It kinda doesn’t make sense.
x¹= x
x² = x*x
x³= x*x*x
etc…
and even with negative numbers you’re still multiplying the number by itself
like (x)-² = 1/x² = 1/(x*x)
In: 1797
although you’ve gotten many good answers, I thought you might find this video on 0⁰ interesting:
[https://www.youtube.com/watch?v=r0_mi8ngNnM](https://www.youtube.com/watch?v=r0_mi8ngNnM)
Here’s my intuitive way of understanding it.
10^(1) = 10
10^(2) = 100
10^(3) = 1,000
etc. The pattern you see is that the number of zeros in the value resulting from raising 10 to some exponent is whatever number that exponent is. So
10^(0) = 1 (zero 0’s).
See, the thing is, I didn’t say what number base I was using; we all presumed this was base 10, but technically speaking, this example could have been base 4 or higher based on the numerals I used. This works for any number base. And since it works for any number base, if you just write that in base ten, then you can put any number for x, and
x^(0) = 0
In the spirit of ELI5, think of it this way
X³ = 1 * X * X * X
X² = 1 * X * X
X¹ = 1 * X
X° = 1
There’s the pattern…
So, the very concept of power in its naive terms is repeated multiplication, right? So, for example, 3⁴ = 3 * 3 * 3 * 3.
When you elevate a number to the power of a sum x^(a+b), what you are basically doing is multiplying the number a times and then multiplying it again b times, which is equivalent to x^a * x^b
This also mean that x⁰ multiplied by x is equal to x^(0+1), which is x¹, which is just x. But if anything multiplied by x is x, then that something is 1. No other number, when multiplied by any number, yields that very number.
The best explanation I’ve ever found for this, and for my students (I’m a math teacher), is that of a number line.
Ask yourself, if you were to draw 3^2 on a number line, you’d be drawing 2 leaps (because this is a power of 2), each representing a multiplication of 3 (because that’s the base of our power).
So the question is – where do these leaps begin? Well they have to start from 1 (because if you start at 0 and multiply, you’ll go nowhere). Any index (power) actually begins at 1, and is a series of multiplications from there.
Example: 4^5. Begin at 1, multiply by 4, 5 times.
So then the answer to your question becomes trivial. Any number, let’s call it x, to the power of 0.
Begin at 1, multiply by x, 0 times. Well because you’re doing it 0 times, you’re not moving from 1. You’re staying exactly there.
That’s why anything to the power 0 is 1. It’s where our indices begin, and we take 0 steps – we stay still!
We want rules like x^(n+m) = x^n * x^m.
That’s all fine and good and easy for positive and negative n, m. To keep the rules, we can simply define x^0 =1 (for nonzero x).
Think about x^0 = x/x. Which is 1 anyway.
Follow the pattern backwards:
* x³= x*x*x divide by x is x²
* x² = x*x divide by x is x
* x¹= x divide by x (x/x) is 1 : Follow the index law
* x^(0) = 1 divide by x is x^(-1) (x/x/x , 1/x)
When an operation inverts on itself, it negates itself into 1 and further increases it’s inverse operation. You are thinking of dividing by 0 which has no value therefore it cannot be calculated.
Why *would* it be non-existent? You’re really close with what you’ve written.
As you move up to higher powers, you’re multiplying by x each step. So x¹•x=x², x²•x=x³, etc. As you move down, you do the opposite, dividing. So x³÷x=x², x²÷x=x¹. Continue, and you get x¹÷x=x⁰. But this is just x/x which is 1.
You can also get there from the negative powers. Dividing still takes you lower (more negative) and multiplying still takes you higher (less negative). Therefore x^-1•x=(1/x)•x=x⁰. This is just another way to write x/x, which equals 1.
The only time that x⁰ doesn’t equal 1 is when x=0, because then we’d have 0/0 which is undefined.
I think there are many answers are so wrong here 🙂 I am a mathematician and if you search nesin math village on google you will see really good things 🙂 The power of a number is inductive relation that is why we have to make those definitions
2^(0) =1
and every number satisfying n>1
2^(n)=2 x 2^(n-1)
So we can define negative powers of a number so that
2^(-n) = (2^(n))^(-1) = 1/2^(n)
if you check these definitions there is no gape about any integer power of a number.
So this is explaining us why 2^(0)=1 is 🙂 it is just by definition.
(I am so sorry for mistakes my english is now very well 🙂
for any real number x, a definition of x^(n), where n is an integer superior or equal to 1, is : x.x.x … where x appears n times
a more general definition of x^(n), where n can be any real number, is: e ^(n ln(x))
(apologies about the parenthesis falling appart, editor issue)
where e is the Euler’s number (approx 2.71828) and ln is the natural logarithm.
This definition encompasses the first one. Meaning that it gives the same result that the first one, in the case of n being an integer superior or equal to 1.
Now, for any x, and for n = 0, x^(0) = e^(0 ln(x)) = e^(0) = 1
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