My simple proof of x^0 = 1

  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
     
  2. jcsd
  3. Kurdt

    Kurdt 4,941
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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.
     
    Last edited: May 30, 2007
  4. Zurtex

    Zurtex 1,123
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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.
     
  5. [tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]


    What could be more simpler than that?
     
  6. Kurdt

    Kurdt 4,941
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    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.
     
  7. I like this proof best. And I also think it is the formal proof.
     
  8. I don't really get it, would you explain more please?
     
  9. Kurdt

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

    [tex] n^0 = e^{0\ln{n}} = e^0 = 1[/tex]
     
    Last edited: May 30, 2007
  10. 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]
     
  11. StatusX

    StatusX 2,567
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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)
     
  12. mathman

    mathman 6,466
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    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
     
  13. This is not a proof, it's a pattern. x^0 = 1 is defined.
     
  14. 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...
     
  15. 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].
     
    Last edited: May 30, 2007
  16. Gib Z

    Gib Z 3,348
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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.
     
  17. Kurdt

    Kurdt 4,941
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    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.
     
  18. D H

    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 [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.
     
    Last edited: May 30, 2007
  19. Gib Z

    Gib Z 3,348
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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 =)
     
  20. 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.
     
  21. Gib Z

    Gib Z 3,348
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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.
     
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