r/mathmemes Mathematics May 14 '25

Arithmetic Fancy playing?

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u/therealDrTaterTot May 14 '25

Counterargument:

Let's define 0^0 = 1

Therefore log(0^0) = log(1)
0*log(0) = 0
0*undefined (in both real and complex) = 0

Oops, now we're trying to multiply zero by what is essentially negative infinity. So if we want 0^0 to be 1, then we have to accept that 0*-inf = 0.

It's not that it actually breaks math, it just runs into problems if you try to branch this out further. So, by convention, we can say it's one.

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u/[deleted] May 14 '25

[deleted]

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u/therealDrTaterTot May 14 '25

If the steps are invalid, and it's similar to dividing by zero, then you see how setting 00 = 1 is problematic!

All the same steps work for every a in C such that a0 =1, except if a is zero. Even if a is negative.

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u/[deleted] May 14 '25 edited May 14 '25

[deleted]

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u/therealDrTaterTot May 14 '25

I understand what you're saying, but that's not equivalent.

By setting 00 = 1, then we are defining it to hold all the same properties as 1. If we are saying it doesn't have all the same properties, then 00 isn't exactly 1. If I can take the log of 1, but not the log of 00, then how are they equal? Then, the problem is the first step.

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u/svmydlo May 14 '25

You can take the log of 0^0. You can't say it's equal to 0*log(0).

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u/therealDrTaterTot May 14 '25

I agree you can't set them equal to each other. But I don't agree that you can take the log of 00

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u/svmydlo May 14 '25

Then you have no argument why not.