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Math Help - x^0 and x^1

  1. #1
    Super Member nycmath's Avatar
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    x^0 and x^1

    I know that x^0 = 1 and x^1 = x where x is any integer. Why is that the case? How can I best understand this concept of zero and one powers without getting too technical?
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  2. #2
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    Re: x^0 and x^1

    This is discussed in countless places on the web. Here for example
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  3. #3
    Super Member nycmath's Avatar
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    Re: x^0 and x^1

    I did not realize that such a basic idea has a complicated definition.
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    Re: x^0 and x^1

    Quote Originally Posted by nycmath View Post
    I did not realize that such a basic idea has a complicated definition.
    math has to be bullet proof. Especially in the base definitions.
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    Re: x^0 and x^1

    Quote Originally Posted by nycmath View Post
    I know that x^0 = 1 and x^1 = x where x is any integer. Why is that the case? How can I best understand this concept of zero and one powers without getting too technical?
    Hi Nycmath,
    I have been trying to learn Latex and I used your post for practice inspiration.
    I finally got it to look almost how I wanted so this is what I wanted to say.

    Unfortunately I don't know how to transfer the program that I built in TEX to this site. Maybe someone can help me with that, so I had to snip the output and attach it as a thumbnail.

    It might be what you were after.
    Attached Thumbnails Attached Thumbnails x^0 and x^1-x-0-x-1.jpg  
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  6. #6
    Super Member nycmath's Avatar
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    Re: x^0 and x^1

    Quote Originally Posted by Melody2 View Post
    Hi Nycmath,
    I have been trying to learn Latex and I used your post for practice inspiration.
    I finally got it to look almost how I wanted so this is what I wanted to say.

    Unfortunately I don't know how to transfer the program that I built in TEX to this site. Maybe someone can help me with that, so I had to snip the output and attach it as a thumbnail.

    It might be what you were after.
    I myself never learned Latex. I do not have a computer or laptop. All my typing is done using my tiny cell phone keyboard.
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  7. #7
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    Re: x^0 and x^1

    You don't have to use a "complicated definition". For this question, it is sufficient to use the definition " a^n, for n positive integer, is a multiplied by itself n times" From that, it follows immediately that x^1= x. To get x^0, we have to extend that original definition to 0, which is not a "positive integer". We are, of course, free to define something however we want, but from that previous definition, it is easy to show that a^{n+ m}= a^na^m. That is such a nice property that we would like to define a^0 in such a way as to have that still true. That is, we want a^{n+ 0}= a^na^0. On the left, n+ 0= n so a^{n+0}= a^n. That mean that, to have a^{n+0}= a^n= a^na^0, we must define a^0= 1. (Part of showing that a^{n+m}= a^na^m requires that a not be equal to 0. So we really have " a^0= 1 as long as a is not 0" and 0^0 is not defined.)
    Last edited by HallsofIvy; December 28th 2013 at 04:18 PM.
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  8. #8
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    Re: x^0 and x^1

    Quote Originally Posted by nycmath View Post
    I know that x^0 = 1 and x^1 = x where x is any integer. Why is that the case? How can I best understand this concept of zero and one powers without getting too technical?
    \displaystyle \begin{align*} x^0 \end{align*} is not \displaystyle \begin{align*} 1 \end{align*} for any integer. \displaystyle \begin{align*} x^0 = 1 \end{align*} for all \displaystyle \begin{align*} x \neq 0 \end{align*}. \displaystyle \begin{align*} 0^0 \end{align*} is undefined.
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  9. #9
    Senior Member sakonpure6's Avatar
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    Re: x^0 and x^1

    Simply: x^3 / x^3  =  27 / 27  = 1 Now x^3/x^3   =  x^(3-3) (according to exponential law)  = x^0 = 1
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    Re: x^0 and x^1

    Quote Originally Posted by HallsofIvy View Post
    You don't have to use a "complicated definition". For this question, it is sufficient to use the definition " a^n, for n positive integer, is a multiplied by itself n times" From that, it follows immediately that x^1= x. To get x^0, we have to extend that original definition to 0, which is not a "positive integer". We are, of course, free to define something however we want, but from that previous definition, it is easy to show that a^{n+ m}= a^na^m. That is such a nice property that we would like to define a^0 in such a way as to have that still true. That is, we want a^{n+ 0}= a^na^0. On the left, n+ 0= n so a^{n+0}= a^n. That mean that, to have a^{n+0}= a^n= a^na^0, we must define a^0= 1. (Part of showing that a^{n+m}= a^na^m requires that a not be equal to 0. So we really have " a^0= 1 as long as a is not 0" and 0^0 is not defined.)
    Thanks for clearing the misunderstandings . Some people of this forum .....tried to give a "proof" ... excellent aproach... I don't know you personally but I have a clear idea that you are a good Mathematician.
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  11. #11
    Senior Member sakonpure6's Avatar
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    Re: x^0 and x^1

    Quote Originally Posted by sakonpure6 View Post
    Simply: x^3 / x^3  =  27 / 27  = 1 Now x^3/x^3   =  x^(3-3) (according to exponential law)  = x^0 = 1
    Assuming x E R , x!= 0 ( x= 3 in this case)
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  12. #12
    Super Member nycmath's Avatar
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    Re: x^0 and x^1

    Thank you everyone for your help.
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