# My simple proof of x^0 = 1

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

Kurdt
Staff Emeritus
Gold Member
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.

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Zurtex
Homework Helper
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.

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

What could be more simpler than that?

Kurdt
Staff Emeritus
Gold Member
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.

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

What could be more simpler than that?
I like this proof best. And I also think it is the formal proof.

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?

Kurdt
Staff Emeritus
Gold Member
I don't really get it, would you explain more please?
If you set n = 0 you get the following:

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

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

StatusX
Homework Helper
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)

mathman
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

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

prasannapakkiam
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...

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

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 $$n = m$$.

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Gib Z
Homework Helper
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.

Kurdt
Staff Emeritus
Gold Member
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...
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.

D H
Staff Emeritus
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.

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Gib Z
Homework Helper
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 =)

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.
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.

Gib Z
Homework Helper
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.

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.
When m isn't equal to n then we are left with a^(m-n). No doubt about that. The problem is that when n=m and we cancel them out, then we end up with 1/1=1. How do you go from there to concluding that a^n/a^n=a^(n-n)?

VietDao29
Homework Helper
When m isn't equal to n then we are left with a^(m-n). No doubt about that. The problem is that when n=m and we cancel them out, then we end up with 1/1=1. How do you go from there to concluding that a^n/a^n=a^(n-n)?
Well, it's just the way we define things. How would you go about computing: 2-3, and stuff like that?
We notice a nice property of exponential:
$$\frac{x ^ 5}{x ^ 3} = \frac{x \times x \times x \times x \times x }{x \times x \times x} = x ^ 2 = x ^ {5 - 3}$$, so we define:

$$x ^ {m - n} = \frac{x ^ m}{x ^ n}$$, i.e, this equation holds true for every m, and n, including m = n, and m < n.

Then, we expand this to the real, and have:
$$x ^ {\alpha - \beta} = \frac{x ^ \alpha}{x ^ \beta}, \ \ \ \ \ \alpha , \ \beta \in \mathbb{R}$$

Well, it's just the way we define things.
My point, exactly. Either we need to define a^0=1 for nonzero a and then derive identities like: a^n/a^m=a^(n-m) or we will have to just assume that a^n/a^m=a^(n-m) and then derive a^0=1 from that as neutrino did.

It has always been my impression that a^0 = 1 in order to make the function a^x continuous at x = 0. Defining a^k/l as the lth root of a raised to the power k determines this definition. As for $$\frac{a^{m}}{a^{n}}$$, it's the same principle. As n, which is kept smaller than m, approaches m, the expression m - n approaches 0 and a^m -n approaches 1 consequently. The only thing to prove here is that, according to our definitions, if

$$a/b > c/d > 0$$ then

$$x^{a/b} > x^{c/d} > 1$$

Once this is proven, we can show that the limit as the rational number k in x^k approaches 0 is 1. It's clear that as k goes to infinity and hence k goes to 0, a^1/k approaches 1. Hence, we could always chose a/b < 1/k for any k hence showing that as ab approaches 0, x^a/b approaches 1. For negative powers, a similar treatment is needed and we need only to show that positive and negative powers make up a continuous and differentiable function at x = 0.

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Hurkyl
Staff Emeritus