why is x^0 = 1?

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

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Originally Posted by sakonpure6

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

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x^y = e^ln(x)y

x⁰ = e^ln(x)*0 = e⁰ = Exp[0] = 1

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

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Originally Posted by sakonpure6

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

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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 $\displaystyle 1=\frac{x^1}{x^1}=x^{1-1}=x^0$

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Originally Posted by sakonpure6

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

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hmm, I'm wondering how to show this w/o using log and exponential functions then.

how about this then?

x^a = x^(0+a) = x⁰ * x^a

therefore x⁰ = 1

Is that any more digestible?

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 :)

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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.

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if z = c * z then c=1 as 1 is the multiplicative identity on the real numbers. ok?

we have x^a = x⁰ * x^a

i.e. z = x^a and c = x⁰

so x⁰ must equal 1

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Originally Posted by romsek

if z = c * z then c=1 as 1 is the multiplicative identity on the real numbers. ok?
we have x^a = x⁰ * x^a
i.e. z = x^a and c = x⁰
so x⁰ must equal 1

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Only if $\displaystyle x\ne 0$.

One can see here how complicated this question has become.

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

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 $\displaystyle \displaystyle \begin{align*} a \neq 0 \end{align*}$.

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

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Originally Posted by MINOANMAN

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

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That is simply not the case.

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

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Originally Posted by Plato

That is simply not the case.

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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..

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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..

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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