Could you provide a mathematical proof for why every nth-root of 2 is irrational, for n greater then or equal to 2?
Sure, here's a proof by contradiction:

Suppose that there exists some integer n greater than or equal to 2 for which 2^(1/n) is rational, i.e., 2^(1/n) = p/q for some integers p and q with no common factors.

Then, we can raise both sides of the equation to the nth power to obtain:

2 = (p/q)^n = p^n/q^n

Multiplying both sides by q^n, we get:

2q^n = p^n

This means that p^n is even, which implies that p must be even (since the only way to obtain an even result from multiplying integers is to have at least one even factor).

Let p = 2k for some integer k. Then, we can substitute this expression for p into our equation above:

2q^n = (2k)^n = 2^n * k^n

Dividing both sides by 2^n, we get:

q^n = 2^(n-1) * k^n

This means that q^n is even, which implies that q must also be even (using the same reasoning as before).

However, this contradicts our assumption that p and q have no common factors, since they both have a factor of 2.

Therefore, we have shown by contradiction that 2^(1/n) is irrational for all integers n greater than or equal to 2.

I hope that helps! Let me know if you have any questions.