Nemra

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

My mistake. Corrected it.

Yes, if I program it as such. But then that's not an addition. It's a custom-made arithmetic operation based on other defined operation(s). I could even make the "addition" process yield different results for "adding" the same or a different thing. Or maybe it'll yield a unicorn.

You mean, the concept of "2" would be the same as "11"? Or would the addition work differently?

1+1=2 (base 10), 1+1=10 (base 2) So, 2 (base 10) = 10 (base 2) However, in the same numeral system, they aren't the same. Have people forgotten that numeral systems exist?

Sorry for my tone.

@mmKay The Consilience Project