Given that a is rational , and b is irrational , can a b be rational?
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Bonus: Prove that this is only possible when a = 0 .
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Assume a = 0 . Let a b = r where r is rational:
a b = r ⟹ b = a r
But a r is rational as a = 0 but this contradicts b being irrational so a = 0 .
Suppose that a is rational and not equal 0; then a = e/f for integers e,f with e not = 0. If ab is rational, b irrational, then ab = k/m, and b = (k/m)*((f/e) = kf/me, a contradiction.
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If a = 0 and b = 2 , , then a b = 0 , which is rational. So the answer is Yes