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?

Results 1 to 12 of 12

9*Thanks*

- Dec 10th 2013, 07:30 PM #1

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- Dec 10th 2013, 07:54 PM #2

- Joined
- Nov 2013
- From
- California
- Posts
- 5,565
- Thanks
- 2348

- Dec 10th 2013, 08:19 PM #3

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- Dec 10th 2013, 08:39 PM #4

- Joined
- Nov 2013
- From
- California
- Posts
- 5,565
- Thanks
- 2348

- Dec 11th 2013, 01:26 AM #5

- Joined
- Nov 2013
- From
- Australia
- Posts
- 261
- Thanks
- 57

## 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, 05:48 AM #6

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7

- Dec 28th 2013, 04:15 PM #7

- Joined
- Apr 2005
- Posts
- 18,948
- Thanks
- 2737

## 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, 05:34 PM #8

- Dec 28th 2013, 07:22 PM #9

- Dec 28th 2013, 07:43 PM #10

- Joined
- Feb 2013
- From
- Saudi Arabia
- Posts
- 445
- Thanks
- 86

## Re: x^0 and x^1

- Dec 28th 2013, 09:12 PM #11

- Dec 29th 2013, 04:16 AM #12

- Joined
- Jan 2013
- From
- New York City
- Posts
- 658
- Thanks
- 7