Nemra

@Salvijus I don't think you understand what I mean.

@Salvijus 1 = b^0 ≡ b^0 = 1

b^m = b^m * b^0, b^m/b^m = b^0, 1 = b^0, You can always multiply a number by 1 because it can't change the values of two sides of the equation. The number 1 is omitted because, when multiplied by some x, the x stays x. You can have an infinite number of x^0 on both sides of the equation, as they result in 1. You can make x^0 as much as complicated.

@Salvijus https://simple.m.wikipedia.org/wiki/Exponentiation Read the sections "Zero rule" and "Negative exponents". I understand that you want to find it at 0 without using known operations and thinking of it as meaning nothing.

@Salvijus They didn't define for that case. They defined x^n for n>0, n ∈ N, N - set of natural numbers, at first.

I don't know. I think it'll make sense when you accept all of the contradictions.

It won't make sense when you only focus on how things are written.

The problem is that you want to avoid contradiction by making everything sound logical to you.

For 0, make it logically sound to you, and you'll see the effects. No one is telling you not to do it. Logically unsound things will always pop up.

@zurew 😁 People don't know that hell is waiting for them when they get introduced to what a matrix is and how it works. 🥲

@Salvijus When you approach 0 using the negative numbers, it's -infinity. When you approach 0 using the positive numbers, it's +infinity.

@Salvijus What is the limit of 1/x as x approaches 0? Does that make sense?

No, I literally showed you how to discover it. You are the one who is changing the representation without knowing that it'll affect the operations.

@Salvijus It sounds illogical to you because it was defined for multiplication in the first place. If you change the representation, then you will always get things that sound illogical to you. Please enlighten me on how division was invented.

Because when you write 0, it means to divide to itself. Why can you makes sense x/x = 1 and not x^0 = 1? Also, when x^n is already as x*x*...*x (n times), Then x^n = x*x^(n-1), x^(n-1) = x^n/x If n = 1, then x^0 = x^1/x = x/x = 1 If n = 0, then x^(-1) = x^0/x = 1/x = 1/(x^1) If n = -1, then x^(-2) = x^(-1)/x = (1/x)/x = 1/(x^2)

Well, math doesn't make sense if you don't understand how the operations work. Someone who doesn't know how it works will find it nonsense. If you want to change how you represent some things, then understand that you're also going to change the operations. I even created new math based on your input with AI, and it transformed into something you didn't expect.

A Demon Pizza

@Salvijus But you were talking about x^0 = 1 not making sense to you.

Ok. x^(-1) = x, x^(-2) = x/x = 1, x^(-3) = x/x/x = 1/x, x^(-4) = x/x/x/x = 1/(x^2), Then x^(-1) = x, x^1 = x^(-1), x^(-2) = 1, x^(-3)*x = 1, x^1 = 1/x^(-3) x^(-4)*x^2 = 1, x^2 = 1/x^(-4), Also, x^(-1)*x^(-1) = x^2, x^(-2) = x^2, x^(-2)*x^(-2) = 1, x^(-4) = 1, x^(-3)*x^(-3) = 1/(x^2), x^(-6) = 1/x^2 x^(-4)*x^(-4) = 1/(x^4), x^(-8) = 1/x^4 What the hell! Do you see how x^(+n) becomes different? And I haven't looked into it deeper yet. Your way doesn't describe reality the way you thought it would.

@Salvijus If 1 = -1, then what other meaning does the sign "-" have besides indicating numbers smaller than 0.

But you are using already-defined operations. When you define that 1 != -1 and when you prove your mentioned expressions' equality, either accept that 1 = -1 at the beginning, which would change the whole math altogether, and you will get a different result that you now haven't even thought about, or you are wrong.

They are not equal, as defining one of them also defines the other.

@Salvijus Sorry, look this one. (Please hide your post which you quote my previous arithmetic process) If x^1 = x and x^(-1) = x, Then, x^1 = x^(-1), log(x^1) = log(x^(-1)), 1*log(x) = (-1)*log(x), 1 = -1, but 1 != -1, so you're wrong.

@Salvijus If you want to invent raising a number to the power of some number, don't use mathematicians' already-defined rules for that process.