Anything to the zero power equals one?

[khaled]

If x = 1 any power = 1 so x to the zero power = 1

If x is not equal to 1 any power is not equal to one except the power of zero

 

2^x doubles so 2 to the power of zero is half of two = 1

 

5^x is 5, 25 125 so increases five fold so 5to the power of 0 is 5/5 = 1

 

If [math]x \neq 1[/math] to any [math]power \neq 0[/math] then [math]x^{power} \neq 1[/math] in Integers !

but this is not applicable for decimal numbers ...

 

decimal numbers have different definitions for powers,

 

[latex]

 

x ^ { \frac{1}{b} } = \sqrt{x}

 

[/latex]

 

[latex]

 

x ^ { \frac{a}{b} } = \left( x ^ { a } \right) ^ { \frac{1}{b} } = \sqrt{ x ^ { a } }

 

[/latex]

Edited February 11, 2011 by khaled

[brockw69]

I too struggled with this [original] concept of x^0=1 as my logic dictates to me that x multiplied by itself naught times would equal naught. i.e. x*0=0

 

x^1=x*x?
x^2=2(x*x)?

 

4^1=4*4=16
4^2=2(4*4) = 32

 

4^0= 4*0=0 I know this is not correct, I simply don't know where the 1 comes from.

[Someguy1]

I understand that anything other than zero to the zero power equals one. But this doesn't seem to make sence to me. Could someone explain how this is, rather than that it is? I'd like as much feedback on this as possible. Thank you.

Here's one way to think about it. 64-32-16-8-4-2-1 are the following powers of 2: 6,5,4,3,2,1,0. In other words, 0 is just the obvious number in the downward sequence of powers. And the pattern continues to the negatives. 2^(-1) = 1/2, 2^(-2) = 1/4, etc. The number in the middle of this process, 1, must be 2^0. It's just completing the pattern in the obvious way.

 

In other words:

 

 

n 2^n

-----------

-2 -- 1/4

-1 -- 1/2

?? -- 1

1 -- 2

2 -- 4

3 -- 8

 

It makes sense to write 2^0 = 1.

 

Of course this isn't a proof, just a heuristic argument.

Edited February 1, 2014 by Someguy1