Why 0 ÷ 0 = 1

[0rbytal]

I started arguing this in high school, and perhaps I am being stubborn and denying the rebuttals and/or maybe I am just searching for someone who agrees with me. So far the only person to agree with me on this is my best friend.

From what I am taught any number divided by zero is undefined.
x÷0=infinity
And any number divided by itself is 1.
x÷x=1

So what about zero? I approached it in terms of quantity:
If I have 3 marbles in my left hand, and I move 3 of those marbles to my right hand, I moved 1 whole of the marbles (3÷3=1). Likewise, If I have zero marbles in my left hand, and I move zero marbles to my right hand, I moved 1 whole of the marbles (0÷0=1).

Also look at it this way: 0÷0 = 0 × (1/0) = zeroes cancel = 1!

If that's not enough find :
lim sin(x)
^{x-›0 X}

By direct substitution you get sin(0)÷0 = 0÷0 but it = 1!

Enough proof?

[matt_mcl]

A quibble:

x÷0=infinity

No, because even if you pile up an infinite number of 0s, they are still 0, not some other x. It's undefined, not infinite.

[Left Hand of Dorkness]

I think you're looking at it wrong. 3/3 means that you're dividing three marbles into three equal groups: how many are in each group? 1, obviously

0/0 means you're dividing 0 marbles into 0 equal groups. How many are in each group? Problem is, since there AREN'T any groups, there's no way to tell how many are in each group.

That's what "undefined" means.

Daniel

[silverfish]

I disagree with you 0rbytal. My argument is this:

Assume 0/0 = 1.
Then multiply both sides by x.
Then x*(0/0) = x
So (0*x)/0 = x
So 0/0 = x
But 0/0 = 1
So x =1. The argument applies for any x you choose, so x = 1, for any x.

With your moving marbles example, yes you do move 1 'whole' lot of you marbles. But also you move 2 * the whole (as the whole is 0), and three times the whole, and so on, so you get a similar result as above.

With the 0/0 = 0*(1/0) example, you have assumed that 1/0 has some meaning, which is doesn't, it is undefined, as matt_mcl points out.

The final example also gives an ambiguity. If you take 2Sin(x) rather than Sin(x), you get the limit to be 2 (assuming your original calculation is correct), but 2Sin(0)/0 = 0/0. as before. The problem is you are assuming that sin(x)/x is continuous for all x.

In the cases where your 'proofs' make sense, they don't lead to just one value of 0/0, which is rather difficult mathematically, as division is supposed to be 'well defined' so it gives only one answer.

[0rbytal]

matt_mcl, I was using a graphical description. When you put y=x÷0 into a calculator, you get a vertical line.

DanielWithrow, thanks for the clarification.

One rebuttal I have heard is "Type it into a calculator and see what you get." Well, of coarse I am going to get undefined because the programming of the calculator check division entries to see if entry1 is divided by zero. If entry1 IS divided by zero it throws the error 'undefined'! So the calculator flaw can be blamed on the programmer.

[Thudlow Boink]

If you think of division as the inverse of multiplication, then for example, 15 ÷ 5 = 3 because you have to multiply 5 by 3 to get 15. Then 0 ÷ 0 could be anything, because any number times 0 = 0. (So, in terms of the OP's marbles example, yeah, you moved 1 "whole" of the marbles, but you also moved 2 wholes of them, and also 3 wholes, and also one half of them, and so on and so on.)

If you want to talk about limits, in calculus, 0/0 is an example of what's called an indeterminate form, meaning that just knowing that the numerator and denominator each approach zero doesn't determine what, if anything, the limit of the whole thing is. Pick your favorite number N. When x = 0, N sin x = 0, and x = 0, but the limit of (N sin x) ÷ x (as x approaches 0) is N.

[iampunha]

Originally posted by silverfish
I disagree with you 0rbytal. My argument is this:

Assume 0/0 = 1.

Paging Dr. Matrix...Dr. Matrix, please answer the white courtesy phone...

[erislover]

if 0/0 = 1
then
1 * 0 = 0, which is true.

So maybe 0/0 = pi
then
pi * 0 = 0 is true.

So which is it, 0rbytal, 0 or pi?

[nogginhead]

Take the limit of x/x as x -> 0.

x/x = 1, so x/x = 1 as x -> 0.

[jshore]

Originally posted by Thudlow Boink
If you want to talk about limits, in calculus, 0/0 is an example of what's called an indeterminate form, meaning that just knowing that the numerator and denominator each approach zero doesn't determine what, if anything, the limit of the whole thing is. Pick your favorite number N. When x = 0, N sin x = 0, and x = 0, but the limit of (N sin x) ÷ x (as x approaches 0) is N.

Yes, this is how it works. 0/0 is defined only in the sense of a limit, i.e., we can look at the limit of the ratio of two continuous functions f(x)/g(x) as x->0 where f(0) = 0 and g(0) = 0. And, it can indeed be any result depending on the functions; it doesn't even have to be finite. For example, while sin(x)/x -> 1 as x -> 0, sin(x)/x^2 -> infinity as x -> 0 (actually, it goes to +infinity when you take the limit from above, usually denoted something like x -> 0+ and it goes to -infinity when you take the limit from below (x -> 0-). And the limit of sin(x)/sqrt(x) as x -> 0 is 0.

[Left Hand of Dorkness]

Gah! Here I am talking about handfuls of marbles, and you guys are bringing up calculus! Whaddya tryin to do, give me a headache?

(And, more importantly, is my handfuls-of-marbles explanation conceptually correct?)

Daniel

[Max Carnage]

0 isn't a number, is it? It's a symbol or a place holder. Heck, it wasn't even around when the number system was first invented. It was invented later. It can't be treated the same as a number in terms of division.

[ultrafilter]

Originally posted by nogginhead
Take the limit of x/x as x -> 0.

x/x = 1, so x/x = 1 as x -> 0.

Now take the limit of 1/x as x -> 0. Then take the limit of x²/x as x -> 0. Which one is right?

0/0 is indeterminate. In order that 0/0 = a, it must be the case that 0a = 0. That works for any real number a.

a/b is defined to be the unique number c such that bc = a. If c is not unique, a/b is indeterminate. If c doesn't exist (as in the case of 1/0), a/b is undefined.

[ultrafilter]

All quotes originally posted by 0rbytal.

I think you're mistaken about a few things here. Let me see if I can pick apart your arguments in an educational manner.

From what I am taught any number divided by zero is undefined.
x÷0=infinity
And any number divided by itself is 1.
x÷x=1

Where were you taught that first equation? x/0 is undefined--it doesn't mean anything.

Also look at it this way: 0÷0 = 0 × (1/0) = zeroes cancel = 1!

Only if 1/0 is defined. It's not.

If that's not enough find :
lim sin(x)
^{x-›0 X}

By direct substitution you get sin(0)÷0 = 0÷0 but it = 1!

What about the limits I mentioned in my last post?

[I, Brian]

Hmm...I'm probably stupid for thinking so - but if I divide 10 marble by nothing then I still have 10 marbles.

And if I have no marbles divided by no marbles then the result is obviously nothing.

[Yeticus Rex]

If you take 0 marbles and divide it amongst 0 people, how the hell do you get ONE marble for each person who wasn't there to get their share?

Therefore, 0/0 = Does a bear shit in the woods? I don't know, ask the friggin' bear! (Hint: Undefined!)

Brian - Your assumption is wrong, you wouldn't have 10 marbles, because you were not in line to get your share! Otherwise, 10 divided by 1 (you Brian, being the one in line) is 10.

[astorian]

Let's put this in very simple terms.

How many nickels can you get for a dollar? Well, 100 / 5, right? That makes twenty.

How many nickels can you get for a quarter? 25 / 5, or 5. Right?

Now, how many nickels can you get for NOTHING? The answer is 0 / 5. In other words, zero.

So far, this all makes sense, right? Now, one last question:

How much nothing can you get for a dollar?

*

You see the dilemma? "How much nothing" is an absurdity, a meaningless concept. But when you try to divide by zero, that's what you get! You can get an INFINITE amount of nothing for any amount!

[Mangetout]

Originally posted by I, Brian
Hmm...I'm probably stupid for thinking so - but if I divide 10 marble by nothing then I still have 10 marbles.

And if I have no marbles divided by no marbles then the result is obviously nothing.

No; if you divide(share out) 10 marbles equally into zero groups, how does it work out for you?

If you have no marbles and you divide them out in to no groups, how many marbles are in each group? - the answer is not zero; it is undefined bcause you don't have any groups to count

If you thought that was fun, try null sometime; it's a strange beastie.

[samarm]

I had a similar question to the OP, and started a thread about it. You may find the debate interesting:

http://boards.straightdope.com/sdmb/...hreadid=158041

[vasyachkin]

dammit orbital, just try and pass your calculus, it will all make sense later.

in case of sin(x)/(x) it is 1. but in other cases it will be other numbers, yet in other cases it really will be undefined, or infinite or zero.

get it now ? of course you dont. how can you get it if you haven't learned it yet ? just be patient relax.

if you want, read about l'hopital's rule ( its in your calculus book, see the index ).

[Ramanujan]

the division operation is just not defined when the denominator is 0. it's just that simple. no way to justify how it is defined is valid. limits are not valid, because taking the limit is not the same operation as division.

most operators we use these days, including multiplication, division, addition, and subtraction, have their definitions in group and ring theory (abstract algebra).

an operator is just a function. like all functions, it needs a definition. the division operator we most often talk about is defined from an ordered pair of reals to a real. if the second of the pair it operates on is 0, there is no number it yields. it's just defined that way.

asking what a number divided by 0 yields is like asking what radiohead marrying a prime number yields. it's just nonsensical. the words and operators just aren't defined to be meaningful with those terms.

[I, Brian]

Ah yes - I can see I was quite wrong - I blame it having been late in the evening for myself!

Yes - undefined - quite right.

[AV8R]

On one of my first calculators (I think it was a TI-55), if you entered 0÷0 the answered returned as 1.
(of course, it was blinking, which indicated error, but I still thought it was kinda cool to have a calculater that produced a wrong answer).

[nogginhead]

--------------------------------------------------------------------------------
Originally posted by nogginhead
Take the limit of x/x as x -> 0.

x/x = 1, so x/x = 1 as x -> 0.
--------------------------------------------------------------------------------

Originally posted by ultrafilter
Now take the limit of 1/x as x -> 0. Then take the limit of x²/x as x -> 0. Which one is right?

Why are 1/x and x²/x relevant? We're not talking about anything divided by 0, only 0.

0/0 is indeterminate. In order that 0/0 = a, it must be the case that 0a = 0. That works for any real number a.

So under this argument, any number is right. Including 1.

a/b is defined to be the unique number c such that bc = a. If c is not unique, a/b is indeterminate. If c doesn't exist (as in the case of 1/0), a/b is undefined. [/b]

Excluding toy math systems (i.e. those without applications outside mathematics) I believe this definition only applies when b=0. So it's a specific rule about what happens when you divide by 0, and is thus an arbirtrary rule. I can just as easily make an arbitrary rule that x/0 = +infty if x>0, = -infty if x < 0, and =1 if x = 0.

We also define 0! =1, for example, because it's convenient.

[december]

Originally posted by nogginhead
I can just as easily make an arbitrary rule that x/0 = +infty if x>0, = -infty if x < 0, and =1 if x = 0.

We also define 0! =1, for example, because it's convenient.

However, defining 0! = 1 really is convenient. The usual formulas work with this definition.
E.g., (n+1)!/n! = n+1, for all positive integer values of n, including n=0.

Defining x/0 = +infty doesn't actually mean anything, because +infty isn't a real number. If you want to extend the real number system to include +infty and -infty, you will lose a good many relationships. Or, you would have to have two classes of numbers, the finite ones and the infinite ones. Then you could try to derive different threorems for various subsets.

The bottom line is that defining 0/1 = +infty is not convenient.

A little learning is a dangerous thing;
drink deep, or taste not the Pierian spring:
there shallow draughts intoxicate the brain,
and drinking largely sobers us again.

Alexander Pope

[imthjckaz]

I was just wondering about the marbles everyone is talking about, because I lost mine and I think that maybe they might be mine.

[nogginhead]

december
The bottom line is that defining 0/1 = +infty is not convenient.

A little learning is a dangerous thing;
drink deep, or taste not the Pierian spring:
there shallow draughts intoxicate the brain,
and drinking largely sobers us again.

Alexander Pope

OK, december, as long as we're sniping with quotations, how about:

Oscar Wilde
There are two ways of disliking poetry; one way is to dislike
it, the other is to read Pope.

and, more to the point

Shakespeare
There are more things in heaven and earth,
Horatio, Than are dreamt of in your philosophy

My point was that if one were to find it convenient to define 0/0 as 1, it would be as acceptable an arbitrary rule as to define it as undefined. Same for x/0=infty, though that wasn't what I was driving at especially.

[Orbifold]

Originally posted by nogginhead
Why are 1/x and x²/x relevant? We're not talking about anything divided by 0, only 0.

Fine. So what's the limit of 0/x as x approaches 0?

So under this argument, any number is right. Including 1.

Yes, any number. That's the point. 0/0 is undefined because there's no unique solution to the equation 0*x=0. An operation which doesn't yield a unique result is not well defined.

Excluding toy math systems (i.e. those without applications outside mathematics) I believe this definition only applies when b=0.

It's the mathematical definition of division, I'm afraid. You believe wrong.

Well, unless you're referring to the more common-sense, "how many A's in a B" definition of division...which, as Mangetout and astorian have pointed out, still doesn't define 0/0.

The definition using limits doesn't let you define it either, 'cause x/x and 0/x don't have the same limit as x approaches 0. And that's it; we're out of definitions of division. Except, of course, the arbitrary one:

So it's a specific rule about what happens when you divide by 0, and is thus an arbirtrary rule. I can just as easily make an arbitrary rule that x/0 = +infty if x>0, = -infty if x < 0, and =1 if x = 0.

Except that as silverfish demonstrated, if you assume 0/0=1 then you can "prove" that x=1 for any x, which makes your arbitrary definition about as useful as hammer made of sponge cake.

I feel kind of silly just coming in here and pointing to everyone else's posts, but it seems the points need to be repeated.

[RSC0318]

I haven't read many of the replys on this post, buthere are my thoughts. Not necessarily going by what I learned in school, but anyways. 0= a number, but it holds no value (i.e. it is neutral)
0/0=0. Yet it is still a number (and without zero, you cannot get to one, because it is at the beginning of the number scale (excluding negative integers in this discussion)). 0/0=1. Does this make sense to anyone else?

[nogginhead]

... officially abandoning line of argument ...

Just to say to orbifold that aspect of the definition I was referring to is the part about uniqueness/definedness, which only exists to deal with the 0 problem.

my point is that just defining something away doesn't make it not exist.

[Orbifold]

That's a reasonable point nogginhead, except that we're not "defining something away". There's no reasonable definition of 0/0 in the first place. Saying that 0/0 is undefined isn't an arbitrary decision, it's a recognition of the fact the definition of division we already have for other numbers just doesn't work for 0/0, and there are good reasons why we can't make it work.

[nogginhead]

Originally posted by Orbifold
That's a reasonable point nogginhead, except that we're not "defining something away". There's no reasonable definition of 0/0 in the first place. Saying that 0/0 is undefined isn't an arbitrary decision, it's a recognition of the fact the definition of division we already have for other numbers just doesn't work for 0/0, and there are good reasons why we can't make it work.

I give in already. It's best to define it as undefined.

[Canadagirl]

Excuse me for asking, but how the hell do you have zero groups of anything, even if it's zero groups of zero? I would definately go with the undefined that I was taught, because it makes sense .

[verybdog]

Originally posted by 0rbytal

x÷0=infinity
And any number divided by itself is 1.
x÷x=1

Also look at it this way: 0÷0 = 0 × (1/0) = zeroes cancel = 1!

If that's not enough find :
lim sin(x)
^{x-›0 X}

By direct substitution you get sin(0)÷0 = 0÷0 but it = 1!

Enough proof?

x÷x=1, yes, but not for 0÷0 which is = 0 therefore not = 1;

0÷0 = 0 × (1/0) = zeroes cancel = 1! ===> No way. You can't treat 0 as an egg that is physical. 0 means nothing, therefore the operation "0 × (1/0)" is basically is "0 x something" which is 0.

When x=0.001, Sinx = 0.84147098, therefore sinx/x = 0.99999983 = 1.

[emarkp]

The amazing thing to me is using a limit to attempt to show what happens when you divide by zero. Especially since limits were introduced to precisely what happens when not at a specific point.

lim x->0 specifically refers to x =/= 0. Sheesh.

[DrMatrix]

emarkp, you bring up a good point. The only reason to bring up limits would be to suggest a reasonable definition of 0/0. Since x/x and 2x/x reach different limits as x approaches 0, this does not get us anywhere.

The bottom line is: a/b is the unique number c that satisfies a = b*c. For b = 0, there is either no value for c (for a non-zero) or no unique number (for a = 0) that satisfies a = b*c.

There is no useful definiton if a/0.

[jshore]

Originally posted by emarkp
The amazing thing to me is using a limit to attempt to show what happens when you divide by zero. Especially since limits were introduced to precisely what happens when not at a specific point.

Well, this is a truly absurb way of looking at things! I would say limits were introduced precisely as a way of finding out what is happening at a specific point when it would be difficult to do so directly; we do it by getting arbitrarily close to the point and see what happens!

Thus limits are extremely useful in this case because while we can't understand 0/0, we can understand it in the sense where the numerator and denominator are continuous functions of x [call them f(x) and g(x)] that approach 0 as x->0. That limit leads us to the conclusion that we should consider 0/0 to be indeterminate we can find functions f(x)/g(x) that give any value in the limit that x->0.

[jshore]

Originally posted by jshore
That limit leads us to the conclusion that we should consider 0/0 to be indeterminate we can find functions f(x)/g(x) that give any value in the limit that x->0.

There should be a "because" before the "we can".

[DrMatrix]

Originally posted by jshore
Well, this is a truly absurb way of looking at things! I would say limits were introduced precisely as a way of finding out what is happening at a specific point when it would be difficult to do so directly; we do it by getting arbitrarily close to the point and see what happens!

No. You are not looking at what happens at a specific point. You are looking at what happens "near" a point. The fact that lim (x->0) sin x/x = 1 does not mean that sin 0/0 = 1. Look at the definition of limit. If the limit is taken as x approaches a, the value at a is specifically excluded.

It is interesting that limits are brought up. The primary motivation for using limits in calculus is to evaluate lim (delta-x->0) delta-y/delta-x. Delta-y and delta-x both approach zero as delta-x approaches 0. The fact that this limit is (with qualifications) defined, but 0/0 is not tells me that limits are no help here.

[RTFirefly]

Let's consider

lim x^3- 8
^x->2 x-2

[RTFirefly]

Ahhh, feathermuck.

We can figure out what the function f(x) = (x³-8)/(x-2) 'should' be from the limit, should we want to define it at x=2: since the limit is 12, defining f(2)=12 turns f into a function that is continuous at 2.

It helps that this particular function has a nice, well-defined limit there.

OTOH, when we're talking about the entire class of indeterminate forms whose numerator and denominator both approach 0 as x gets near to some value, such well-definedness goes right out the window. Each such expression has its own limit (if it has a limit at all), and the values of these limits are all over the place. So as others have pointed out here, there is no single value that 0/0 'ought' to be, hence calculus is no help, except in a negative sort of way - reaffirming our intuition that there's no appropriate definition for 0/0.

[jshore]

Originally posted by DrMatrix
No. You are not looking at what happens at a specific point. You are looking at what happens "near" a point. The fact that lim (x->0) sin x/x = 1 does not mean that sin 0/0 = 1. Look at the definition of limit. If the limit is taken as x approaches a, the value at a is specifically excluded.

Yes, I know what a limit is. But, the point is that you are considering what is happening arbitrarily close to a point which gives one an idea of what one might want to define as the value at the actual point...if one wants to make the function continuous at that point, for example (as RTFirefly points out).

The fact that this limit is (with qualifications) defined, but 0/0 is not tells me that limits are no help here.

I still disagree. I think we learn from limits an important negative conclusion, namely that it doesn't really make sense to define 0/0 to be 0 or 1 or anything like that since the limit of the ratio of continuous functions that both approach 0 can in fact be any value.

I agree with what RTFirefly has said except as a matter of connotation because I think this "negative sort of way" in which calculus is of help is important enough that I would not say that "calculus is of no help, except...". I think it has helped us to see another reason why it is most sensible to consider 0/0 to be indeterminate and this is important enough to phrase in a more positive light like "Calculus is of help here because..." But, this issue of connotation is at some point getting down to philosophy and interpretation more than mathematics.

[yme]

Using WindowsNT calculator program, calc.exe. It gives me this when dividing 0/0: Cannot divide by zero. Or maybe I need a Windows Update.

[verybdog]

Originally posted by RTFirefly
Let's consider

lim x^3- 8
^x->2 x-2

This one got a limit. there's a factor (x-2) that can be taken out.

[ultrafilter]

RT mentioned that. Go back and read his post again.

[verybdog]

Originally posted by 0rbytal

lim sin(x)
^{x-›0 X} = 1

By direct substitution you get sin(0)÷0 = 0÷0 but it = 1!

lim sin(x)
^{x-›0 X}
= 1

Funny, this lim(sinx/x) =1 thing is discovered by experiment rather than by substitution. If using substitution, it would be no limit, as initially thought.

[verybdog]

Originally posted by ultrafilter
RT mentioned that. Go back and read his post again.

I didn't see his post.

[jshore]

Originally posted by verybdog
Funny, this lim(sinx/x) =1 thing is discovered by experiment rather than by substitution. If using substitution, it would be no limit, as initially thought.

You are misusing terminology here. The fact that you cannot find out what sin(x)/x is when x=0 by substitution does not mean that lim(sin(x)/x) is not well-defined. And, you don't have to get the limit "by experiment"; you can get it using calculus techniques (for example, either by l'Hopital's rule or the power series expansion for sin).

[verybdog]

Hear, hear.

Which one was the first - the known result of Lim sinx/x and the known name of l'Hopital?

[Johnny L.A.]

How much nothing can you get for a dollar?

Everyone send me dollars to my PayPal account. I'll send you nothing in return.