1.7 Proof Methods and Strategy

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

 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?

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

1.7 Proof Methods and Strategy

 Quote by modzz 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. Lets 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 $$y_2|y_1=> y_1=ky_2, k \in Z$$
 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.