# My simple proof of x^0 = 1

1. ### Joza

139
I always wondered, how can any number raised to the power of 0 be 1.

So, I came up with this! ( * = multiplication sign)

1 * 4 * 4 * 4 = 4^3

1* 4 * 4 = 4^2

1 * 4 = 4^1

Therefore, 1 = 4^0

2. ### Kurdt

4,941
Staff Emeritus
One can define exponentials in terms of the family of standard functions and it falls naturally that anything to the power zero is one. Exponentials are defined as follows:

$$n^x = e^{x\ln{n}}$$

Now put x = 0 and see what happens.

Last edited: May 30, 2007
3. ### Zurtex

1,123
Not such a convincing proof when I saw that, given thanks to my wonderful school system I'd learnt of exponentials as a continuation of powers and got taught how to get the taylor series of ex and stuff without respect to where e came from in the first place.

4. ### neutrino

$$1 = \frac{a^n}{a^n} = a^{n-n} = a^0$$

What could be more simpler than that?

5. ### Kurdt

4,941
Staff Emeritus
I hear you! I was exactly the same until I started university and finally learned things in the "proper" order. I am still very opposed to the way they teach A-level maths in the UK and that is one of the reasons.

6. ### Shing

140
I like this proof best. And I also think it is the formal proof.

7. ### Shing

140
I don't really get it, would you explain more please?

8. ### Kurdt

4,941
Staff Emeritus
If you set n = 0 you get the following:

$$n^0 = e^{0\ln{n}} = e^0 = 1$$

Last edited: May 30, 2007
9. ### bomba923

736
Shing,
$$\forall a \ne 0,\;a^n = a^{n - 1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}$$
Given that a1=a,
$$a^0 = a^{ - 1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1$$

10. ### StatusX

2,567
Here's my proof that 4^0=0:

4^3=4*4*4+0
4^2=4*4+0
4^1=4+0
4^0=0

Can you explain why your proof is better than mine? (it is, but you haven't shown why)

11. ### mathman

6,432
This is a proof by analogy, but not by logic. You can just as well say:

4^3=1*4*4*4
4^2=1*4*4
4^1=1*4
4^0=1

12. ### Werg22

This is not a proof, it's a pattern. x^0 = 1 is defined.

13. ### prasannapakkiam

0
Neutrino's proof is the basic elementary method. However, with Kurdt's method; I accept it. However one must be careful when taking logarithms, as this restricts the limit to which a function may take, as I found the hard way...

For example: x^x = y. Techniqually the domain has integers below 0. But when converting it to y=e^(xln(x)), it has immediately taken those integers away...

Anyway, I think that here it is valid. But how can this be proven for sure? Or is it completely unneccessary to consider here...

14. ### gunch

54
Can you prove that $$\frac{a^n}{a^m} = a^{n-m}$$ when $$n = m$$? Usually the proof is only valid when they aren't equal and then people define $$x^0=1$$ such that the property remains valid for $$n = m$$.

Last edited: May 30, 2007
15. ### Gib Z

3,348
Excuse me? Why do we have to work on the case n=m specifically when the proof works for all n and m? n=m is a nice case that lets us prove this nicely.

16. ### Kurdt

4,941
Staff Emeritus
Thats a good point. My example is the opposite way round which I shall correct. But of course it doesn't solve the problem of something like the function you have stated.

### Staff: Mentor

0^0 is either undefined or arbitrarily defined to be some value, typically one. It is clear when examining a plot of z=x^y that the value of z at (x,y)=(0,0) depends on the direction via which one approaches this (x,y) pair. All of the previous posts have approached this point in a direction that makes the value appear to be one. I could just as easily make it appear to be zero by looking at $$\lim_{x\to0}0^x$$ which is obviously zero. I could just as easily make it any complex number!

Given any complex number $a$, it is easy to come up with a form that reaches $0^0[/tex] as some limit and evaluates to [itex]a$. Suppose $0<||a||<1$. Using $1=n/n[/tex], $$a=(a^{n/n})=(a^n)^{(1/n)}\equiv x^y$$ where [itex]x\equiv a^n, y\equiv 1/n$

Note that both $x$ and $y$ approach zero as $n\to\infty$. Thus $$0^0=\lim_{n\to\infty}(a^n)^{(1/n)} = a$$

==================

0^0 is typically defined to be one for the sake of simplicity. This definition eliminates zero as a special case for power series, binomial expansions, etc. It is just a convention however, not a proof.

Last edited: May 30, 2007
18. ### Gib Z

3,348
DH I think you made the mistake I was about to, for this thread. This thread isnt about the commonly talked about 0^0 which one expects to find, its just about the exponent of zero in general. His original post said he proved that 4^0 = 1, which we know on solid ground =)

19. ### gunch

54
What proof? Assuming we are dealing with real numbers then the proof usually makes the assumption that a^0=1 when dealing with the case n=m. For such proofs neutrino's proof would use circular reasoning and therefore not be valid.

In my experience either a^n/a^m=a^(n-m), a^0=1, a^n=a*a^(n-1) or some similar identity is just taken as an axiom and the rest are proved from that (can you show me a proof of one of those without use of the others?) a^n=a*a^(n-1) may seem intuitive, and it definitely is for most value of n, but when n=1 it isn't really clear why it should hold.

Actually the conclusion of the original post is false as others have noted. 0^0 is generally not considered to be 1, but the original post said any number. That means anything which would prove it needs to be incorrect.

20. ### Gib Z

3,348
I'm talking about that proof that you see in your Elementary Algebra textbook...the one where a^m/a^n is expressed as m terms of a divided by n terms of a, then we cancel them out till we are left with a^(m-n). No assumption that a^0=1 needed there.