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danparsonson 2 days ago

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

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drfloyd51 2 days ago | parent | next [-]

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.

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mojomark 2 days ago | parent [-]

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.

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reichstein 2 days ago | parent | next [-]

Multiplication is the generalization of repeated addition. Repeated addition can be generalized to fractional values.

In the integers, it is the same thing, with a rule for negative times.

In the rational numbers, you can generalize to adding a number to itself a rational number of times. Adding a/b to itself p/q times is defined as adding a to itself p times, with a divisor that is adding b to itself q times.

So adding the rational 5 (5/1) with itself 2_and-1/3 (4/3) times just means (5+5+5+5) / (1 + 1 + 1) = 20/3, a rational that we can also write as 6 -and-2/3. Only addition of integers is used.

Which is the same as rational multiplication (a/b)(p/q) = (ap)/(b*q).

But it's true that you can't say that rational multiplication is the same as _integer_ repeated addition.

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goodmythical 2 days ago | parent | prev | next [-]

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

	▲	mojomark 2 days ago \| parent \| next [-]
		By dividing by 3 you are introducing a scaler into the operation. Multiplication also flips signs (direction). That's why a negative * negative = positive, whereas a negative + negative = negative. The addition operator does not flip direction. Multiplication is not generally the same as multiple addition operations. However, it has the same result for positive whole numbers. I am not writing this to flippant. I'm a 40-something year old engineer and I litterally thought the same thing until a year ago-ish. I was trying to build a function (I can't recall what exactly) for a project and I said, well instead of multiplying these terms (which wasn't efficient for what I was doing or something), I'll just cycle through and add the terms X-number of times. ChatGPT said "uh, no, that's not the same thing," to which I said NFW and proceeded down a deep rabbit hole on the difference. I'm glad I did because I also now have a much better intuition for imaginary numbers, which I was severely lacking before.
	▲	red369 2 days ago \| parent \| prev \| next [-]
		:) I like it when people who are clearly smart do things like this! Makes me feel better. Reminds me of this story: https://news.ycombinator.com/item?id=5663709
	▲	srean 2 days ago \| parent \| prev \| next [-]
		That's Grothendieck multiplication.
	▲	alpaca128 2 days ago \| parent \| prev [-]
		9+3=12, but otherwise I agree

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anthk 2 days ago | parent | prev [-]

>you can't add a number to itself 1/3 times,

Learn logarithms.

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jibalt 2 days ago | parent | prev | next [-]

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.)

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spieglt 2 days ago | parent | prev [-]

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.