# Why is $\frac{x}{0}$ undefined? [duplicate]

Possible Duplicate:
Division by $0$

The way I see things (and I know most will scoff) is some math doesn't align to common sense. Why does $x^{0}=1$? You are taking x zero times. Why not $x^{0}=0$?

$\frac{x}{0}$ is undefined. x is divided into zero parts. Why not zero?

Thus are my standing thoughts. The only exception is I can not explain $\frac{0}{0}$. Maybe infinity, or $\mathbb{R}$?

There is obviously something I am missing. What is wrong with my logic here?

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## marked as duplicate by Austin Mohr, Asaf Karagila, Zev ChonolesFeb 6 '12 at 7:14

$x^{0}=x^{1-1}=\frac{x^{1}}{x^{1}}=\frac{x}{x}=1$ for $x\neq 0$. –  Thomas E. Feb 6 '12 at 7:02
1-1 needs not simplification before application to x? –  Sean Pedersen Feb 6 '12 at 7:06
$\frac x 0=y$ would mean $0\times y=x$. –  azarel Feb 6 '12 at 7:08
I don't think I quite understand your comment. We know that $0=1-1$ and we only used some basic properties of exponents that are not dependent on the choice of $x$, as long as $x\neq 0$. –  Thomas E. Feb 6 '12 at 7:09
What I meant is doesn't 1-1 resolve to zero before being applied to x as an exponent? –  Sean Pedersen Feb 6 '12 at 7:13