Division by ZERO!!!

**Shahil** · Aug20-04, 11:09 AM

Hey guys!!!

Was just pondering...

ALL my mathematical subject lecturers go on and on and on about trying to avoid division by zero. Fine - after all, it makes logical nonsense to divide by zero but my pondering thoughts led me somewhere...

Is it possible to do some maths with the division by zero?? You know - like a NEW number system. So far - I haven't found one. Can anyone help me here??

**Muzza** · Aug20-04, 11:12 AM

http://www.math.su.se/~jesper/research/wheels/

**humanino** · Aug20-04, 11:17 AM

it depends what you mean by division by zero. Sometimes $LaTeX Code: \\epsilon^2=0$ is taken as $LaTeX Code: \\epsilon$ divides zero. In that case, you can perform symbolic calculations with $LaTeX Code: \\epsilon$ exactly as you do with imaginary numbers. Indeed, $LaTeX Code: \\epsilon$ is part of the generalized imaginary numbers. It has applications in geometry.

**Shahil** · Aug20-04, 11:23 AM

So does that mean 0/0 is a complex number???

**humanino** · Aug20-04, 11:40 AM

No no no ! Generalized complex numbers !

See, "i" is just defined as the root of an equation which has none in R(eals). Namely

. The idea is : I define the new "imaginary" number "j" as the root of a convenient equation. That could be

for instance. Or

as well ! Depending on how "relevant" your initial equation is, you get more or less interesting results.

**humanino** · Aug20-04, 11:42 AM

I don't remember quite well, because I never used them. I think the $LaTeX Code: \\epsilon$ example has to do with a geometry in which lines can be parallel or anti-parallel. They have orientation. I can't elaborate on this.

**ArmoSkater87** · Aug21-04, 06:09 AM

Originally Posted by Shahil

Is it possible to do some maths with the division by zero?? You know - like a NEW number system. So far - I haven't found one. Can anyone help me here??

Im not sure what u mean by a new number system, but no, you cant use dividing by zero in math because it leads to ilogical anwers. Let me show you an example...
a = b
a^2 = b^2
a^2 - b^2 = a^2 - ab
(a+b)(a-b) = a(a-b)
a + b = a
a + a = a
2a = a
2 = 1

Obviously this cant be right? 2 doesnt equal 1, so how can this be?
Well, in the math you do a step where you divide the factor (a-b) from both sides. Since a=b, then (a-b) = 0, and dividing by zero isnt allowed.

**Alkatran** · Aug22-04, 06:19 PM

Originally Posted by Shahil

Hey guys!!!

Was just pondering...

ALL my mathematical subject lecturers go on and on and on about trying to avoid division by zero. Fine - after all, it makes logical nonsense to divide by zero but my pondering thoughts led me somewhere...

Is it possible to do some maths with the division by zero?? You know - like a NEW number system. So far - I haven't found one. Can anyone help me here??

There would have to be a set of rules which stopped 0/0 from returning any result you want. For example, you could say that x*0 <> 0, 0/x <> 0, and 0/0 = 1. If we can't reduce numbers using 0 then we don't have any problems.

Take ArmoSkater87's example:
a = b
a^2 = b^2
a^2 - b^2 = a^2 - ab
(a+b)(a-b) = a(a-b)
a + b = a
a + a = a
2a = a
2 = 1

Step 4 to Step 5 would violate x*0 <> 0 (or in this case, (a+b)*0 <> 0), stopping the error from ever being reached.

**Alkatran** · Aug23-04, 07:06 PM

I just wanted to know if there was any flaws in my above post. I don't want to waste my time thinking about this.

**HallsofIvy** · Aug23-04, 09:26 PM

The only flaw is that any structure in which 0/0 is defined is either the trivial structure containging ONLY 0, or is not a field. Yes, you can define structures that are not fields and they may be interesting, but they are a little too general.

**Alkatran** · Aug23-04, 11:08 PM

Originally Posted by HallsofIvy

The only flaw is that any structure in which 0/0 is defined is either the trivial structure containging ONLY 0, or is not a field. Yes, you can define structures that are not fields and they may be interesting, but they are a little too general.

Can you give an example, following the rules I gave (x*0 <> 0, 0/x <> 0, 0/0 = 1) that doesn't make sense?

**Hurkyl** · Aug23-04, 11:43 PM

Can you give an example, following the rules I gave (x*0 <> 0, 0/x <> 0, 0/0 = 1) that doesn't make sense?

Yes.

0 = x*0 - x*0 = x * (0-0) = x * 0
Which contradicts x*0 <> 0
1 = 0 / 0 = (0+0) / 0 = 0/0 + 0/0 = 1 + 1 = 2

The laws of arithmetic (specifically the ring axioms) prove that x*0 = 0. (My first example above is the proof) Division is defined as the inverse of multiplication.

If you're working with some system where x*0 is not equal to zero, or where division is not the inverse of multiplication, then you are not doing arithmetic; you're working on some other, strage, system.

**mathwonk** · Aug24-04, 01:20 AM

the only sense in which division by zero works in my experience is when we consider a function like [(x-a)(f(x))]/(x-a), and plug in x=a, geting f(a).

I.e. whenever we take the limit, of [g(x)-g(a)]/(x-a), in some sense we are dividing zero by zero.

**Alkatran** · Aug24-04, 08:41 AM

Originally Posted by Hurkyl

Yes.

0 = x*0 - x*0 = x * (0-0) = x * 0
Which contradicts x*0 <> 0
1 = 0 / 0 = (0+0) / 0 = 0/0 + 0/0 = 1 + 1 = 2

The laws of arithmetic (specifically the ring axioms) prove that x*0 = 0. (My first example above is the proof) Division is defined as the inverse of multiplication.

If you're working with some system where x*0 is not equal to zero, or where division is not the inverse of multiplication, then you are not doing arithmetic; you're working on some other, strage, system.

Nice post. I was aware that x*0 = 0 is not an error, I was just saying if you put some rules in, operations on 0 wouldn't result in asymptotes.

Another good example would have been log[0](x) = ? (0 to the power of ? gives x) etc etc... You'd actually have to say any operation involving 0 is unsolvable unless it is cancelled by another 0.

Question: Some of the equations involving 0 contradict 1 = 1, but they are assumed wrong because 1=1 is one of the basic laws of math. I would argue that x*0 - x*0 = x * (0-0) is false (IN THE RULED SYSTEM) because to get from x*0 - x*0 you had to do an operation involving 0.... but this is all nonsense anyways. There's no point in having none of the 0 operations solvable instead of half just to give operations involving 0 twice sensable.

**matt grime** · Aug24-04, 08:45 AM

"I would argue that x*0 - x*0 = x * (0-0) is false"

eh? when did 0 not become equal to zero?

**Hurkyl** · Aug24-04, 03:04 PM

Some of the equations involving 0 contradict 1 = 1, but they are assumed wrong because 1=1 is one of the basic laws of math.

Well, your premise is inaccurate.

These arguments aren't "assumed wrong": they are invalid. The reason they are invalid is not because they "contradict 1 = 1", but because there is a step that does not logically follow from previous steps.

For example, in the argument in #7, "(a + b) = a" does not logically follow from "(a+b)(a-b) = a(a-b)", or from any combination of previous steps.

This flaw is the sole reason the argument is deemed invalid.

Incidentally, there is a "law" of arithmetic that says

If uw = vw, then u = v or w = 0 (or both).

I would argue that x*0 - x*0 = x * (0-0) is false

And your argument is invalid. Not only is no law, theorem, axiom, or anything that says "An equation involving operations with 0 is false".

There is a very specific collection of axioms that are true for arithmetic, period. These axioms vary slightly for different systems (e.g. integers, complexes), but the point is that arithmetic is very clear and precise.