Can someone tell me if my proof about irrational ^ irrational = rational is valid?

Question

So I'm trying to solve the problem irrational ^ irrational = rational. Here is my proof Let $i_{1},i_{2}$ be two irrational numbers and r be a rational number such that $$i_{1}^{i_{2}} = r$$ So we can rewrite this as $$i_{1}^{i_{2}} = \frac{p}{q}$$ Then by applying ln() to both sides we get $$i_2\ln(i_1) = \ln(p)-\ln(q)$$ which can be rewritten using the difference of squares as $$ i_2\ln(i_1) = \left(\sqrt{\ln(p)}-\sqrt{\ln(q)}\right)\left(\sqrt{\ln(p)}+\sqrt{\ln(q)}\right)$$ so now we have $$i_1 = e^{\sqrt{\ln(p)}+\sqrt{\ln(q)}}$$ $$i_2 = \sqrt{\ln(p)}-\sqrt{\ln(q)}$$ because I've found an explicit formula for $i_1$ and $i_2$ we are done.

So I'm new to proofs and I'm not sure if this is a valid argument. Can someone help me out?

Answer 1

As your proof is currently set up, you would need to show that your explicit formula gives you irrational numbers $i_1,i_2$ for at least one pair of integer values $p,q$. While it is certainly believable (and in fact true) that this is the case, it requires proof.

You're approaching this problem from the wrong mindset. All we need is an example of a pair of irrational numbers satisfying "irrational ^ irrational = rational". It is not necessarily useful to find a "general solution" of any kind.

The classic proof the statement can be summarized as follows:

If $\sqrt{2}^{\sqrt{2}}$ is rational, then we're done. If not, then note that $\left[\sqrt{2}^{\sqrt{2}}\right]^{\sqrt{2}}$ is rational.

Answer 2

How do you know that the values you end up with are irrational? If you are going to use the irrationality of e and natural logarithms, you might aswel just use $e^{\ln (2)}=2$ as a counterexample and you're done. The famous proof given in another answer uses only $\sqrt{2}$ because it's easy to prove that that is an irrational number

Answer 3

This is not a valid proof. You've assumed the conclusion:

Let $i_{1},i_{2}$ be two irrational numbers and r be a rational number such that $$i_{1}^{i_{2}} = r$$

You can't just start by assuming that such numbers exist. At best, what you have would be a proof that if such numbers exist, then they must have the forms you've derived - except that you also have a mistake later, where you assume that because two products are equal, their factors must be equal:

which can be rewritten using the difference of squares as $$ i_2\ln(i_1) = \left(\sqrt{\ln(p)}-\sqrt{\ln(q)}\right)\left(\sqrt{\ln(p)}+\sqrt{\ln(q)}\right)$$ so now we have $$i_1 = e^{\sqrt{\ln(p)}+\sqrt{\ln(q)}}$$ $$i_2 = \sqrt{\ln(p)}-\sqrt{\ln(q)}$$

Answer 4

When writing a proof - any proof - the first and most important step is to decide what you are proving. If the thing you end up with is not exactly the thing you were proving, you're not done with the proof.

In this case, it's not clear to me what you're trying to prove - are you trying to prove that $i_1^{i_2}$ is always rational for every irrational $i_1,i_2$? Hopefully not, because that isn't true - but even if it were, your proof only supplies one example. Are you trying to show that for every rational $r$ there exists irrationals $i_1,i_2$ so that $i_1^{i_2} = r$? If so, then you have a problem - in your first sentence you assumed the thing you were trying to prove!

Your reasoning itself is good, assuming you're shooting for the second conclusion, but this looks more like the work you should use to get the proof, not the proof itself. If I were you, I'd have the proof look sort of like this:

Let $r$ be any rational. Then $r = \frac{p}{q}$ for some integers $p,q$. Let $i_1 = e^{\sqrt{ln(p)} + \sqrt{\ln(q)}}$ and $i_2 = \sqrt{\ln(p)} - \sqrt{\ln(q)}$. Then... (insert calculations here)... so $i_1^{i_2} = r$. Therefore, for any $r$, there exist $i_1$ and $i_2$ so that $i_1^{i_2} = r$.

Now, there's still a problem - how do you know that your $i_1$ and $i_2$ are irrational? For example, if $r = 1$, then you end up with $i_1 = 1$ and $i_2 = 0$ no matter which $p$ and $q$ you choose. You may be able to do some sort of clever argument here, but I'd recommend trying a different technique.