# My simple proof of x^0 = 1

by Joza
Tags: proof, simple
 P: 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
 Emeritus Sci Advisor PF Gold P: 4,980 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.
HW Helper
P: 1,123
 Quote by Kurdt 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: $$x^n = e^{n\ln{x}}$$ Now put n = 0 and see what happens.
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.

 P: 2,046 My simple proof of x^0 = 1 $$1 = \frac{a^n}{a^n} = a^{n-n} = a^0$$ What could be more simpler than that?
Emeritus
PF Gold
P: 4,980
 Quote by Zurtex 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.
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.
P: 140
 Quote by neutrino $$1 = \frac{a^n}{a^n} = a^{n-n} = a^0$$ What could be more simpler than that?
I like this proof best. And I also think it is the formal proof.
P: 140
 Quote by Kurdt 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: $$x^n = e^{n\ln{x}}$$ Now put n = 0 and see what happens.
I don't really get it, would you explain more please?
Emeritus
PF Gold
P: 4,980
 Quote by Shing I don't really get it, would you explain more please?
If you set n = 0 you get the following:

$$n^0 = e^{0\ln{n}} = e^0 = 1$$
 P: 736 Shing, $$\forall a \ne 0,\;a^n = a^{n - 1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}$$ Given that a1=a, $$a^0 = a^{ - 1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1$$
HW Helper
P: 2,567
 Quote by Joza 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
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)
P: 6,109
 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)
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
 P: 1,520 This is not a proof, it's a pattern. x^0 = 1 is defined.
 P: n/a 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...
P: 54
 Quote by neutrino $$1 = \frac{a^n}{a^n} = a^{n-n} = a^0$$ What could be more simpler than that?
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 $$n = m$$.
HW Helper
P: 3,348
 Quote by gunch 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 $$n = m$$.
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.
Emeritus
 Mentor P: 15,202 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.