Why Is 0 Divided by 0 Undefined?

[lifelearner]

Homework Statement

Prove that 0/0 is undefined

Homework Equations

See above

The Attempt at a Solution

Let 0/0 = a

Then 0/a*0 = 1

But a*0 = 0. So 0(1/0) = 1. But 0 times anything is zero, so 0 = 1, which is not true.

 

[skeptic2]

Let 5/5 = a

Then 5/a*5 = 1?

 

[lifelearner]

Yes, iff a=1

 

[Martin Rattigan]

If you have defined 0/0 then it's defined. If you haven't it's undefined.

To prove that 0/0 is undefined would involve going back through all your past notes and pointing out that everything in them is not a definition of 0/0.

It's probably quicker to say:

Define 0/0=1
Therefore 0/0 is defined.
Therefore the result to be proved is false.

But what is probably required is to point out that there are multiple solutions to 0x=0, e.g. x=0 and x=1 (in which case I would say the question is badly worded).

 

[Mathnomalous]

Can't one just say "nothing can't be divided by nothing because there's nothing to divide"?

 

[WhoWee]

Can't one just say "nothing can't be divided by nothing because there's nothing to divide"?

No - just do what you're told.

 

[Keldon7]

If you have defined 0/0 then it's defined. If you haven't it's undefined.

To prove that 0/0 is undefined would involve going back through all your past notes and pointing out that everything in them is not a definition of 0/0.

It's probably quicker to say:

Define 0/0=1
Therefore 0/0 is defined.
Therefore the result to be proved is false.

But what is probably required is to point out that there are multiple solutions to 0x=0, e.g. x=0 and x=1 (in which case I would say the question is badly worded).

I believe the word undefined is well defined.

 

[BerryBoy]

I think this question is a bad question for those learning maths, the sinc function for example ($\rm{sinc}\,(x) = \frac{sin x}{x}$) at $x = 0$ becomes:

$$\rm{sinc}\, 0 = \frac{0}{0} = 1$$

but this value is certainly well-defined (obtainable with L'Hopital's rule). I think the quantity zero needs careful treatment as it can be misleading to say the anything divided by zero is undefined.

 

[Hurkyl]

Ah, but the sinc function is not sin(x)/x -- the sinc function is the continuous extension of sin(x)/x.

 

[resaypi]

suppose f(a) = 0/0
as x goes to a lim f(x) can yield any real number, so it would not be wise to define 0/0 some real number

 

[Martin Rattigan]

I believe the word undefined is well defined.

No doubt meant in jest, but in fact it isn't as well defined as all that. It rather depends on who is defining or not defining.

For example, in some expositions of symbolic logic, "$\vee$" may be an undefined idea while $A\Rightarrow B$ is defined as $\neg A\vee B$, while in others "$\Rightarrow$" may be an undefined idea while $A\vee B$ is defined as $\neg A\Rightarrow B$.

Unfortunately there is no well defined authority to arbitrate on what is and isn't defined (though some American standards bodies have taken upon themselves this task in certain cases).

 

[lifelearner]

Is there any way to formulate a proof by contradiction?

EDIT: The following site seems to have an interesting approach to this query:

http://www.friesian.com/zero.htm

 

[Martin Rattigan]

Yes. My original suggestion about going back through your past notes could be a proof by contradiction.

That is to say.

Assume 0/0 is not undefined. Then 0/0 is defined.

This means that a statement in the previous exposition of the subject (your notes) defines 0/0. (*)

The statements in your notes are:

1. ...
2. ...

...

n. ...

(You will have to provide the details here)

Then (assuming each is true).

1. doesn't define 0/0.
2. doesn't define 0/0.

...

n. doesn't define 0/0.

Therefore no statement in your notes defines 0/0. (**)

(*) and (**) are a contradiction, therefore the original assumption that 0/0 is not undefined is false.

Therefore 0/0 is undefined.

 

[HallsofIvy]

If 0/0 were "defined" then it would be defined to be some specific number. That is there would some specific x such that $\frac{0}{0}= x$. But saying "$\frac{a}{b}= x$" is the same as say a= bx and 0= 0x is true for all x.

By the way, many texts use the term "undefined" to refer to fractions such as $\frac{a}{0}$ where a is non-zero and the term "undetermined" to refer to 0/0.

The reason is that a/0 is undefined because a= 0x is not true for any x (and, in particular, $\lim_{h\to 0}\frac{f(x)}{g(x)}= L$, where f(x) is a function that goes to a nonzero limit and g(x) is a function that goes to 0 as h goes to 0, is not true for any number L) while 0/0 is "undetermined" because 0= 0x is true for all x (and, in particular, $\lim_{h\to 0} \frac{f(x)}{g(x)}= L$ may be true and we can find f and g making L any specificied number).

 

[BerryBoy]

Ah, but the sinc function is not sin(x)/x -- the sinc function is the continuous extension of sin(x)/x.

Exactly my point, the fact that sin(x)/x for a particular value of x may not be defined can be misleading. Especially when, in reality, we usually deal with continuous variations where undefined values suddenly are defined!??

Apologies lifelearner, we're digressing from the problem.

 

[Martin Rattigan]

There is absolutely nothing to stop you defining 0/0 as some specific number. Some computer languages do exactly that. What you can't then do is say $\frac{a}{b}=x$ iff $bx=a$; this would not follow from your definition.

It generally would follow from the definition in other cases, indeed it would usually be a prerequisite for the definition in other cases.

So for the real numbers you would prove that if $b\neq 0$, then $bx=by$ implies $x=y$. You would then be at liberty to define $\frac{a}{b}$, when $b\neq 0$, as the unique number $x$ satisfying $bx=a$. Without the proof of uniqueness the definition is invalid, which is why you can't extend it to the case where b is 0.

At this point $\frac{a}{b}$ would be undefined whenever $b=0$, but it would naturally follow immediately from the definition that when $\frac{a}{b}$ is defined $\frac{a}{b}=x$ iff $bx=a$.

You could then further define 0/0 as some specific number, say

$0/0=_{df}1$

but you could not then say that when $\frac{a}{b}$ is defined $\frac{a}{b}=x$ iff $bx=a$, because this would no longer be true. E.g. $0\times 2=0$, but $\frac{0}{0}=1\neq 2$.

When $\frac{a}{0}$ is defined in the extended complex numbers (for $a\neq 0$), it is then no longer universally true that when $\frac{a}{b}$ is defined $\frac{a}{b}=x$ iff $bx=a$.

Although you are at liberty to define 0/0 as any number you like or anything else for that matter it's probably not a good idea because $\frac{a}{b}=x$ iff $bx=a$ can then fail. But having said it's a bad idea doesn't mean that nobody has ever done it, so if you were to take "undefined" to mean "nobody has ever defined it", you would be hard pressed prove the original statement.

What I was trying to say in my first post is that you shouldn't be in the business of trying to prove or disprove whether or not something is defined. It is defined if you have accepted a definition of it and undefined otherwise; but this may change tomorrow after you've read or been taught a bit more (or adopted definitions of your own).

 

[jamalahmed68]

See how 0/0 is undefined
0/0.1=10
0/0.01=100
0/0.001=1000
0/0.00000000001=100000000000
since 0.00000000001>0
if it is divided by 0 then it's value approaches to infinity..

 

[jamalahmed68]

See how 0/0 is undefined
0/0.1=10
0/0.01=100
0/0.001=1000
0/0.00000000001=100000000000
nevertheless 0.00000000001>0
if it is divided by 0 then it's value approaches to infinity..

 

[HallsofIvy]

Hardly worth posting twice! If, instead of that particular sequence you use .1/.1= 1, .01/.01= 1, .001/.001= 1, ... which converges to 1. Or .2/.1, .02/.01, .002/.001, ... which converges to 2.

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$
does not, in general, go to infinity.

 

[Martin Rattigan]

See how 0/0 is undefined
0/0.1=10
0/0.01=100
0/0.001=1000
0/0.00000000001=100000000000
nevertheless 0.00000000001>0
if it is divided by 0 then it's value approaches to infinity..

Or even if you get the original arithmetic right i.e.:

0/0.1=0
0/0.01=0
0/0.00000000001=0

it converges to 0.

 

[Mark44]

If 1/0 had a defined value, say a, you could write 1/0 = a. Since a * 0 is 0 for any real or complex number, a * 0 can never be 1. For this reason, 1/0 is undefined, as is b/0 for any nonzero number b.

0/0 has problems for a different reason, since you could define 0/0 to be k, where k could be any number at all. If 0/0 = k, then clearly k * 0 = 0. Unlike 1/0, which is not equal to any number, 0/0 could be arbitrarily defined as any number. For this reason it is called indeterminant.

 

[Martin Rattigan]

$1/0$ has the defined value $\infty$ in the extended complex number system $\mathbb{C}^*$ in most analysis books. On the other hand $0\times\infty$ remains undefined, so you don't have in this case that when $a/b$ is defined then $b\times(b/a)=a$.

You may accept Mark44's reasons in the first paragraph of the previous post as good reasons for not defining, $1/0$ or disagreeing with the people who did define it or disagreeing with what they defined it to be. What you can't do is take it as an incontrovertible proof that they didn't define it at all.

 

[jamalahmed68]

See how 0/0 is undefined
0/0.1=10
0/0.01=100
0/0.001=1000
0/0.00000000001=100000000000
nevertheless 0.00000000001>0
if it is divided by 0 then it's value approaches to infinity..

sorry for this
let
x=.01 then 1/x=100
x=.0001 then 1/x=10000
x=.00001 then 1/x=100000
as x approaches to zero 1/x tends to infinity
since then there are two case
1: 0*a=0 for all a belonges to real numbers
2: 1/0 * a= infinity for all a blonges to real number
so 0/0 is undefined ...

 

[The Chaz]

Hardly worth posting twice! If, instead of that particular sequence you use .1/.1= 1, .01/.01= 1, .001/.001= 1, ... which converges to 1. Or .2/.1, .02/.01, .002/.001, ... which converges to 2.

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}$$
does not, in general, go to infinity.

I'm not sure where all the confusion lies...
This quote (^^^) is exactly the intent of the question, which I have reworded:
"Show that any definition of 0/0 is inconsistent".

 

[Martin Rattigan]

... But what is probably required is to point out that there are multiple solutions to 0x=0, e.g. x=0 and x=1 (in which case I would say the question is badly worded).

Or, presumably, (^^^) as I first posted. With the rewording, no problem.

 

[Hurkyl]

"Show that any definition of 0/0 is inconsistent".

You can't, because the claim is false. Your calculator doesn't vanish in a puff of 'logic' if you ask it to divide zero by zero now, does it? And then there's things like wheel theory...

 

[The Chaz]

Right. It wouldn't be hard to show that a "function" that divides by zero would not be "well-defined".

 

[Martin Rattigan]

I'd assume the question as reworded meant, "Show that any definition of 0/0 is inconsistent ... with the definition of a/b when b$\neq$0." Which could be taken (without too much free interpretation) to mean, "show that the x such that 0x=0 does not exist". (I.e. $\neg(\exists x\in\mathbb{R})(0x=0\wedge(\forall y\in\mathbb{R})(0y=0\Rightarrow y=x))$.)

 

[Martin Rattigan]

Right. It wouldn't be hard to show that a "function" that divides by zero would not be "well-defined".

z/OS assembler language provides an arithmetical divide operation that divides by zero. The answers are n/0=0 if n$\neq$0, 0/0=1. That doesn't in any way conflict with "standard" definitions of division (though the results when dividing by anything other than 0 generally do), it merely extends the definitions to cover cases not normally defined. The division function is perfectly well defined (though not, I should admit, on $\mathbb{R}$).

 

[The Chaz]

You can't, because the claim is false. Your calculator doesn't vanish in a puff of 'logic' if you ask it to divide zero by zero now, does it? And then there's things like wheel theory...

wheel theory ... where "x - x" doesn't equal ZERO?? Yeah, I'd say you're too smart for this discussion! The assumption is that this is from a lower-level undergrad (or even high school) class.

And my calculator can't prove anything. Dividing by zero is undefined on it because the programmers couldn't define it consistently.

Surely you're all aware of my math limits at this point... but another thing puzzles me:
"n/0=0 if nLaTeX Code: \\neq 0, 0/0=1" hmm. not sure how that will come out.
n/0 = 0 if n !=o; 0/0 =1
Assume n != 0.
1 = 0/0 = (0*n)/0 = 0*(n/0) = 0*0 = 0. hmm

 

[Martin Rattigan]

(0*n)/0 = 0*(n/0)

This step is incorrect. With IBM's implementation (a*b)/c is not necessarily a*(b/c) - in fact this fails not only when c=0.

It is necessary to live with the consequences of what has been defined. But what has been defined in any given circumstance is not generally a matter for proof. IBM give you some very good manuals. These tell you what is defined and what the definition are.

 

[The Chaz]

That's all fine and good (and of some interest, though it's a strictly decreasing function over time), but with the OP in mind -
WHAT does this have to do with the QUESTION! Is this kid in an IBM class? Is he even a kid (or male)? ;)
I imagine a h.s. or undergrad student thinking "what the $%^ are they talking about?!". Can we help this guy get some closure, or just regurgitate an endless stream of advanced topics? I personally don't mind talking about this stuff, but we should do that elsewhere.

 

[Martin Rattigan]

I would imagine that OP had got all he wanted to from the thread long before you reopened it.

 

[The Chaz]

I would imagine that OP had got all he wanted to from the thread long before you reopened it.

Let's hope so! Funny word choice, though. "reopened" implies closure, which wasn't apparent (at least to me) in the 1-hour gap between my first post and the post immediately preceding that...

 

[lifelearner]

Thank you all for your input. This question actually appears in Calculus by Varberg, Purcell, and Rigdon and I encountered while I was going through the pages to refresh my calculus.