3 ms·

Why can't we divide by zero?

37 points by jwilliams 18y ago

speek 18y ago
because the math police will come after us.
- eipiman 18y ago
  Suppose 5/0 = infinity. What does it buy you? At the very least, one would expect that 5/0 = infinity implies that 5 = 0 * infinity, but that is not the case. And if you insist that it is, does 4/0 = infinity then imply that 4 = 0 * infinity, and hence 4 = 5?
  - speek 18y ago
    Infinity is not a number... its a concept. Say you were to count every integer, you would count to infinity. Now what if you were to count every number between 0 and 1? Would it still be infinity? I've always thought in degrees of infinity, but either way its still not a number. If you kept the degrees of infinity separate, you could divide 5 by zero and it would not be the same as dividing 4 by zero. If you wanted to play with limits, you could say the limit of 5/x as x approaches 0 is infinity, but we don't really have a name for what x/0 would be. Someone did come up with nullity, but that's not good enough. Plus, you can't have Zero of something. Having implies a positive net, thus it's not correct to say that you have zero of something... you could say you don't have something, but they're semantically different. Zero is just another concept.
    - abstractbill 18y ago
      Take a look at Cantor's work, if you're interested in this stuff. The number of integers is known as a countable infinity. The number of numbers between 0 and 1 is called an uncountable infinity. The proof that these are not the same (Cantor's diagonalization argument) is my favorite proof - it's really neat. It's quite clear and maybe even obvious in retrospect (I like to explain it to friends who are curious about higher mathematics), but probably not something you would have come up with if you weren't Cantor! You can ask if there's an infinity between these two, and that's actually unprovable. You can prove that, without adding more axioms to mathematics, it's impossible to prove one way or another whether there's an infinity between them. Sometimes I really miss this stuff.
      - Retric 18y ago
        I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. PS: This is why I went to CS there is no way to verify that a given proof or program is accurate but that does not stop a program from being useful. Edit: Restating the second argument: Start with all zero's, count in binary from the left, 1000.. > 0100.. > 1100.. > ... and the diagonal becomes 1111111 which is basically just a higher order of infinity like N = the sum of all prior numbers.
        hhm 18y ago
        I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. It doesn't work like that. You want to show that you can't enumerate all reals, so what do you do? You start enumerating all reals. Then you build a new number, whose Nth bit is different from the Nth bit of the Nth real number. You can do that because you have enumerated them, and we have never talked about an infinite N so we are never flipping the last bit of the last number nor anything similar. We are only flipping the Nth bit of the Nth number. Number 819439243? We flip the 819439243th bit. That's all we do. What do you do then? Well, you compare that new number you created, with the first number, the second number, and so on. And you see that this number is different to all the other numbers. What does it mean? That it doesn't matter what your enumeration scheme is, there is at least one number that will always be out of your enumeration scheme, and so your enumeration scheme will fail (so this is an absurd also, as we were supposing that this specific enumeration scheme -that could be any enum scheme- was complete). So you don't have as many integers as reals, so QED. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. As you see, it doens't matter, as in this proof you never need to do the sum of all numbers.
        Retric 18y ago
gills 18y ago
A little off topic, but interesting practice for 0. What are the mathematical ramifications of 0% capital reserves? This is a good chance to remind the readers that the bailout bill contains a clause to accelerate a 2006 law to 10/01/2008, which allows the Federal Reserve to set reserve requirements for a bank to 0%.
mystagogue 18y ago
I took a couple of numerical analysis classes from this guy (Peter Alfeld) at the University of Utah and he is a very entertaining teacher.
DanielBMarkham 18y ago
The real number system is discontinuous over division. Interestingly enough, you can divide by zero. But if you do, then you can't define integers.
- abstractbill 18y ago
  For more on this, take a look at the http://en.wikipedia.org/wiki/Riemann_sphere http://en.wikipedia.org/wiki/Riemann_sphere
  - hhm 18y ago
    How does this relate to the original comment? What I want to know is how you can divide by 0, so that then you can't define integers.
    - abstractbill 18y ago
      I think DanielBMarkham is being rather loose with terminology here (or perhaps talking about something other than the Riemann sphere, but it sounds like that's what he's talking about). The problem isn't that you "can't define integers" when you introduce an infinity. The problem is (as the article I linked to says) the resulting system does not form a field (i.e. the usual arithmetic operations don't behave like you would expect anymore). Follow the link to fields if you want to know more.
      - hhm 18y ago
        Oh, I thought he was saying something like you could have real numbers with division by zero as far as you didn't want to have a special class of numbers, the integer numbers. So I was very curious about that, as I studied fields and all that a long while ago and that sounded like a rare concept.
      - hugh 18y ago
        Wait, but the integers don't form a field anyway (not closed under division).
marvin 18y ago
I'm still not convinced that you can't just say that x/0 equals infinity. Certainly, infinity is not a number..but it has the same properties as a transcendental number in the sense that it is impossible to describe it accurately with finitely many digits. You wouldn't be able to use this definition to build other theories, but it would be nice to have a settled definition. Saying that x/0 is undefined still seems to me like a bit of a hack. I agree that 0/0 should be undefined. There are infinitely many possibilities for what it could be, depending on the context.
- hhm 18y ago
  Infinity is a much studied concept in mathematics; the problem is not that we don't know what infinity is but that we know it's not part of the set of integer numbers, real numbers, etc. But there are many ways to work with infinities, and even to include it in the number sets (for example, in the case of hyper-reals).
- derefr 18y ago
  Is it at least true that (0 < |0/0| < Infinity); that is, that 0/0 is a finite, non-zero number?
  - newt0311 18y ago
    No. 0/0 is completely undefined. For example, consider limit as x approaches 0 of (-x)/x. This approaches (-0)/0 = 0/0 but simplifying the limit given an answer of -1. So 0/0 is completely dependent on the characteristics of the convergence and had no meaning in isolation. In fact, the field of real (and complex) numbers does not even include 0 in the multiplicative group. Therefore, dividing by 0 is nonsensical as it has no inverse as it is not in the multiplicative field. According to the basic axioms of groups theory, given a field F with a the identity of the additive group in the field and b some element of the multiplicative group, it can be proven that mul(a,b) = a where mul is the multiplication operation. This explains why 0 times any other number is 0. If you would like a copy of the proof, just drop me an email. Sorry for the rant, math major.
    - 18y ago
jkent 18y ago
I explained this as: Take a number, e.g. 5. Then take a small number, like 0.1. Add 0.1 to itself, and repeat, you'll eventually reach 5. But if you take 0, and keep adding 0 to it, you'll never reach 5. So no amount of 0's will go in to 5. That logic breaks down if you do 0/0 though. My students never picked up on that one but they believed me!
- etal 18y ago
  Picturing the graphs of y = 1/x and y = cx/x worked for me. You can make the "reasonable" value at x=0 be Inf, -Inf, or anything in between.
etal 18y ago
There was a professor who tried to get around this a couple years ago by defining a special value called "nullity." http://en.wikipedia.org/wiki/James_Anderson_%28computer_scientist%29 http://en.wikipedia.org/wiki/James_Anderson_%28computer_scie...
- orib 18y ago
  Which, when you look at it, is just another name for "undefined". It didn't solve anything, and seemed to me like a bad troll on the mathematical community.
  - etal 18y ago
    Indeed. A little knowledge is a dangerous thing.
kylec 18y ago
I think of it like that blank Scrabble piece - it can be anything until you define what it is, then it has to be that value from now on (in the calculation).
rms 18y ago
You just need to define the undefinable. http://en.wikipedia.org/wiki/Hyperreal_number http://en.wikipedia.org/wiki/Hyperreal_number
- rms 18y ago
  Also, the naive perspective on the .999999999999_=1 debate happens because some people intuit the hyperreals. It's in the hyperreal number system that there can be an infinitely small difference between .99_ and 1, because that infinitesimal is defined to exist where it isn't defined in the real number system.
mattmaroon 18y ago
Ok, so why does 2+2=4?
- maxklein 18y ago
  Because if I had two apples and you gave me two oranges, I would have 4 fruits. If I asked you how many pair groups of apples you had, you'd say 2. If I asked you how many groups of no apples you had, you are free to choose any number, depending on how imaginative you were.
huhtenberg 18y ago
> Why can't we divide by zero? By definition (of the division operation).
maxklein 18y ago
Because 0 is not a real number like the rest. It's like having a long row of cowboy boots and a pair of highheels in the middle. 0 is different from the rest.
snorkel 18y ago
I don't care how your define 1/0 as long as my program keeps running. I always found it to be inexcusable that a run-time environment would halt a program that tried to do N/0 as if it allowing it would cause a complete and utter loss of data and property. Is returning NaN, null, or undefined instead really so hard? Lazy frigging compilers, I hate you!

37 points by jwilliams 18y ago

speek 18y ago

because the math police will come after us.

eipiman 18y ago
Suppose 5/0 = infinity. What does it buy you? At the very least, one would expect that 5/0 = infinity implies that 5 = 0 * infinity, but that is not the case. And if you insist that it is, does 4/0 = infinity then imply that 4 = 0 * infinity, and hence 4 = 5?
- speek 18y ago
  Infinity is not a number... its a concept. Say you were to count every integer, you would count to infinity. Now what if you were to count every number between 0 and 1? Would it still be infinity? I've always thought in degrees of infinity, but either way its still not a number. If you kept the degrees of infinity separate, you could divide 5 by zero and it would not be the same as dividing 4 by zero. If you wanted to play with limits, you could say the limit of 5/x as x approaches 0 is infinity, but we don't really have a name for what x/0 would be. Someone did come up with nullity, but that's not good enough. Plus, you can't have Zero of something. Having implies a positive net, thus it's not correct to say that you have zero of something... you could say you don't have something, but they're semantically different. Zero is just another concept.
  - abstractbill 18y ago
    Take a look at Cantor's work, if you're interested in this stuff. The number of integers is known as a countable infinity. The number of numbers between 0 and 1 is called an uncountable infinity. The proof that these are not the same (Cantor's diagonalization argument) is my favorite proof - it's really neat. It's quite clear and maybe even obvious in retrospect (I like to explain it to friends who are curious about higher mathematics), but probably not something you would have come up with if you weren't Cantor! You can ask if there's an infinity between these two, and that's actually unprovable. You can prove that, without adding more axioms to mathematics, it's impossible to prove one way or another whether there's an infinity between them. Sometimes I really miss this stuff.
    - Retric 18y ago
      I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. PS: This is why I went to CS there is no way to verify that a given proof or program is accurate but that does not stop a program from being useful. Edit: Restating the second argument: Start with all zero's, count in binary from the left, 1000.. > 0100.. > 1100.. > ... and the diagonal becomes 1111111 which is basically just a higher order of infinity like N = the sum of all prior numbers.
      - hhm 18y ago
        I never really accepted Cantor's idea. First off if you have a sequence ...1 and ...0 at infinity they are the same number so saying you flip the last bit does not demonstrate that the above number is not in your set. It doesn't work like that. You want to show that you can't enumerate all reals, so what do you do? You start enumerating all reals. Then you build a new number, whose Nth bit is different from the Nth bit of the Nth real number. You can do that because you have enumerated them, and we have never talked about an infinite N so we are never flipping the last bit of the last number nor anything similar. We are only flipping the Nth bit of the Nth number. Number 819439243? We flip the 819439243th bit. That's all we do. What do you do then? Well, you compare that new number you created, with the first number, the second number, and so on. And you see that this number is different to all the other numbers. What does it mean? That it doesn't matter what your enumeration scheme is, there is at least one number that will always be out of your enumeration scheme, and so your enumeration scheme will fail (so this is an absurd also, as we were supposing that this specific enumeration scheme -that could be any enum scheme- was complete). So you don't have as many integers as reals, so QED. Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity. As you see, it doens't matter, as in this proof you never need to do the sum of all numbers.
        Retric 18y ago

gills 18y ago

A little off topic, but interesting practice for 0. What are the mathematical ramifications of 0% capital reserves? This is a good chance to remind the readers that the bailout bill contains a clause to accelerate a 2006 law to 10/01/2008, which allows the Federal Reserve to set reserve requirements for a bank to 0%.

mystagogue 18y ago

I took a couple of numerical analysis classes from this guy (Peter Alfeld) at the University of Utah and he is a very entertaining teacher.

DanielBMarkham 18y ago

The real number system is discontinuous over division. Interestingly enough, you can divide by zero. But if you do, then you can't define integers.

abstractbill 18y ago
For more on this, take a look at the http://en.wikipedia.org/wiki/Riemann_sphere http://en.wikipedia.org/wiki/Riemann_sphere
- hhm 18y ago
  How does this relate to the original comment? What I want to know is how you can divide by 0, so that then you can't define integers.
  - abstractbill 18y ago
    I think DanielBMarkham is being rather loose with terminology here (or perhaps talking about something other than the Riemann sphere, but it sounds like that's what he's talking about). The problem isn't that you "can't define integers" when you introduce an infinity. The problem is (as the article I linked to says) the resulting system does not form a field (i.e. the usual arithmetic operations don't behave like you would expect anymore). Follow the link to fields if you want to know more.
    - hhm 18y ago
      Oh, I thought he was saying something like you could have real numbers with division by zero as far as you didn't want to have a special class of numbers, the integer numbers. So I was very curious about that, as I studied fields and all that a long while ago and that sounded like a rare concept.
    - hugh 18y ago
      Wait, but the integers don't form a field anyway (not closed under division).

marvin 18y ago

I'm still not convinced that you can't just say that x/0 equals infinity. Certainly, infinity is not a number..but it has the same properties as a transcendental number in the sense that it is impossible to describe it accurately with finitely many digits. You wouldn't be able to use this definition to build other theories, but it would be nice to have a settled definition. Saying that x/0 is undefined still seems to me like a bit of a hack. I agree that 0/0 should be undefined. There are infinitely many possibilities for what it could be, depending on the context.

hhm 18y ago
Infinity is a much studied concept in mathematics; the problem is not that we don't know what infinity is but that we know it's not part of the set of integer numbers, real numbers, etc. But there are many ways to work with infinities, and even to include it in the number sets (for example, in the case of hyper-reals).
derefr 18y ago
Is it at least true that (0 < |0/0| < Infinity); that is, that 0/0 is a finite, non-zero number?
- newt0311 18y ago
  No. 0/0 is completely undefined. For example, consider limit as x approaches 0 of (-x)/x. This approaches (-0)/0 = 0/0 but simplifying the limit given an answer of -1. So 0/0 is completely dependent on the characteristics of the convergence and had no meaning in isolation. In fact, the field of real (and complex) numbers does not even include 0 in the multiplicative group. Therefore, dividing by 0 is nonsensical as it has no inverse as it is not in the multiplicative field. According to the basic axioms of groups theory, given a field F with a the identity of the additive group in the field and b some element of the multiplicative group, it can be proven that mul(a,b) = a where mul is the multiplication operation. This explains why 0 times any other number is 0. If you would like a copy of the proof, just drop me an email. Sorry for the rant, math major.
  - 18y ago

jkent 18y ago

I explained this as: Take a number, e.g. 5. Then take a small number, like 0.1. Add 0.1 to itself, and repeat, you'll eventually reach 5. But if you take 0, and keep adding 0 to it, you'll never reach 5. So no amount of 0's will go in to 5. That logic breaks down if you do 0/0 though. My students never picked up on that one but they believed me!

etal 18y ago
Picturing the graphs of y = 1/x and y = cx/x worked for me. You can make the "reasonable" value at x=0 be Inf, -Inf, or anything in between.

etal 18y ago

There was a professor who tried to get around this a couple years ago by defining a special value called "nullity." http://en.wikipedia.org/wiki/James_Anderson_%28computer_scientist%29 http://en.wikipedia.org/wiki/James_Anderson_%28computer_scie...

orib 18y ago
Which, when you look at it, is just another name for "undefined". It didn't solve anything, and seemed to me like a bad troll on the mathematical community.
- etal 18y ago
  Indeed. A little knowledge is a dangerous thing.

kylec 18y ago

I think of it like that blank Scrabble piece - it can be anything until you define what it is, then it has to be that value from now on (in the calculation).

rms 18y ago

You just need to define the undefinable. http://en.wikipedia.org/wiki/Hyperreal_number http://en.wikipedia.org/wiki/Hyperreal_number

rms 18y ago
Also, the naive perspective on the .999999999999_=1 debate happens because some people intuit the hyperreals. It's in the hyperreal number system that there can be an infinitely small difference between .99_ and 1, because that infinitesimal is defined to exist where it isn't defined in the real number system.

mattmaroon 18y ago

Ok, so why does 2+2=4?

maxklein 18y ago
Because if I had two apples and you gave me two oranges, I would have 4 fruits. If I asked you how many pair groups of apples you had, you'd say 2. If I asked you how many groups of no apples you had, you are free to choose any number, depending on how imaginative you were.

huhtenberg 18y ago

> Why can't we divide by zero? By definition (of the division operation).

maxklein 18y ago

Because 0 is not a real number like the rest. It's like having a long row of cowboy boots and a pair of highheels in the middle. 0 is different from the rest.

snorkel 18y ago

I don't care how your define 1/0 as long as my program keeps running. I always found it to be inexcusable that a run-time environment would halt a program that tried to do N/0 as if it allowing it would cause a complete and utter loss of data and property. Is returning NaN, null, or undefined instead really so hard? Lazy frigging compilers, I hate you!