Is LEM Actually Invoked in the Classic Irrational Exponent Argument

Question

I am almost positive this has been discussed on here but I can't seem to find it after an hour of searching. Please redirect if so.

There is a classic non-constructive argument we have probably all seen to the assertion that "There exists irrational $x,y$ such that $x^y$ is rational. The argument asserts that the number $\sqrt{2}^\sqrt{2}$ must be rational or irrational (exclusive or), and either case yields the result, thus the assertion is true by exhaustion of the cases.

My question is: does the statement "for any real number $x$, $x$ is either rational or irrational" actually assume the Law of the Excluded Middle? The reason I think it does not is because the literal definition of irrational is that it is $not$ rational, so if the statement '$x$ is rational' is either true or false by the principle of bivalence, but being false means 'there does not exist $a,b$ such that $x = \frac{a}{b}$' which is the definition of irrational. In this situation, it seems it can be justified without LEM that irrational numbers form a disjoint union of real numbers. Or am I making a jump in interpretting 'what it means' for a statement to be false? Perhaps what allows me to even quantify or interpret what it means for the statement to be false is where I am using LEM? If someone did not accept LEM could they raise a logical flaw with the argument given above?

This is kind of abstract and more rooted in logic/philosophy and I think (at least in most US universities) the approach to teaching collegiate mathematics tends to yield confusion about logic. Logic seems to be a bigger bubble that encompasses mathematics so I may not understand the more abstract setting in which these concepts make sense.

Answer 1

As it stands, the statement “for any real number $x$, $x$ is either rational or irrational” does not assume law of excluded middle.

The statement “for any real number $x$, $x$ is either rational or not-rational” would assume law of excluded middle, in which case ‘not-rational’ might or might not coincide with ‘irrational’ and if it coincided, we would call ‘irrational’ instead of ‘not-rational’. But it is conceivable that it might well coincide with the union of two sets numbers, suppose irrational numbers and what we would call "anti-rational" numbers.

To express symbolically, the former statement tells that

$$\forall x(x\in\mathbb{R}\rightarrow(x\in\mathbb{Q}\vee x\in\mathbb{P}))$$

This does not exemplify the law of excluded by itself, for the predicates $\mathbb{Q}$ and $\mathbb{P}$ are distinct in the formal language of logic. However, the later statement

$$\forall x(x\in\mathbb{R}\rightarrow(x\in\mathbb{Q}\vee \neg(x\in\mathbb{Q})))$$

or equivalently,

$$\forall x(x\in\mathbb{R}\rightarrow(x\in\mathbb{Q}\vee x\not\in\mathbb{Q}))$$

exemplifies the law of excluded middle, for the predicates at stake are one and the same in the formal language of logic.

Compare these to the imaginary example

$$\forall x(x\in\mathbb{R}\rightarrow(x\in\mathbb{Q}\vee x\in\mathbb{P}\vee x\in\mathbb{Q}^{anti}))$$

It is a fact of mathematics, not logical necessitation, that there is no such set as $\mathbb{Q}^{anti}$.

Anyway, the idea that the proof is non-constructive does not hinge on this. For further discussion, I cite the proof referred from Michael Dummett's Elements of Intuitionism (1st edition, p. 10 and 2nd edition, p. 6, see the ensuing discussion)

Theorem 1.1 There are solutions of $x^{y} = z$ with $x$ and $y$ irrational and $z$ rational.

Proof. $\sqrt{2}$ is irrational, and $\sqrt{2}^{\sqrt{2}}$ is either rational or irrational. If it is rational, put $x = \sqrt{2}, y = > \sqrt{2}$ so that $z = \sqrt{2}^{\sqrt{2}}$, which, by hypothesis, is rational. If, on the other hand, $\sqrt{2}^{\sqrt{2}}$ is irrational, put $x = \sqrt{2}^{\sqrt{2}}$ and $y =\sqrt{2}$, so that $z = (\sqrt{2}^{\sqrt{2}})^\sqrt{2} = \sqrt{2}^{\sqrt{2}} = 2$, which is certainly rational. Thus in either case a solution exists.

Notice that the proof does not discern which one from the pair of numbers $(x, y$ falls under the predicate of being a rational number (equivalently, an irrational number). That makes the proof non-nconstructive.

To recapitulate, the world may offer us two options or more than two options. If two options A and B are presented, then B naturally coincide with not-A. But the law of excluded middle dictates merely two logical options, however the world turns out to be.

It is a mathematical fact (i.e., of the world) that there are two complementary sets of numbers, rational and irrational. It is not a logical truth. But since mathematics offers two choices, it coincides with the law of excluded middle. Compare this to the proof: The offered options are logical, not mathematical, for we do not know which is which.