# X^0 = 1?

1. Apr 19, 2010

### Deicider

1. The problem statement, all variables and given/known data

the title says it all,its very simple question but i can't get it.

2. Relevant equations

3. The attempt at a solution
10^2=100 =10*10
10^1=10 = 10*1
10^0=1(?) =10*0=0 ...1?
I never asked anyone this,and never questioned it just followed what teacher said but now i wanna some answers : D

2. Apr 19, 2010

### Dick

If you want the laws of exponents to work correctly you need x^0*x^n=x^(0+n)=x^n. Since x^0 multiplied by x^n is x^n, x^0 must be 1.

3. Apr 19, 2010

### Deicider

2^0*2^2 =2^(0+2)=2^2=4
if
2^0=2*0=0 and 2^2=2*2=4
then
0+4=4
isnt the same? i think it works with being zero too.
i know am wrong cause ,obviously, all mathematicians(and non) would agree with you,but with my 'denial' i just want to understand it better and so far i presume that this has to do with "bending" a property of numbers to fit in exponent laws as u said.
or smth like that

4. Apr 19, 2010

### Staff: Mentor

But 2^0 $\neq$ 0

5. Apr 19, 2010

### Dick

Defining 2^0=0 doesn't work! 2^0*2^2 should be 2^(0+2)=2^2=4. 0*4 is NOT 4. You can't just change '*' into '+'.

6. Apr 19, 2010

### Deicider

i just found a pretty good example:
"""""""""
3^1, 3^2, 3^3, 3^4, ....
3 , 9 , 27 , 81 , ....

So what is the pattern in the bottom sequence? Well, every time you move to the right in the list you multiply by 3, and every time you move to the left in the list you divide by 3. So we could take the bottom sequence and keep going to the left and dividing by 3, and we'd have the sequence that looks like this:

..., 3^-3, 3^-2, 3^-1, 3^0, 3^1, 3^2, 3^3, 3^4, ....

..., 1/27, 1/9, 1/3 , 1 , 3 , 9 , 27 , 81 , ....
"""""""""
as goes beneath ^1 still sucks intuitively : D

7. Apr 19, 2010

### elect_eng

Dick's explanation is correct of course.

If it's still not sinking in, then think of something like 2^x as a function. Plot the function and look at the value of the function at x=0. I've attached a plot to make it easy to see.

#### Attached Files:

• ###### PlotPower0.pdf
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8. Apr 19, 2010

### sylas

Try this.

write down 21, 22, 23, 24 and so on. You should get

2, 4, 8, 16, 32, 64, ...

Going backwards, you divide by 2, don't you.

64, 32, 16, 8, 4, 2, ...

So extend it.

64, 32, 16, 8, 4, 2, 1, 0.5, 0.25, 0.125, ....

This helps give the intuition that 20 = 1; it also shows what you expect for negative powers!

Cheers -- sylas

9. Apr 19, 2010

### icystrike

$$1 = \frac{x}{x} = (x)(x^{-1}) = x^{1+(-1)} = x^{1-1} = x^{0}$$

Conversely you can say that $$x^{0} = 1$$

10. Apr 20, 2010

### Post

Use the division rules of exponents. (x^a)/(x^b) = x^(a-b)

if a = b then a - b = 0, and any number divided by itself is 1. Done

11. Apr 21, 2010

### HallsofIvy

But you switched from * to +! 0*4= 0, not 4.

No, no bending involved. It only requires recognizing that addition is not multiplication!

12. Apr 21, 2010

### Martin Rattigan

$x^0=1$ may not be something you prove at all.

Before you could prove it you have to know what $x^y$ means. That comes from a definition.

In this case there are a series of definitions for $x^y$, usually starting with the case when $y$ is a natural number (the ones you count with starting 0,1,2) and adding further definitions to cope with negative integers, fractions, complex numbers etc., at each stage making sure that the definition added doesn't conflict with the ones you have already made.

The usual approach to defining $x^y$ when $y$ is a natural number is to define $x^0$ to be $1$ for any number $x$ and then define $x^n$ to be $x^{n-1}x$ for any $n>0$. Having defined the expression for $y=0$, the second part successively defines it for 1, 2, 3 and eventually for all natural numbers.

With this approach, $x^0=1$ exactly because you say so and for no other reason. The proof, such as it, is simply that $x^0=1$ by definition.

Of course there is no harm in speculating on why people define it that way and what the consequences are.

13. Apr 21, 2010

### Susanne217

I am not trying to be a smart ***, but as I said in another post. This is a pre-Calculus forum which means its suppose to be problems which doesn't involve the usage of Real Analysis.

You just to accept that that is the definition. If not grab a Real Analysis book and its explained to you there :)