# x^0 and x^1

• Dec 10th 2013, 07:30 PM
nycmath
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?
• Dec 10th 2013, 07:54 PM
romsek
Re: x^0 and x^1
This is discussed in countless places on the web. Here for example
• Dec 10th 2013, 08:19 PM
nycmath
Re: x^0 and x^1
I did not realize that such a basic idea has a complicated definition.
• Dec 10th 2013, 08:39 PM
romsek
Re: x^0 and x^1
Quote:

Originally Posted by nycmath
I did not realize that such a basic idea has a complicated definition.

math has to be bullet proof. Especially in the base definitions.
• Dec 11th 2013, 01:26 AM
Melody2
Re: x^0 and x^1
Quote:

Originally Posted by nycmath
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.
• Dec 11th 2013, 05:48 AM
nycmath
Re: x^0 and x^1
Quote:

Originally Posted by Melody2
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.
• Dec 28th 2013, 04:15 PM
HallsofIvy
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 "$\displaystyle a^n$, for n positive integer, is a multiplied by itself n times" From that, it follows immediately that $\displaystyle x^1= x$. To get $\displaystyle x^0$, 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 $\displaystyle a^{n+ m}= a^na^m$. That is such a nice property that we would like to define $\displaystyle a^0$ in such a way as to have that still true. That is, we want $\displaystyle a^{n+ 0}= a^na^0$. On the left, n+ 0= n so $\displaystyle a^{n+0}= a^n$. That mean that, to have $\displaystyle a^{n+0}= a^n= a^na^0$, we must define $\displaystyle a^0= 1$. (Part of showing that $\displaystyle a^{n+m}= a^na^m$ requires that a not be equal to 0. So we really have "$\displaystyle a^0= 1$ as long as a is not 0" and $\displaystyle 0^0$ is not defined.)
• Dec 28th 2013, 05:34 PM
Prove It
Re: x^0 and x^1
Quote:

Originally Posted by nycmath
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 \displaystyle \begin{align*} x^0 \end{align*} is not \displaystyle \displaystyle \begin{align*} 1 \end{align*} for any integer. \displaystyle \displaystyle \begin{align*} x^0 = 1 \end{align*} for all \displaystyle \displaystyle \begin{align*} x \neq 0 \end{align*}. \displaystyle \displaystyle \begin{align*} 0^0 \end{align*} is undefined.
• Dec 28th 2013, 07:22 PM
sakonpure6
Re: x^0 and x^1
Simply: $\displaystyle x^3 / x^3 = 27 / 27 = 1$ Now $\displaystyle x^3/x^3 = x^(3-3) (according to exponential law) = x^0 = 1$
• Dec 28th 2013, 07:43 PM
MINOANMAN
Re: x^0 and x^1
Quote:

Originally Posted by HallsofIvy
You don't have to use a "complicated definition". For this question, it is sufficient to use the definition "$\displaystyle a^n$, for n positive integer, is a multiplied by itself n times" From that, it follows immediately that $\displaystyle x^1= x$. To get $\displaystyle x^0$, 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 $\displaystyle a^{n+ m}= a^na^m$. That is such a nice property that we would like to define $\displaystyle a^0$ in such a way as to have that still true. That is, we want $\displaystyle a^{n+ 0}= a^na^0$. On the left, n+ 0= n so $\displaystyle a^{n+0}= a^n$. That mean that, to have $\displaystyle a^{n+0}= a^n= a^na^0$, we must define $\displaystyle a^0= 1$. (Part of showing that $\displaystyle a^{n+m}= a^na^m$ requires that a not be equal to 0. So we really have "$\displaystyle a^0= 1$ as long as a is not 0" and $\displaystyle 0^0$ 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.
• Dec 28th 2013, 09:12 PM
sakonpure6
Re: x^0 and x^1
Quote:

Originally Posted by sakonpure6
Simply: $\displaystyle x^3 / x^3 = 27 / 27 = 1$ Now $\displaystyle x^3/x^3 = x^(3-3) (according to exponential law) = x^0 = 1$

Assuming x E R , x!= 0 ( x= 3 in this case)
• Dec 29th 2013, 04:16 AM
nycmath
Re: x^0 and x^1
Thank you everyone for your help.