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Simple Proofs

  1. Nov 1, 2005 #1
    How do you prove these?
    Let 0 < a, an element of R, and x, y an element of R. Then:
    (a^x)(a^y) = (a^(x+y))
    a^(-x) = 1/(a^x)
    Thanks.
     
  2. jcsd
  3. Nov 1, 2005 #2
    A hint before midnight would be great!
     
  4. Nov 1, 2005 #3
    Or maybe they are not so simple?
     
  5. Nov 1, 2005 #4
    what class is this for? There is prolly many ways to prove this but some of the techniques might not be appropriate for all classes.

    how about write
    (aaaaa....a)(aaaaaaa......a) so we have a times itself x+y times
    ....x times.......... y times

    =a^(x+y)
     
    Last edited: Nov 1, 2005
  6. Nov 1, 2005 #5
    for the second part start with a^(-x) and multiply by a well placed 1.
    Since 1(anything)=anything
    so try multiplying by 1 where your 1= (a^x)/(a^x)
     
  7. Nov 1, 2005 #6

    lurflurf

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    Homework Helper

    The second follows easily from the first by the case y=-x
    To handle the first you need to consider the definition you are using for a^x
    for instance one possible definition is
    a^x:=exp(x*log(a))
    in which case the problem is reduced to showing
    exp(x+y)=exp(x)*exp(y)
    for this one must considerthe definition of exp(x)
    one possible definition being
    exp(x) is the unique function for which
    exp(x) is a real number if x is a real numberjk
    exp(x)*exp(y)=exp(x+y) if x and y are real numbers
    limit x->0 [exp(x)-1]/x=1
    This is a nice definition for this problem, of course if other definitions are used you will need to prove your statement other ways.
    The other common definition of a^x is to let a^x be defined for rational numbers r then
    a^x:=lim n->infinity a^r_n
    where r_n is a rational sequence for which
    lim n->infinity r_n=x
    it is also possible to stay with
    a^x:=exp(x*log(a))
    and use other definitions for exp and log
     
  8. Nov 1, 2005 #7
    (a^x)(a^y) = exp(x*ln(a))exp(y*ln(a)) = exp((x+y)*ln(a)) = exp(ln(a)^(x+y)) = a ^ (x+y)

    Okay there. Now the second one?

    exp(-x*ln(a)) = exp(ln(a)^-x) = a ^-x = 1/a^x

    I'm not too sure about the second one...
     
    Last edited: Nov 1, 2005
  9. Nov 1, 2005 #8

    lurflurf

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    Homework Helper

    The second is a special case of the first
    a^x*a^-x=a^(x+(-x))=a^(x-x)=a^0=1
    hence
    a^-x-1/a^x
    just remember what is needed to rigorize this
    1)take as a definition
    a^x:=exp(x*log(a))
    2)show that definition is equivelent other definitions if needed
    3)define exp and log as desired
    4)show these definitions are equivelent other definitions if needed
    5)show that exp(x+y)=exp(x)*exp(y)
    6)show that 5 and 1 together can be used to prove the desired result
     
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