Explain x^(zero)

Let x = any number.

Why does x^(0) = 1?

 

Why does x^(0) = 1?

proof:

let have x^n/x^n
so if you divede a number by it self, you get quotient equal to 1
=>x^n/x^n=1...........eq.1

however, using exponent rules we can write x^n/x^n as
x^n/x^n=x^(n-n)
x^n/x^n=x^0.........eq.2

eq.1 and eq.2 have same left sides, then right sides must be same too

so, x^0=1

the reason that any number to the zero power is one is because any number to the zero power is just the product of no numbers at all, which is the multiplicative identity, 1

 

You said:

". . .any number to the zero power is just the product of no numbers at all, which is the multiplicative identity, 1."

Can you provide at least two samples of your statement?

 

It is easy to calculate x^n with n a positive integer: x times itself n times. From that we get the general rule that x^(m+n)= (x^m)(x^n). That's a very nice general rule so while we are free to define x^n for n NOT a positive integer, we want to define it in such a way that x^(m+n)= (x^m)(x^n) is still true when m or n is not a positive integer. In particular, with n= 0, m+ 0= m so we want x^m= x^(m+ 0)= (x^m)(x^0). Dividing both sides by x^m (which we can only do is x is not 0) we get x^0=1 as long as x is not 0. We leave 0^0 undefined as explained in a previous post.

 

Good study notes.

 

Notice my condition "(which we can only do if x is not 0)"

Your first statement in your first post in this thread was "Let x = any number.
Why does x^(0) = 1?" and MathLover1 said "any number to the zero power is one" but neither of those is true! 0 to the zero power is "undefined". It has NO value and certainly is not 1.

 

How about this?

Let x = any number EXCEPT for 0.
Now, x^(0) = 1.