Nemra

@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.

2^(-1) = 1/(2^1) = 1/2 = 2^(-1) 2^(-2) = 1/(2^2) = 1/4 = 4^(-1) 2^(-3) = 1/(2^3) = 1/8 = 8^(-1)

You play with numbers and do operations to understand them. Sometimes you have to do it differently. You can't know what 2^0 is immediately because you're multiplying. The same way, you wouldn't understand division without understanding multiplication. You know what 2^3 is because you defined it already.

That doesn't translate into a number raised to the power of some number.

So, 2^0 means 2/2, which is 1.

@Salvijus Why are you even writing x*x like x^2? Why not x[2] or something else?

@Salvijus If I showed you 2^3, you wouldn't understand it if you didn't know multiplication. Raising a number to the power of 0 means to divide it by itself.

I showed you what means to raise a number to the power of zero.

How did you learn multiplication?

You have to understand what it means to raise a number to the power of zero. In this case, you are literally dividing apples by themselves.

x = x, x = 1*x, x/x = 1, x != 0 (x^1)*(x^(-1)) = 1, x != 0 x^(1+(-1)) = 1, x != 0 x^0 = 1, x != 0