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Question 243988: prove cube root 6 is irrational
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Prove  is irrational.
Assume is rational. That means  where  is reduced to lowest terms, that is  and  have no common integer factors.
It follows then that:
Since at least one factor of the RHS is even, the entire RHS must be even. Since the RHS is even, the LHS must therefore also be even. Since the product of two odd numbers is always odd, it follows that:
Since is even, it follows that 
So:
^3\ =\ 6b^3)
Now, since at least one factor of the LHS is even, the LHS must be even. Therefore the RHS must be even and  must be even. But if is even, then must be even.
Since the the original assumption leads to the conclusion that both  and are even, contradicting the part of the original assumption that and have no common integer factors, the assumption that is rational must be false.
Therefore is irrational. QED.
John
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