Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?
This is from MIT's Mathematics for Computer Science
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Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method. – DanielWainfleet Apr 26 '19 at 01:48
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1What makes you think this is a mistake? – Servaes Apr 26 '19 at 07:31
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No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.
Randall
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No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.
John Doe
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