My simple proof of x^0 = 1

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  • #1
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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
 

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  • #2
Kurdt
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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:

[tex] n^x = e^{x\ln{n}} [/tex]

Now put x = 0 and see what happens.
 
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  • #3
Zurtex
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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:

[tex] x^n = e^{n\ln{x}} [/tex]

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.
 
  • #4
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[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]


What could be more simpler than that?
 
  • #5
Kurdt
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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.
 
  • #6
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[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]


What could be more simpler than that?
I like this proof best. And I also think it is the formal proof.
 
  • #7
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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:

[tex] x^n = e^{n\ln{x}} [/tex]

Now put n = 0 and see what happens.
I don't really get it, would you explain more please?
 
  • #8
Kurdt
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I don't really get it, would you explain more please?
If you set n = 0 you get the following:

[tex] n^0 = e^{0\ln{n}} = e^0 = 1[/tex]
 
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  • #9
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Shing,
[tex]\forall a \ne 0,\;a^n = a^{n - 1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}[/tex]
Given that a1=a,
[tex]a^0 = a^{ - 1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1[/tex]
 
  • #10
StatusX
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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)
 
  • #11
mathman
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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
 
  • #12
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This is not a proof, it's a pattern. x^0 = 1 is defined.
 
  • #13
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...
 
  • #14
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[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]


What could be more simpler than that?
Can you prove that [tex]\frac{a^n}{a^m} = a^{n-m}[/tex] when [tex]n = m[/tex]? Usually the proof is only valid when they aren't equal and then people define [tex]x^0=1[/tex] such that the property remains valid for [tex]n = m[/tex].
 
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  • #15
Gib Z
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Can you prove that [tex]\frac{a^n}{a^m} = a^{n-m}[/tex] when [tex]n = m[/tex]? Usually the proof is only valid when they aren't equal and then people define [tex]x^0=1[/tex] such that the property remains valid for [tex]n = m[/tex].
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
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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.
 
  • #17
D H
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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 [tex]\lim_{x\to0}0^x[/tex] which is obviously zero. I could just as easily make it any complex number!

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

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

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

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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  • #18
Gib Z
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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
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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.
 
  • #20
Gib Z
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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.
 
  • #21
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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)?
 
  • #22
VietDao29
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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:
[tex]\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}[/tex], so we define:

[tex]x ^ {m - n} = \frac{x ^ m}{x ^ n}[/tex], 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:
[tex]x ^ {\alpha - \beta} = \frac{x ^ \alpha}{x ^ \beta}, \ \ \ \ \ \alpha , \ \beta \in \mathbb{R}[/tex]
 
  • #23
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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.
 
  • #24
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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 [tex]\frac{a^{m}}{a^{n}} [/tex], 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

[tex]a/b > c/d > 0 [/tex] then

[tex]x^{a/b} > x^{c/d} > 1[/tex]

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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  • #25
Hurkyl
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Not to be a spoilsport, but none of these are proofs; they are simply demonstrations why we might desire things to behave in a certain way.

A proof would invoke a definition of exponentiation. Once a definition is chosen, the issue is quite clear-cut.
 

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