# Simple Proof That Division By Zero Is Impossible

## Main Question or Discussion Point

x / y = z, so z * y = x
1 / 0 = x, so x * 0 = 1
But 0 does not equal 1, so x / 0 is unsolvable.
Oh, and I'm new to forums so if this shouldn't be here you can delete it.

Related Linear and Abstract Algebra News on Phys.org
Perhaps you want to look at:

$\underbrace{lim}_{i->0} \frac{Sin(x)}{x} = 1$

Simon Bridge
Homework Helper
You mean:
$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$$

... mathematicians can set up numbers to do all kinds of things.

Hurkyl
Staff Emeritus
Gold Member
x / y = z, so z * y = x
1 / 0 = x, so x * 0 = 1
But 0 does not equal 1, so x / 0 is unsolvable.
Oh, and I'm new to forums so if this shouldn't be here you can delete it.
This argument makes the assumption that, whenever x/y is defined, the equation y(x/y) = x holds. It also makes the assumption that x0 is always defined, and it is 0.

Both of these assumptions are, indeed, true for the real number system, and the complex number system. But other important number systems do not have these properties.

RamaWolf is talking about something different (although some people have trouble distinguishing the two).

HallsofIvy
Homework Helper
x / y = z, so z * y = x
1 / 0 = x, so x * 0 = 1
But 0 does not equal 1, so x / 0 is unsolvable.
Oh, and I'm new to forums so if this shouldn't be here you can delete it.
Yes, that's a perfectly valid proof that the number 0 (or the additive identity in any field with more than one member) does not have an inverse.

Perhaps you want to look at:

$\underbrace{lim}_{i->0} \frac{Sin(x)}{x} = 1$
True but has nothing to do with the topic here.

Oh, and I've been thinking: does 0 / 0 = 1, or is it undefined?

Oh, and I've been thinking: does 0 / 0 = 1, or is it undefined?
In most number systems 0/0 is undefined. Only a few exotic number systems have defined it and that is outside my scope.