Final answer:
Rational numbers when operated on within addition, multiplication, or powers result in rational numbers. x+y or xy can be irrational or rational regardless of x and y being irrational. If x is rational and y is irrational, then xy is irrational. The proof of irrationality cannot apply to 3 as it is a rational number.
Step-by-step explanation:
We will address the parts of the question one by one:
Proof that if x is rational, then x²+x−1 is rational:
If x is rational, we can write x as a fraction of two integers a/b where a and b are integers and b ≠ 0. Then x² + x - 1 = (a/b)² + (a/b) - 1 = (a²/b²) + (a/b) - (b/b) = (a²+ab-b²)/b², which is also a ratio of two integers, hence rational.
Is x+y irrational if x and y are irrational?
This is not necessarily true; consider −π (irrational) + π (irrational) = 0, which is rational.
Is xy irrational if x and y are irrational?
Not necessarily true again; consider −√2 (irrational) * √2 (irrational) = -2, which is rational.
Proof that if x is rational and y is irrational, then xy is irrational:
Assume the contrary, that xy is rational. As x is rational, let's consider it as a nonzero fraction a/b. Then a/b * y = k/l for some integers k, l with l ≠ 0. This implies y = (bk)/(al), which would make y rational, a contradiction. Hence, if x is nonzero rational and y is irrational, xy must be irrational.
Proving that 3 is irrational:
This statement is incorrect since 3 can be expressed as the ratio of two integers (3/1), hence it is rational. The proof for irrationality fails here.