Prove square root of 3 is irrational I have read several articles on math.stackexchange.com, and also this article: https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm
I still can't quite understand why one of the numbers can't be even.
Especially this part:
"Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms."
Isn't 2/3 or 3/2 smallest possible terms as well?

Your citation establishes that $a$ and $b$ must either both be even, or both be odd:


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*$b$ is either even or odd (2 cases)


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*If $b$ is odd, then $b^2$ is odd.  Hence $3b^2$ is odd, being the product of two odd numbers.  But $3b^2 = a^2$, and so $a^2$ is odd.  And this means $a$ must be odd, since the square of an even number would be even.

*If $b$ is even, then $b^2$ is even, so $3b^2$ is even, so $a^2$ is even, so $a$ is even.


