# Thread: why is x^0 = 1?

1. ## why is x^0 = 1?

I do not understand!!! Can some one please explain it to me.

2. ## Re: why is x^0 = 1?

Originally Posted by sakonpure6
I do not understand!!! Can some one please explain it to me.
xy = eln(x)y

x0 = eln(x)*0 = e0 = Exp[0] = 1

3. ## Re: why is x^0 = 1?

I am sorry, but this is a bit too advanced for me, mind explaining what e ^ ln(x)y means?

4. ## Re: why is x^0 = 1?

Originally Posted by sakonpure6
I do not understand!!! Can some one please explain it to me.
Here is a less obscure reason.
First of all, there is some debate about what $\displaystyle 0^0$ could mean.

So let's say that $\displaystyle x\ne 0~.$
Then $\displaystyle \frac{x}{x}=1$ but that means $\displaystyle \frac{x^1}{x^1}=1$.

So by the laws of exponents we have $\displaystyle 1=\frac{x^1}{x^1}=x^{1-1}=x^0$

5. ## Re: why is x^0 = 1?

Originally Posted by sakonpure6
I am sorry, but this is a bit too advanced for me, mind explaining what e ^ ln(x)y means?
hmm, I'm wondering how to show this w/o using log and exponential functions then.

xa = x(0+a) = x0 * xa

therefore x0 = 1

Is that any more digestible?

6. ## Re: why is x^0 = 1?

How did we conclude x^0 dont we generaly add the exponents when multiplying them? And how did you conclude the1? I really appreciate it.

@Plato, Ooohh, I see. Thanks

7. ## Re: why is x^0 = 1?

Originally Posted by sakonpure6
How did we conclude x^0 dont we generaly add the exponents when multiplying them? And how did you conclude the1? I really appreciate it.
if z = c * z then c=1 as 1 is the multiplicative identity on the real numbers. ok?

we have xa = x0 * xa

i.e. z = xa and c = x0

so x0 must equal 1

8. ## Re: why is x^0 = 1?

Originally Posted by romsek
if z = c * z then c=1 as 1 is the multiplicative identity on the real numbers. ok?
we have xa = x0 * xa
i.e. z = xa and c = x0
so x0 must equal 1
Only if $\displaystyle x\ne 0$.

One can see here how complicated this question has become.

9. ## Re: why is x^0 = 1?

I didn't want to open up that can of worms.

10. ## Re: why is x^0 = 1?

From the index law \displaystyle \displaystyle \begin{align*} \frac{a^m}{a^n} = a^{m - n} \end{align*}, in the case where \displaystyle \displaystyle \begin{align*} m = n \end{align*} we have

\displaystyle \displaystyle \begin{align*} LHS &= \frac{a^m}{a^n} \\ &= \frac{a^m}{a^m} \\ &= 1 \end{align*}

but we also have

\displaystyle \displaystyle \begin{align*} RHS &= a^{m - n} \\ &= a^{m - m} \\ &= a^0 \end{align*}

Thus \displaystyle \displaystyle \begin{align*} a^0 = 1 \end{align*}.

This is of course provided that \displaystyle \displaystyle \begin{align*} a \neq 0 \end{align*}.

11. ## Re: why is x^0 = 1?

its only by definition.... no need for ..proofs...they do not mean anything...

12. ## Re: why is x^0 = 1?

Originally Posted by MINOANMAN
its only by definition.... no need for ..proofs...they do not mean anything...
That is simply not the case.

13. ## Re: why is x^0 = 1?

Whoops. Too fast. Already done by Plato and Romsek (integer exponents). Missed it. Oh well, I'm not the only one.
My apologies.

14. ## Re: why is x^0 = 1?

Originally Posted by Plato
That is simply not the case.
yes it is the case mathematics is not just computations...it is philosophy..x^0 is meaningless in mathematics unless you define it...you cannot just supply "proofs"...x^0 is by definition =1 as it is 0! =1 .... No Further comments..

15. ## Re: why is x^0 = 1?

Originally Posted by MINOANMAN
yes it is the case mathematics is not just computations...it is philosophy..x^0 is meaningless in mathematics unless you define it...you cannot just supply "proofs"...x^0 is by definition =1 as it is 0! =1 .... No Further comments..
The debate as to whether 0^0 is equal to 1 or 0 is philosophy. x^0 = 1 (for non-zero x) is not philosophy...It can be unambiguously derived from the laws of exponents, as has been shown in several posts in this thread.

-Dan

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