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points by danparsonson 3 hours ago | 0 comments

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.)