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
an example is probably the best in this case
lets take x = 2
2^3 = 8
lets go down a power
2^2 = 4 which is also (2^3)/2
as you can see, you can, we can “subtract” one power by dividing by the base. Lets keep going
2^1 = 2 which is also (2^2)/2
2^0 is (2^1)/2 which is 1
This should apply to any positive X value
Any number is equal to 1 times itself. 2=2 * 1. 3=3 * 1 etc.
So 2^2 = 1 * 2 * 2, 2^3 = 1 * 2 * 2 * 2 etc.
any number to the power of x is equivalent to multiplying 1 by your number x times.
So anything to the zero power would be multiplying 1 by your number zero times which is just 1.
To make it make sense when we transition to negative exponents.
3^3 = 3•3•3 = 27
3^2 = 3•3 = 9 (27/3)
3^1 = 3 = 3 (9/3)
3^0 = 1 (3/3)
3^-1 = 1/3
3^-2 = 1/9 ((1/3)/ 3)
3^-3 = 1/27 ((1/9)/3)
We don’t really need to prove things about operations. We made them up. We define them. The reason x to the power of 0 is 1 is because we say so. So there’s nothing really to prove necessarily, you just need to understand why that definition is useful.
When we define exponentiation we start by thinking of it as repeated multiplication. However, this definition really only works when the exponent is a positive integer. What does it mean to multiply something by itself 3.2 times or -2 times or even something like pi times? That’s much harder to define.
So we do a bunch of stuff to fill in what exponentiation means when you have different types of numbers in the exponent. For a negative power we think of that as a reciprocal being raised to a positive power. For a rational power we think of that as a number being raised to the power of the numerator and a root of the denominator. Irrational numbers have a definition to, although getting into that is a lot harder.
When we come up with all these things we get a curve. It just so happens that that curve always goes straight through 1 as the exponent goes to 0 no matter what the base of our exponent is (The notable exception being 0^x). So it is very useful to define x to the power of 0 to be 1.
Leaving aside 0 powers for a moment, what is x^y / x^z ?
Well, let’s look at a few examples.
2^10 / 2^8 = 1024 / 256 = 4 = 2^2 = 2^(10 – 8)
3^3 / 3^1 = 27 / 3 = 9 = 3^2 = 3^(3 – 1)
In general, we find that x^y / x^z = x^(y – z)
So what if z = 0?
Well x^y / x^0 = x^y, therefore x^0 *has* to be 1, regardless of what x is. If it wasn’t, then our rule for dividing powers wouldn’t work.
>where a formula would apply. This always helps, but i can’t find one here.
When dividing two numbers, you can express them as exponents and subtract the exponents.
e.g. 8 / 4 = 2
But we could say
2^3 / 2^2 = 2^1
The general formula is: x^a / x^b = x^(a-b)
–
A number divided by itself equals 1.
x / x = 1 (for x not equal to 0).
Well, if that is true, then we can pick any (non-zero) number we want.
Instead of just “x” let’s try some power of x. Let’s use x^n, where n is just some (non-zero) number.
x^n / x^n = 1.
But, from the previous formula for dividing exponentials:
x^n / x^n = x^(n-n)
So x^n/x^n equals two things, and so those two things must also be equal.
Therefore x^(n-n) = 1
But we can simplify “n-n” to 0.
So x^0 = 1.
Exponentiation is repeated multiplication. Raising the exponent by 1 is the same as multiplying the number by the base. In the same way, everytime the exponent is lowered by 1, you divide the total by the base.
2³ = 8 (2×2×2)
2² = 2³ ÷ 2 = 8 ÷ 2= 4
2¹ = 2² ÷ 2 = 4 ÷ 2= 2
Well,
2⁰ = 2¹ ÷ 2
= 2 ÷ 2
= 1
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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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