>>14340263A constructivist might accept the set of all natural numbers larger than 10.
It can be constructed by taking the set of all natural numbers (axiom of infinity, which constructivism doesn't have a default position on) and then filtering it with the sentence "x > 10" (axiom schema of specification).
But a finitist probably wouldn't since it's an infinite set.
For the other way around, consider the claim, there exists a pair of irrational numbers x and y, such that x^y is a rational number.
Side note: Finitists can believe in algebraic numbers in the same way that the can believe in natural and rational numbers (just not the set of all algebraic numbers). Since some algebraic numbers or not rational, it still makes sense to talk about irrational numbers in a classical finitst context.
A finitist might accept the following proof:
Either root 2 to the root 2 is rational or it's not (law of excluded middle)
If root 2 to the root 2 is rational, then we're done!
If root of 2 to the root of 2 is irrational, then that irrational number to the root 2 is equal to 2, a rational number and we're done.
Therefore, there exists a pair of irrational number x and y such that x^y is a rational number.
But a contructivist wouldn't accept that proof since it uses law of excluded middle. After this proof, you still don't know which pair of irrational numbers work. You haven't constructed the solution.
They would still accept the claim though because there are other ways to prove it without using LEM. For example, e^ln(2) is a pair of irrational numbers that work. That's a direct proof.
I unfortunately can't think of any claims that finitits would accept and not constructivists.
If anyone can come up with an example I would be interested in knowing about it!
The only things that I know of that can't be proven without LEM are things that require the axiom of choice and infinite sets, which finitists would accept in the first place.
