Because it's just wrong. hahajk said that taking the 2nd root spends a 2, not that "roots are invalid". (FWIW, I'm not necessarily endorsing the view that the square root symbol implicitly spends a 2, I'm just saying that the counterarguments given are logically wrong ... actually more like semantically wrong.)

The comment below by drfloyd51 is similarly wrong ... hahajk didn't say anything about "giving exceptions to powers in the rules".

And the comment by speglt is also wrong:

> If taking a square root is an inherent use of a two, then dividing a number by another number should also be a use of a two, because there are two operands.

The number of operands isn't relevant; the degree of the root is.

And from SideQuark we get

> One could even claim division is raising to the -1 power, thereby disqualifying it in exactly the same manner you disqualify sqrt.

But hahajk said nothing about anything being disqualified. (Not to mention that division is not raising to the -1 power ... raising what? n/d = n * d^-1, but this is completely irrelevant to hahajk's argument.)