Hacker News

points by JumpCrisscross 4 hours ago | 0 comments

> it's not an operator like + or !

Squaring is to multiplication as is the latter to addition. If roots are invalid then so should be multiplications.

Exponentiation is to multiplication as multiplication is to addition. Squaring is to multiplication as doubling is to addition.

[–] SamBam 2 hours ago

GP didn't say roots are invalid, just that they burn a two.

If we agree that 2^2 uses two twos (and, indeed, that 2^3 isn't valid, because it uses a 3), then what's the equivalent of a square root?

Of course to my simple math brain, a square root is ^1/2, which includes the digit 1. But I agree there is someone 2-ish about the square root.

[–] direwolf20 25 minutes ago

It can be written as ²√2 and there's also ³√2 and so on

[–] danparsonson 3 hours ago

Would you elaborate on that? That doesn't make any sense to me

[–] drfloyd51 3 hours ago

Multiplication is repeated addition.

2x3= 2+2+2 AxB= A+A B times.

Raising to a power is repeated multiplication.

2^3= 2x2x2 A^B = AxA B times

Roots are the inverse of powers

Division is the inverse of multiplications.

Given this… the OC’s comment about giving exceptions to powers in the rules is nonsense because they are still operations.

[–] mojomark 2 hours ago

Multiplication is not the same as repeated addition, which you can observe as soon as you try the two different operations on fractions or negative numbers. For example, you can't add a number to itself 1/3 times, but you can certainly scale a number by 1/3.

Multiplication is a scaling operation.

[–] goodmythical 1 hour ago

Can you not? 9+(9(1/3)) [adding nine to itself a third of a time] is nine plus one third of nine {or nine divided by three} which is one of the three in three times three that equals nine which is three therefore 9+(9(1/3))=11 because 9+3=11

[–] spieglt 3 hours ago

Multiplication is adding numbers to themselves. Exponentiation is multiplying numbers by themselves. Ditto division/taking roots. Squaring a number is just the instance of exponentiation that happens to have two operands. 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.

[–] jibalt 2 hours ago

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.)