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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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$x^{0}=x^{1-1}=\frac{x^{1}}{x^{1}}=\frac{x}{x}=1$ 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
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marked as duplicate by Austin Mohr, Asaf Karagila, Zev Chonoles Feb 6 '12 at 7:14

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1 Answer

X/n, where n is any number, is neither defined, because you haven't defined any. By dividing you "say": "How many times is the denominator the numerator?". Zero can fill any real number infinite times.

I suppose you wanted to say what's the limit when the denominator tends to 0. It fully depends on the ratio of the slopes of both functions.If the denominator is x, then the limit is 1. If the denominator is x^2, it's 0. That's why it's undefined

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Thanks. You were a great help. –  Sean Pedersen Feb 6 '12 at 7:19
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