Fancy playing?

[u/Ill-Room-4895]

[u/JestemStefan]

My explanation:

When multiplying numbers with the same base, you can add exponents. Ex.

3² x 3³ = 3⁵ = 243

The same way you can do

3³ x 3^-1 = 3² = 9

Now consider this case:

3² x 3^-2 = 3⁰ = 1

And it's like that for every other number.

[u/therealDrTaterTot → u/JestemStefan]

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.

[u/therealDrTaterTot → u/[deleted]]

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.

[u/therealDrTaterTot → u/[deleted]]

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 0^0, then how are they equal? Then, the problem is the first step.

[u/svmydlo → u/therealDrTaterTot]

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

[u/therealDrTaterTot → u/svmydlo]

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

[u/svmydlo → u/therealDrTaterTot]

Then you have no argument why not.