# Homework Help: Solution to 0/0, i.e., dividing by zero

1. Mar 22, 2008

### RogerDodgr

1. The problem statement, all variables and given/known data
0/0=?

(This actually came up in a homework assignment). My teacher dismissed it as d.n.e. but I actually believe it has the solution set of 'all real numbers'; it has pertinence to a specific problem in calc III that I find interesting. I have read some contradictory statements on the web.

3. The attempt at a solution
I'm thinking of division as 'how many times a denominator can go into a numerator':
e.g. 8/2 = 4 because 2 goes into 8 four times.

Following this logic, how many times can zero go into zero? one,two, three... how ever many times you want. Therefore 0/0 has the solution set of 'all real numbers'.

Is there an error in my logic?

2. Mar 22, 2008

### Air

Erm...I would have thought that you cannot divide by zero hence this may be a trick question.

3. Mar 22, 2008

### RogerDodgr

It's not a trick question. I honestly am not sure that I am right; that's why I asked :)
Is there a flaw in my logic?

4. Mar 22, 2008

### arunbg

IMO 0/0 is akin to asking how many times can "nothing" go into "nothing", and the answer would be that the value is indefinite(not defined). Compare this with the division of other real numbers by 0, then we could say "nothing" goes any number of times into "something"(though for negative numbers this seems dubious, we might say how many times the absence of nothing goes into the absence of something), thus giving a value that tends to + or - infinity.

Last edited: Mar 22, 2008
5. Mar 22, 2008

### HallsofIvy

No 0/0 is not "asking how many times can "nothing" go into "nothing". Division has a very specific mathematical definition, not necessarily the way you learned division in elementary school. In most definitions, division means "divide by the multiplicative inverse". Since 0 does not have a multiplicative inverse, you cannot divide by it- not even 0.

Yes, we can argue that a/b= x means that a= bx. And since 0= 0*x for any x, you could argue that "0/0" can be any number. But that assumes that you can divide by 0- and you can't. Technically, your teacher is right. If you really want to argue with your teacher, go ahead. But I wouldn't recommend it.

In terms of limits, if, looking at numerator and denominator separately, you arrive at a/0, then the limit does not exist. If you arrive at 0/0, then the limit might or might not exist- you have to look more carefully. But that's limits.

6. Mar 22, 2008

### Schrodinger's Dog

In simple language he means it is not defined, thus it does not exist. Although saying that in some forms of maths it exists, I doubt you're at that level though. So your teacher is correct.

In calculus it d.n.e. without redefining the laws of calculus in a way where division 0 by 0 exists.

Consider what 0 represents, now in set theory or number theory you might define it in such a way that it exists, but that is nothing like calculus, and to be frank it's also pretty useless, and it's also pretty useless in the maths you are doing.

In short in no field of maths is the term 0/0 actually a useful explanation, except in some really odd ways of defining zero, which are pure maths but not useful in terms of maths in your course.

Last edited: Mar 22, 2008