I tried to google a proof, and i found that it comes from the formula: x^1*(1/x^1) where the two x^1s cancel eachother out. I wasn’t satisfied with the answer, so i am still somewhat confused.
Usually when there’s some math i don’t get, i try to imagine a scenario irl, where a formula would apply. This always helps, but i can’t find one here.
In: 5
If we divide exponents with different base numbers, we keep the base number and subtract the exponential numbers from one another, so effectively (2^3) / (2^2) = (2^(3-2)) = (2^1).
So if we have X^0, we can demonstrate some problem where we make the exponents subtract to 0…. well shiiiiieeettt If just above we showed that (2^3) / (2^2) = (2^1) because when we share a base number the problem becomes (2^(3-2)) = 2^1. AND If a number minus itself is equal to zero (e.g. Y-Y=0) we can work backwards from X^0
X^0 = X^(Y-Y)
X^(Y-Y) = (X^Y) / (X^Y)
And what do we know about a number divided by itself… 3/3=1, 2/2=1, 4/4 = 1.Therefore (X^Y) / (X^Y) = 1
soooooo
X^0 = X^(Y-Y) = (X^Y) / (X^Y) = 1
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