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1.7 Proof Methods and Strategy

  1. Sep 20, 2008 #1
    1. Prove or disprove that if a and b are rational numbers, then a^b is also rational??




    PLEASE HELP......
     
  2. jcsd
  3. Sep 20, 2008 #2
    well, you can easily find a counter example...

    let

    [tex] b=\frac{1}{2}; a=2=> a^b=2^{\frac{1}{2}}=\sqrt{2}[/tex]

    But we know that [tex]\sqrt{2}[/tex] is not rational. Or if you don't want to take this for

    granted, then all you have to do is prove that [tex] \sqrt{2}[/tex] isn't rational.

    Or are you asked to do this differently, like in a more general form?
     
    Last edited: Sep 20, 2008
  4. Sep 21, 2008 #3
    I really don't know..but if u can do it in general form that would be great.
     
  5. Sep 21, 2008 #4
    Well, usually it is sufficient to find a counterexample to show that something does not hold in general.

    because in your case you can think of it this way: let

    [tex]c=\frac{x_1}{x_2}, b=\frac{y_1}{y_2}, x_1, x_2, y_1,y_2 \in Z^+[/tex] I am first working only with positive integers, but if u want to prove for any integer, then you have to work in cases.

    Also, let [tex] gcd(x_1,x_2)=gcd(y_1,y_2)=1[/tex] Now,

    [tex]c= a^b=(a)^{\frac{y_1}{y_2}}[/tex] Now we want to show that c is irrational. Lets suppose the contrary, suppose that c is rational so we can rewrite c as:

    [tex] c=\frac{x_1}{x_2}, gcd(x_1,x_2)=1[/tex] This way:


    [tex]\frac{x_1}{x_2}=(a)^{\frac{y_1}{y_2}}=>(\frac{x_1}{x_2})^{y_2}=a^{y_1}=>(x_1)^{y_2}=a^{y_1}(x_2)^{y_2}[/tex] so we notice that

    [tex] (x_1)^{y_2}\in a^{y_1}Z[/tex] now we want to know what happenes with x_1.

    Suppose that x_1 is not in [tex] a^{y_1}Z[/tex] So, this means that

    [tex] x_1 \in m+ a^{y_1}Z[/tex] for [tex] m=1,2,.....,a^{y_1}-1[/tex] Now let m =1, for our case, so

    [tex] x_1=1+a^{y_1}=>(x_1)^{y_2}=(1+ka^{y_1})^{y_2}[/tex] we notice that when we expand the RHS all terms besides the first one will have an [tex] a^{y_1}[/tex] so we can factor this one out, which means that also

    [tex](x_1)^{y_2} \in m+ a^{y_1}Z[/tex] which is not true, so the contradition derived means that
    [tex]x_1\in a^{y_1}Z=>x_1=a^{y_1}k, k \in Z [/tex]

    Now,


    [tex](x_1)^{y_2}=a^{y_1}(x_2)^{y_ 2}=>(a^{y_1}k)^{y_2}=a^{y_1}(x_2)^{y_ 2}=>(x_2)^{y_2}=a^{y_1(y_2-1}}k^{y_2}[/tex]

    By doing the same reasoning we come to the point where

    [tex]x_1=a^rk_1,x_2=a^rk_2, =>gcd(x_1,x_2)=a^r[/tex] which contradicts the fact that [tex] gcd(x_1,x_2)=1[/tex] this way we have proved that

    [tex]c= a^b=(a)^{\frac{y_1}{y_2}}[/tex]
    cannot be rational.

    Indeed it can be rational only if [tex] y_2|y_1=> y_1=ky_2, k \in Z[/tex]
     
  6. Sep 21, 2008 #5
    I don't know whether what i did above makes sens to you, but in any case if i were you, i would take the counterexample as a means of showing that in general a^b, cannot be rational.
     
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