1.7 Proof Methods and Strategy

[modzz]

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




PLEASE HELP...

 

[sutupidmath]

well, you can easily find a counter example...

let

b=\frac{1}{2}; a=2=> a^b=2^{\frac{1}{2}}=\sqrt{2}

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

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

Or are you asked to do this differently, like in a more general form?

 

[modzz]

well, you can easily find a counter example...

let

b=\frac{1}{2}; a=2=> a^b=2^{\frac{1}{2}}=\sqrt{2}

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

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

Or are you asked to do this differently, like in a more general form?

I really don't know..but if u can do it in general form that would be great.

 

[sutupidmath]

I really don't know..but if u can do it in general form that would be great.

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

c=\frac{x_1}{x_2}, b=\frac{y_1}{y_2}, x_1, x_2, y_1,y_2 \in Z^+ 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 gcd(x_1,x_2)=gcd(y_1,y_2)=1 Now,

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

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


\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} so we notice that

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

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

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

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

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

Now,


(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}

By doing the same reasoning we come to the point where

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

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

Indeed it can be rational only if y_2|y_1=> y_1=ky_2, k \in Z

 

[sutupidmath]

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.