# Prove or disprove non-constructively there exist irrationals $a, b, c$ such that $a^{b^c}$ is rational.

Consider the interesting question:

Do there exist irrationals $$a$$ and $$b$$ such that $$a^b$$ is a rational? Alternatively, prove or disprove that there exist irrationals $$a$$ and $$b$$ such that $$a^b$$ is a rational.

There has been an interesting nonconstructive proof:

Let $$x:=\sqrt{2}^\sqrt{2}$$. $$x$$ is either a rational or an irrational. (Law of excluded middle is used here.) If assuming that $$x$$ is rational, than its existence gives a proof. If assuming that $$x$$ is irrational, then we put $$a:=x:=\sqrt{2}^\sqrt{2}$$ and $$b:=\sqrt{2}$$ and we have $$a^b=(\sqrt{2}^\sqrt{2})^\sqrt{2}=\sqrt{2}^2=2$$, a rational. So, no matter we assuming x is a rational or not, there is always a proof.

A more interesting question:

Do there exist irrationals $$a$$ and $$b$$ and $$c$$ such that $$a^{b^c}$$ is a rational? Alternatively, prove or disprove that there exist irrationals $$a$$ and $$b$$ and $$c$$ such that $$a^{b^c}$$ is a rational.

Can we obtain a non-constructive proof, in which law of excluded middle is used, again?

Note: We follow the Tower rule ( https://brilliant.org/wiki/exponential-functions-properties/#the-tower-rule ) to compute $$a^{b^c}=a^{(b^c)}$$

• probably the (positive) solution to the equation $$x^{x^{1/x}}=2$$ is not rational – Masacroso Nov 30 '18 at 14:24
• Every constructive proof induces a non-constructive: Assume that no such numbers exist, use the constructive proof $\rightarrow$ contradiction. – gammatester Nov 30 '18 at 15:09
• Non-constructive prrof is not proving by contradiction. It is that: whether you assuming such a, b, c exist or not , they all induce the same conclusion. (x is either a rational or an irrational.) – user561154 Nov 30 '18 at 15:12
• This is a weird definition of non-constructive. – gammatester Nov 30 '18 at 15:15
• en.wikipedia.org/wiki/Law_of_excluded_middle#Examples – user561154 Nov 30 '18 at 15:16

I'm not exactly sure if this is what you want, but off the top of my mind, by using $$a^{\log_a b} = a$$, you could come up with many examples of a rational number for $$a^{b^c}$$ with $$a$$, $$b$$, and $$c$$ being irrational, such as

$$\large{(\sqrt 2)^{e^{\ln 2}}}$$

or more generally,

$$\large{(\sqrt[n] x)^{a^{\log_a n}}}$$

for irrational values of $$\sqrt[n] x$$, $$a$$, and $$\log_a n$$.

• Nice observation, which is more general than mine. $+1$ – GNUSupporter 8964民主女神 地下教會 Nov 30 '18 at 14:58
• Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used. – user561154 Nov 30 '18 at 15:08
• Well, you could just use the expressions given by either answer to conclude that there there exists irrational numbers for $a$, $b$, and $c$, such that $a^{b^c}$ becomes rational. – KM101 Nov 30 '18 at 15:27
• @KM101 I don't quite understand we you mean. the currently 2 answers here solve my problem by giving explicit expressions, so they are constructive. I am curious about non-constructive. en.wikipedia.org/wiki/Law_of_excluded_middle#Examples "x is either a rational or an irrational". – user561154 Nov 30 '18 at 15:41
• I'm not really into these different types of proofs, so I may have misunderstood what exactly you wanted. – KM101 Nov 30 '18 at 15:44

Take $$a=c=\sqrt2, b={\sqrt2}^{\sqrt2}$$.

$$a^{b^c}={\sqrt2}^{({\sqrt2}^{\sqrt2})^{\sqrt2}}={\sqrt2}^{(\sqrt2)^2}=2$$

• Thank you! :-) ut in the origin the problem was solved by a non-constructive proof. It may be cooler if someone give a non-constructive proof, in which law of excluded middle are used. – user561154 Nov 30 '18 at 15:11
• In your current way you need to firstly prove that $\sqrt{2}^\sqrt{2}$ is an irrational, so you have an example of irrationals $a$ and $b$ and $c$. – user561154 Nov 30 '18 at 23:38
• In your current (constructive) way..., I mean. – user561154 Nov 30 '18 at 23:45
• @JZH Like the other answerer, I've no interest in finding a "non-constructive proof". Add a bounty if you like. Here's a non-constructive proof that you asked in the comments. Consider $b^2=2^{\sqrt2}=r/s$, with $r,s$ integers. We have a positive irrational exponent of $2$ on the left-hand side, but rational exponents of any prime factors on the right-hand side. – GNUSupporter 8964民主女神 地下教會 Dec 1 '18 at 9:03