Final answer:
Using proof by contradiction, we've shown that if x² is irrational, then x must be irrational. Similarly, if n² is odd, then n must be odd, as the square of an even number results in an even number, which is a contradiction to the odd condition.
Step-by-step explanation:
To prove by contradiction that if x² is irrational, then x is irrational, we start by assuming the opposite; that is, we assume that x is rational. If x were rational, it could be expressed as a fraction of two integers, say ²a/b², where a and b are integers and b is not zero. Now, if x² is irrational, that means it cannot be expressed as a fraction of two integers, which contradicts our initial assumption that x can be expressed as such a fraction. Hence, we must conclude that if x² is irrational, x must also be irrational.
To prove that for every integer n, if n² is odd, then n is odd, we again use proof by contradiction. Assume n is even. An even number can be expressed as 2k, where k is an integer. Squaring an even number would result in (2k)² = 4k², which is clearly even because it is a multiple of 4. This contradicts the statement that n² is odd. Therefore, if n² is odd, n cannot be even, and thus must be odd.