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To prove that 3√2 is irrational, assume the product is rational and set it equal to a/b , where b is not equal to 0: 3√2=a/b . Isolating the radical gives √2=a/3b . The right side of the equation is choose (rational or irrational) . Because the left side of the equation is choose (rational or irrational) , this is a contradiction. Therefore, the assumption is wrong, and the number 3√2 is choose (rational or irrational).
