E Q and t<_ x, where x is real.

    Show Sup(B(r))=b^r where r E Q

    r>_t so b^r >_ b^t

    Since Q is dense in R, for all y < r, there is a c E Q such that y<c<r, implying b^y<b^c<b^r so b^y is not an upperbound. Note all reals > 1 can be expressed b^y for some real y.

    Hence b^x=Sup(B(x)) for real x.

    So now to the proof in the title. b^x*b^y =Sup(B(x))Sup(B(y)) for real x,y

    B(x+y)={b^t:t E Q, t<x+y}. I see I have to show Sup(B(x))Sup(B(y))=Sup(B(x+y))
    but can't see how to proceed. Thanks
    The idea here is that the textbook defines b^x (for a real number x) to be b^x \overset{\text{d{e}f}}= \sup(B(x)). The point of the preliminary proof is to check that this coincides with the existing definition of b^x in the case where x is rational.

    So now you have two real numbers x and y, and you want to show that \sup(B(x+y)) = \sup(B(x))\sup(B(y)). Notice first that if r, s are rational numbers with r<x and s<y then r+s<x+y. You also know that b^rb^s = b^{r+s} \leqslant \sup(B(x+y)) (because the index laws are known to hold for rational powers). Taking the sup over all such r and s, you deduce that \sup(B(x))\sup(B(y)) \leqslant B(x+y).

    To get the reverse inequality, you need to show that if t is any rational number less than x+y, then it is possible to express t in the form t=r+s, with r<x and s<y. It then follows that b^t = b^rb^s \leqslant \sup(B(x))\sup(B(y)). Taking the sup over t, you get \sup(B(x+y)) \leqslant \sup(B(x))\sup(B(y)).

    Notice that if t < x+y then t y < x. Therefore there is a rational number r with t - y < r < x. If you define s = t-r then s<y. That justifies the assertion at the start of the previous paragraph (about expressing t in the form r+s).
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  3. #3
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    Re: b^(x+y)=b^x*b^y

    thanks. I did get half of that myself. I think you know the book this is from. Do you think it is difficult? Anyway in another part he states that A and B have a one to one correspondance iff they have the same number of elements. Surely the if part is incorrect.
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