Why is division by zero undefined?

[Beer w/Straw]

This question seems to befall everyone at one point or another. So much so I begin to get deliberately silly when it is asked http://www.wolframalpha.com/input/?i=Abs[1/0]

Anyway, I'm wondering if there is a sticky present on these forums that addresses it specifically. Something besides mathworld or wiki.

Thanks.

 

[DivisionByZro]

Here's a quick rundown: You can't divide by zero because if you re-arrange the following:
<br /> \frac{x}{y} = w<br />
Multiplying each side of the equation by y:
<br /> y\cdot w = x<br />

And we are looking at the case where y=0 and x=anything, then:

<br /> \frac{3}{0} = w<br />
Which we put as:
<br /> 0\cdot w = 3 <br />

So here we are asking, "what number, w, times 0 (zero), will give 3?" (or anything non-zero)

But, any number times 0 must be zero (by definition). So it doesn't make sense to divide by zero. Moreover, this is called a one-to-many operation, because instead of 3 we could have chose any other number, and we would still be in the same boat. So it's not hard to see why we leave division by zero undefined, most of the time it simply does not make much sense.

Here's another explanation by one of our mentors:
https://www.physicsforums.com/showthread.php?t=530207

 

[Beer w/Straw]

Thanks, I just found that sticky.

My bad for not doing a little searching first.

 

[D H]

This question seems to befall everyone at one point or another. So much so I begin to get deliberately silly when it is asked http://www.wolframalpha.com/input/?i=Abs[1/0]

There is not a simple question of 1/0 being undefined. There is no ambiguity to abs(1/0).

The reason 1/0 is undefined is because \lim_{x\to0} \frac 1 x does not exist. It's either +∞ or -∞, depending on the direction from which x approaches zero. Extending to the complex domain, \lim_{z\to0} \frac 1 z, doesn't help. Now you get a number with infinite magnitude but unknown direction.

On the other hand, \lim_{z\to 0} \left|\frac 1 z\right| does exist. It is +∞, and this is exactly what Mathematica reports.