This is discussed in countless places on the web. Here for example

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- Dec 10th 2013, 08:30 PM #1

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- Dec 10th 2013, 08:54 PM #2

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- Dec 10th 2013, 09:19 PM #3

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- Dec 10th 2013, 09:39 PM #4

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- Dec 11th 2013, 02:26 AM #5

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## Re: x^0 and x^1

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.

- Dec 11th 2013, 06:48 AM #6

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- Dec 28th 2013, 05:15 PM #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 " , for n positive integer, is a multiplied by itself n times" From that, it follows immediately that . To get , 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 . That is such a nice property that we would like to define in such a way as to have that still true. That is, we want . On the left, n+ 0= n so . That mean that, to have , we must define . (Part of showing that requires that a not be equal to 0. So we really have " as long as a is not 0" and is not defined.)

- Dec 28th 2013, 06:34 PM #8

- Dec 28th 2013, 08:22 PM #9

- Dec 28th 2013, 08:43 PM #10

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## Re: x^0 and x^1

- Dec 28th 2013, 10:12 PM #11

- Dec 29th 2013, 05:16 AM #12

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