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False. Let's consider two irrational numbers, $2+\sqrt{3}$ and $2-\sqrt{3}$. Their product is simply $(2+\sqrt{3})*(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4-3 = 1$ which is an integer.

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Pooja Khatri asked Apr 4, 2019
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