skullpatrol
  • skullpatrol
Why can you never divide by zero in the set of real numbers? http://matheducators.stackexchange.com/questions/5648/dividing-by-zero
Mathematics
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schrodinger
  • schrodinger
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hartnn
  • hartnn
Think Practically first. If you want to distribute 20 apples among 5 kids, you divide 20 by 5. What are you even trying to do by distributing 20 apples among 0 kids? (Keeping all with yourself? :P) I don't see any practical application of division by 0. mathematically, division by 0 leads to \(\infty\) (positive or negative) which does not belong to set of real numbers. Adding \(\infty\) to Real number's set \(\mathbb R \) breaks its field structure. And just because \(\mathbb R \) is a field, it brings lot of nice properties to real numbers like addition, subtraction....
skullpatrol
  • skullpatrol
Is there a mathematical answer to this question without talking about infinity?
skullpatrol
  • skullpatrol
Thinking practically, if I take away 20 apples from 5 kids, then I can represent this situation as –4 apples per kid. But, as you said, this problem makes no sense if there are no kids to take apples away from :-)

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inkyvoyd
  • inkyvoyd
There are certain properties of the real numbers that hartnn is referring to. In division/multiplication, one of them is the existence of an inverse. But consider that when you wish to define division by zero, you need a number to multiply by to get the same result. That number can (or so I think) be shown not to exist in the reals
ganeshie8
  • ganeshie8
\(\dfrac{x}{0} := \text{ unique number y such that } 0*y=x \) we can consider two cases case1 : \(x\ne 0\) \(0*y=x\) has no solutions because \(0*y\) always equals \(0\) and the right hand is not \(0\) case2 : \(x=0\) \(0*y=0\) has infinitely many solutions because every number \(y\) satisfies this eqn. In both cases, we don't have an unique \(y\). Therefore it makes sense to keep the expression \(\dfrac{x}{0}\) undefined.
bubblegum.
  • bubblegum.
i am not sure about this we can write \(n!=n(n-1)!\) \(0!=0(-1)!\) \((-1!)=complex ~infinity\) http://www.wolframalpha.com/input/?i=%28-1%29%21 and \(0!=1\) sooo \(\large\frac{1}{0}=complex~infinity\) 1 and 0 both are real and the division gave a complex number
anonymous
  • anonymous
No. The property that \(n! = n(n-1)!\) implicitly assumes that n > 1. 0! is simply defined to equal 1, and the concept of factorial has no meaning over negative or non-integer numbers. @ganeshie8 already gave the correct answer.
skullpatrol
  • skullpatrol
http://math.stackexchange.com/questions/26445/division-by-0
thomas5267
  • thomas5267
http://ee.usc.edu/stochastic-nets/docs/divide-by-zero.pdf Math is ultimately a game with rules that we make. By defining the multiplicative inverse of 0, we end up with whole host of problems and ugliness that we don't want to deal with. As a good mathematician, the right thing to do is sweep it right under the carpet and pretend it doesn't exist. Not that dividing by 0 has any use in real numbers anyway.
SolomonZelman
  • SolomonZelman
I would define zero as (for instance) \(\color{#000000 }{ \displaystyle \lim_{ x\to \pm\infty }{(1/x)} }\) (you can use + or -) And then, \(\color{#000000 }{ \displaystyle k/0=k/\lim_{ x\to \pm\infty }{(1/x)} =\lim_{ x\to \pm\infty }{\frac{k}{(1/x)}} =\lim_{ x\to \pm\infty }{kx} \quad \Longrightarrow {\rm Diverges} }\) But 0/0 doesn't go according to this logic.
SolomonZelman
  • SolomonZelman
I remember being in a disscussion somewhere here on OpenStudy related to this topic, and this is what \( \color{#0049ff}{\bf Hero~offered.}\) I will bring this idea up in this post. (not in exact words, but just to pass the idea over). When you divide, \(\large\color{#000000 }{ \displaystyle \color{teal}{14}\div \color{red}{7}=\color{blue}{2} }\) You can represent this differently, via subtraction, \(\large\color{#000000 }{ \displaystyle \color{teal}{14}~\underbrace{-\color{red}{7}-\color{red}{7}}_{\color{blue}{2}~{\rm times}}=0 }\) Or alternatively, \(\large\color{#000000 }{ \displaystyle \color{teal}{14}~\underbrace{-\color{blue}{2}-\color{blue}{2}-\color{blue}{2}-\color{blue}{2}-\color{blue}{2}-\color{blue}{2}-\color{blue}{2}}_{\color{red}{7}~{\rm times}}=0 }\) HOWEVER; If you set, \(\large\color{#000000 }{ \displaystyle \color{teal}{n}\div \color{red}{0}=\color{blue}{x} }\) \(\large\color{#000000 }{ \displaystyle \color{teal}{n}~\underbrace{-\color{red}{0}-\color{red}{0}-\color{red}{0}}_{\color{blue}{x}~{\rm times}}=0 }\) (You can write it differently, 14-2•(7)=0 and n-x•(0)=0 ... and if you write it like this you can tell that the quotient (result of division) doesn't have to be an integer... in any case the essence is that ...) NOW, WHAT I IMPLY FROM THIS (WONDEFUL) IDEA: To satisfy this subtraction, \(\large\color{#000000 }{ \displaystyle \color{teal}{n} }\) must =0. (Or else you don't get 0 when you subtract) And when we check n÷0=x we said n=0, so 0÷0=x And any number x can satisfy \(\large\color{#000000 }{ \displaystyle \color{teal}{0}~\underbrace{-\color{red}{0}-\color{red}{0}-\color{red}{0}}_{\color{blue}{x}~{\rm times}}=0 }\) So, as regards to division by 0, the only possible case that hasn't yet been proven (in this post) to be undefined, is: 0÷0=0 That is useless, but not that I see why 0÷0=0 is wrong.
thomas5267
  • thomas5267
Every reals except 0 have a unique multiplicative inverse. 0 divided by x is 0 if x!=0, hence 0 does not have a unique multiplicative inverse. This is ugly and since 0 divided by 0 is of no use, the right thing to do is sweep it right under the carpet.
crabbyoldgamer
  • crabbyoldgamer
If x/0 = y, then y*0 = x. x=1. You won't find a y such that y*0 =1. That's why division by 0 isn't defined. The simpler answer is because mathematicians aren't idiots.
mathmale
  • mathmale
Please check that statement, Crabby : You won't find a y such that y*0 =1. I don't agree with it.
SolomonZelman
  • SolomonZelman
I suppose the only reason 0/0 is undefined, because there isn't such a thing that you can divide \(b\) by \(c\), but not by \(a\). (In other words, I am attempting to claim that the divident is not any more unique, although I would also disagree with that, because 0 is the only case when you divide by all none-zeros (don't know about 0) to get the same result.)
SolomonZelman
  • SolomonZelman
0/0=0 should be defined, but I guess thomas5267 was right. "sweep it under the carpet"
thomas5267
  • thomas5267
I am not even sure whether whether 0/0=0. I think there is one way to divide zero item into sets of zero item. Or I am just really tired.
tkhunny
  • tkhunny
In the 3rd grade understanding of division, we have: 12/6 = 2 BECAUSE 6 + 6 = 12 12/3 = 4 BECAUSE 3 + 3 + 3 + 3 = 12 If 0/0 = 0, that would have to be because ___ = 0 Oddly, we also have: 0 + 0 = 0 and 0 + 0 + 0 + 0 + 0 = 0 This rather loosely suggests that 0/0 is either nothing at all (not even so much as zero), or anything at all. 1) This is why we invented indeterminance. 2) Still, when coding machines, 0/0 has to be something. Your implementation must decide. There is no definitive conclusion.

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