Anything multiplied by 0 is 0 right so why does x number raised to the power of 0 = 1? isnt it x^0 = x*0 (im turning grade 10 and i asked my teacher about this he told me its because its just what he was taught 💀)
In: Mathematics
X^1 = X
X^2 = X * X
X^3 = X * X * X
…
To get up, you multiply by X.
So, to get down, you divide by X.
X^1 = X
X^0 = X / X = 1
There’s a pattern of multiplication and division with exponents. 5^3 is 5x5x5. 5^2 is 5×5. 5^1 is 5. We’re dividing by 5 each time, so 5^0 is…1. This happens with any whole number. Divide again. 1/5 = 5^-1. 5^-2 = 1/5^2.
Because when you go below the power of 1 it becomes a division rather than a multiplication. So where x¹ is just the base value of x, when you go to x⁰ you are dividing x by itself. A number divided by itself is always 1.
2^3 = 2 x 2 x 2 = 8
2^2 = 2 x 2 = 4
2^1 = 2
2^0 = 2/2 = 1
Notice that each time you decrease the exponent by 1, you’re effectively dividing by the base number, since to remove a multiplication operation you must divide. Once you get to an exponent of 0, you’re simply just dividing by the base number, which always equals 1
If X^2 = X*X.
X^2 = X^1 * X^1 <=> X^1+1 .
X^2 / X^1 = X^1 * X^1 / X^1 <=> X^2-1 = X^1 .
X = X
So far so good, powers work and we showed that X^-1 = 1/X, as well as X^a * X^b = X^a+b
Then
X = X <=> X^1 * X^-1 = X^1 * X^-1 <=> X^1-1 = X/X.
X^0 = 1.
No, x^0 has nothing to do with x*0. Think about the powers of 2
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
The exponent is the number of twos in that multiplication. What would make this work if there were _zero_ twos? 1, as in
1 = 1
1 * 2 = 2
1 * 2 * 2 =4
1 * 2 *2 * 2=8
Etc. also, remember that negative exponents cause you to put a 1/ over the result, as in:
2^-1 = 1/2
2^-2 = 1/4
If you start with the higher powers and go down, you’ll see it’s like dividing by 2, and it will be easy to see what should go there:
32
16
8
4
2
?????
1/2
1/4
1/8
1/16
1/32
x^(3) = x * x * x
x^(2) = x * x
x^(3) = x * x^(2)
Right?
Written in general
x^(n) = x * x^((n-1))
So then, applying the above logic
x^(1) = x * x^(0) , zero being (1-1)
If x^(0) was zero then x^(1) would also be zero.
This math teacher explains it in a really cool (and ELI5) way: https://youtu.be/X32dce7_D48?si=KHWgVDY2U5NRkhpN
[I like the Wikipedia explanation](https://en.m.wikipedia.org/wiki/Exponentiation)
From the *definition* of exponentiation you get a rule that makes a lot of sense and then use algebra to demonstrate a rule that is less apparent.
So your instructor was right. He was told what it was. He just didn’t bother to remember the demonstration.
Ok so
x^0 = x^(n-n), since n-n is of course = 0
Know, for properties of however those are called in English, I think exponentials, x^a / x^b = x^(a-b)
So you can now imagine how
x^0 = x^(n-n) = x^n / x^n =
Since a number divided by himself is 1
x^0 = x^(n-n) = x^n / x^n = 1
EDIT:
You can now interrogate yourself on how
x^a / x^b = x^(a-b)
Since x^a means you will multiply x for it self a times, you will get a thing like
x^4 / x^3 = (x x x x) / (x x x) = x = x^1 = x^(4-3)
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