What's the difference between 0/0 and 1/0?

https://blog.plover.com/math/division-by-zero.html

by korethr • 7 years ago

194 101 7 years ago

blog.plover.com

saagarjha2• 7 years ago

Last year a new Math Stack Exchange user asked What's the difference between 0/0 and 1/0? I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”.

Never change, Stack Exchange.

kouh• 7 years ago

There is something inextricably human about this sort of anecdote though, like modern reenactments of the drowning of Hippasus by the Pythagoreans after bringing up irrational numbers.

edoo• 7 years ago

He showed that the square root of 2 was an irrational.

jandrese3• 7 years ago

It's more hostile than Wikipedia to new users. Ask a tough question and there's a decent chance of getting the dreaded 1 downvote and lock.

btrettel• 7 years ago

This hasn't been my experience. I've asked questions that in retrospect were fairly stupid and I've only been treated with respect.

wuliwong2• 7 years ago

I found wikipedia's interface to be pretty unfriendly for me as a new user. Stackoverflow was a lot easier for me to get started with. Maybe that lower barrier to entry leads to more random posts from people that are "low quality?"

Side note: I find a similarity between SO and HN. Both can feel pretty hostile and over-the-top with penalizing you for not saying things "the right way." But as much as it does annoy me sometimes, these are two of the most useful websites for my purposes.

gowld• 7 years ago

https://math.stackexchange.com/questions/26445/division-by-0

has a long discussion of this problem.

JdeBP• 7 years ago

It's Mathematics Stack Exchange, not Stack Overflow.

saagarjha2• 7 years ago

Fixed, thanks.

RyJones3• 7 years ago

It's pretty much the same across all of the properties, though. I quit using them; they're run poorly and, especially in system administration or programming areas, the quality is uselessly low

felipellrocha• 7 years ago

I just now assume that whenever I post there, the first few comments will be people berating me out for going out of my comfortable zone to ask/answer a question.

BMorearty• 7 years ago

For a second there I assumed you were talking about HN.

sockpuppet999• 7 years ago

I've felt the same way here many times. Although I'm trying to keep it under control now. I understand the mods motivation a little bit better

dreamcompiler• 7 years ago

Almost every useful answer I've ever found on SE or SO has been marked down with some snarky judgmental horseshit meta-comment. At this point, if I don't see such markdowns, I assume the answer isn't very useful until proven otherwise.

ultrarunner• 7 years ago

I wonder if this is a repeatable phenomenon wherein a universally-useful resource is more likely to be accessed by a wider gamut of people, thus essentially guaranteeing a judgmental response. That would mean those snarky responses really are a mark of a good and useful question.

gsich• 7 years ago

it's the wiki-syndrome.

duxup• 7 years ago

Yup, first answer on Stack Overflow sometimes is the most technically correct, often refined to it's smallest elements, but also not nearly as ready to use as the third or fourth.

The third or fourth cover actual use case like solutions, of course that leaves more room to quibble and vote down, but you learn more from it.

It's the difference between answering a trivia question about a car vs actually fixing a car.

TrevorJ• 7 years ago

The worst part about it is, all these deleted/locked questions rank really high on google so they pollute the search results.

LoSboccacc• 7 years ago

as most of these discussion they go from "these number are impossible!" to "but wait! there's algebra!" without really providing a framing context of what's going on.

this is what's going when the article say "slides smoothly in toward 2"

https://www.google.com/search?q=(x^2%2B2x)/x

and this is another common example, which "slides" toward a different number (which is why when you see the division alone is "undefined" and only acquires value in context)

https://www.google.com/search?q=sin(x)/x

but beyond the "just apply limits" party trick it's the implication however that's the most interesting part, and I'll use the words from here[1]:

Things that appear to be zero may be nonzero in a different dimension (just like i might appear to be 0 to us, but isn’t)

[1] https://betterexplained.com/articles/why-do-we-need-limits-a...

lucasgonze• 7 years ago

I'm inspired by how clean his reasoning is. Something basic but fuzzy which we refer to all the time - "divide by zero is undefined" - actually contains meaning! This makes me happy.

ggm• 7 years ago

The one thing his article didn't say for the arithmetically challenged person (me) is very much about 1/0 - he certainly talked about anything/0 and function()/0 but dived off why 1/0 and 0/0 have subtle difference.

It's a good read but he missed his clear mission. One extra paragraph?

mannykannot• 7 years ago

The middle section is the author's answer: they are different because they fail to yield a definite answer in different ways and for different reasons. You will find some alternative ways of looking at it in the comments here.

marknadal• 7 years ago

The main argument I've seen against division by zero is because:

"Then 2 + 2 could equal 7!"

If you go through the examples of how they derive that, you can actually fix it quite easily by assigning each 0/0 pair to different variables. This prevents the underhand commutative rule violation.

Then all that you are left with is linear algebra. Whether that equation is solvable depends upon how many constraints you have. But it at least doesn't cause math to violate itself.

throwawaymath3• 7 years ago

That doesn't work. You're changing the setting of the problem from scalars to vectors in order to avoid breaking the uniqueness properties of fields and rings. The solution set for a system of linear equations is not unique (if it is at all) in the same sense that e.g. additive and multiplicative identity elements are unique.

maxk42• 7 years ago

My layman's interpretation:

Taking the author's argument that a/b = the number x, for which x * b = a, it's easy to draw this out to it's logical conclusion:

For numerators (that is: a) other than 0, you get impossible equations. That is: 1 / 0 = "the number which, when multiplied by 0 is 1." In equation form: x * 0 = 1 Well no such number exists. Therefore x can be said to be "no number."

But for the case where a and b are both zero, the equation becomes: x * 0 = 0. And in this case, the answer is any number, since any number multiplied by zero = 0.

So we can think of 0 / 0 as "any number" and non-0 / 0 as "no number".

That means assigning each instance of 0/0 to a variable is a very good solution. Because we use a letter to represent an unknown number in basic algebra, but in linear algebra we also use a letter to represent a value which could be any number. As in the equation for a line: y=mx+b, where x and y can both take on any value.

mjd3• 7 years ago

I wrote a rather detailed article on this exact topic. You might enjoy it.

https://math.stackexchange.com/a/125208/25554

tim333• 7 years ago

Well, in javascript 0/0 == NaN while 1/0 == Infinity.

An interesting thing is they have the value -0 so you can set a=-0 and 1/a returns -Infinity.

strainer• 7 years ago

This is not only javascripts behavior, it is the IEEE standard for floating-point arithmetic which most modern hardware and software strictly adheres to.

dylan604• 7 years ago

This should be part of an update to the Destroy All Software's Wat video: https://www.destroyallsoftware.com/talks/wat

garmaine• 7 years ago

Except that is defined, standard IEEE 754 behavior and is true on nearly all programming languages.

stOneskull• 7 years ago

-0 is my favorite number

ikeboy• 7 years ago

When f(0)≠0, we can state decisively that there is no such Q

This is false. This relies on an assumption not stated - that f is continuous.

For a simple counterexample:

f(x) = 1, x=0

f(x) = x, x!=0

g(x) = x

Limit of the ratio as x approaches 0 is 1.

TomMckenny2• 7 years ago

I wrote an answer I thought was pretty good, but the question was downvoted and deleted as “not about mathematics”.

Why post on some else's site? Especially quality work that took real effort. I'm guilty of it too (although nothing so cool is this article).

We complain about the web becoming dangerously centralized compared to the past. Much of this is because of free high grade labor that adds value to an already dominant organization.

jobigoud• 7 years ago

The usual answer to that is "discoverability". You can either post your content to someone else's website and have it read/viewed/watched, or post it where virtually no one will see it so you might as well not post it.

TomMckenny2• 7 years ago

Yes but discoverability to what end? In one case the reader then thinks and says "I learned this on Stack Exchange" rather than a possibility of "I learned this from Mark".

In one case an organization that is much complained about becomes even more of a necessity and so has less incentive to fix itself. In another the reverse happens.

In hyperbole: in one case you may as well not post it, in the other case you may as well have not written it.

8note• 7 years ago

my calculus answer is that usually you see 0/0 when looking for limits

1/0 definitely has no real value 0/0 might have one, but you've framed the question poorly -- if you rewrite the problem in a different way, you might find the value of 0/0 for your case eg, by applying L'Hôpital's rule

ddxxdd2• 7 years ago

Someone needs to copy this poster's simplistic style, and use it to explain to me why Laplaces equation of electrostatic potential is solved with Bessel functions times an exponential times a sine/cosine wave (edit: in cylindrical coordinates).

codethief4• 7 years ago

I suppose you've already looked at:

https://en.wikipedia.org/wiki/Cylindrical_harmonics

ddxxdd2• 7 years ago

No, but that page has all the info that Jackson's Classical E&M textbook already explains.

I was being slightly facetious, but any average person will tune out a complete derivation after 15 integral signs, 50 terms, and 100 factors unless there is a pleasing narrative woven throughout the discussion.

mjd3• 7 years ago

I haven't done electrostatic potential yet, but I have been blogging for 13 years. Maybe you would enjoy one of my other articles.

I like this one: https://blog.plover.com/math/60-degree-angles.html

whatshisface• 7 years ago

Here's the answer: an exponential times a sine/cosine wave is what you already have in rectangular coordinates. The Bessel functions show up because they are the cylindrical coordinate equivalent of sine, cosine and exponential. "What does that mean," you say? It is quite simple, as sine, cosine, the exponential function and all Bessel functions are defined as the solutions to specific "fundamental" differential equations. The basic thread between solving Laplace's equations in any coordinate system is, you separate variables and break down the equations in other ways until you have factored them into a set of standard ("prime?" I don't know if that's a good word, I'm not a mathematician) DEs that somebody like Bessel has tabulated the numerical solution of.

tzs• 7 years ago

Well, 1/0 - 0/0 = (1-0)/0 = 1/0, so the difference is 1/0. :-)

aijony• 7 years ago

or -1/0

ngcc_hk• 7 years ago

Or whatever/0

For the article what it try to do is using limit to see whether the difference converge to a finite number

diff(x) = one(x)/zero’(0) - zero(x)/zero’’(0)

Where one and all zero function approach 1 and 0, smooth etc

Find diff(x) when lim x-> 0

codethief4• 7 years ago

The author's explanation is based on how the rationals are defined explicitly but this leaves the question open whether the rationals couldn't be extended somehow, so that the inverse of 0 is suddenly defined in this extension. More generally, couldn't there be other fields where 0 has a multiplicative inverse?

Of course the answer to both questions is negative but it's the consequence of a much more general fact: In any field the additive zero element can never have a multiplicate inverse. (If it did, the field's other algebraic operations would suddenly lead to contradictions.) And in this sense, there is also no difference between the expressions 1/0 = 1 · 0⁻¹ and 0/0 = 0 · 0⁻¹ because 0⁻¹ simply doesn't exist.

I've always found this explanation much more illuminating. No matter what you do, you can never have 0⁻¹!

sunstone• 7 years ago

Yes this seems like the right answer, both results are undefined so they are the same but not equal.

daveFNbuck• 7 years ago

You can have a multiplicative inverse of 0 if 0 = 1. This results in a field with only a single element.

gizmo6863• 7 years ago

Unless your definition of field include the non-triviality axiom that 0 != 1.

Certhas• 7 years ago

Not to be confused with the field with one element :P

https://en.wikipedia.org/wiki/Field_with_one_element

throwawaymath3• 7 years ago

The definition of a field explicitly requires the additive and multiplicative identities to be distinct elements. You can't induce a field structure on a singleton set. You can construct a trivial ring, but the trivial ring does not comprise a trivial field because it fails to be an integral domain.

To be fair this is mathematical pedantry when we're talking about trivial objects. But in principle it's meaningful because every field you can construct will lose critical properties if you allow one element to be both the additive and multiplicative identity. Unfortunately authors don't always make it clear that you need both existence and uniqueness when they state the field axioms.

mjd3• 7 years ago

I did actually write a separate article on this exact subject, which I think is also pretty good.

https://math.stackexchange.com/a/125208/25554

codethief4• 7 years ago

Thanks for the link! I liked your explanation that

[the] warped numbers are like a hole that you can fall into but you can't climb out of, and 0/0 is a deeper hole inside the first hole.

pstch• 7 years ago

Although they aren't a field as such (if I'm remembering well, because they don't form a division algebra), sedenions (16-tuples, and also further applications of the Cayley-Dickson construction) do have zero divisors (84 of them).

This makes giving any kind of sense to 0^(-1) even harder for me (when 0 is a sedenion).

chess93• 7 years ago

No denominators for zero divisors in the localization of any ring!!! [English: No denominators where you can multiply the denominator by anything nonzero to get zero.]

Technically, many definitions of localization do allow for zero/zero divisors to be included (we can use any multiplicatively closed set) however by the definition of localization all of the elements in our localized ring become equal to each other.

The mathematical meaning of a/b = c/d in a general ring is that there exists some value s in the set of legal denominators where s(ad - c*b) = 0. If 0 is an allowable denominator then a/b = c/d for every a,b,c,d.

kccqzy• 7 years ago

I've sometimes used in Alexandroff one-point compactification in scenarios where you really "need the zero to have an inverse" and it works well in certain applications.

graycat• 7 years ago

To answer the question, move to the extended real numbers as in, say, H. Royden, Real Analysis.

In short, the extensions work and at times are convenient but work less well than the rules of the real numbers; the extensions are a bit tricky and have to be careful.

pugworthy• 7 years ago

In a kind of both joking way, and in a serious as a programmer way, either way is the same - and not exceptional. I mean beyond that it will raise an exception. Either way, you divided by zero.

gizmo6863• 7 years ago

But to really understand the difference you probably need to use the calculus approach

I am probably showing my bias as an algebraist here, but there is a purely algabraic way of seeing the difference between 0/0 and 1/0, which is (in my view) more intuitive.

The standard way of constructing the field of fractions is to start with the integers (or your choice of base ring), and define an equivalence relation on the set:

    Q = { (x,y) | x,y are integers, y!=0 }

Given by:

    (a,b) ~ (c,d) iff ad = bc

Intuitively, the pair (a,b) is the fraction a/b, and the equivalence relationship is defining that: a/b = c/d is equivalent to ad =bc (assume b,d are non-zero). From that you can define multiplication and division in the expected way and verify that the resulting structure is a field that behaves as you would expect the rationals to behave.

One thing to notice in the above view is that the prohibition on dividing by 0 is entirely artificial. By removing 6 characters from the above definition, we arrive at the following structure:

    W = { (x,y) | x,y are integers } 
    (a,b) ~ (c,d) iff ad = b

Let [W] be the set of equivalence classes on W, and [Q] be the set of equivalence classes on Q.

If we define addition and mulitplication on [W] and [Q] in the "obvious" way, then [Q] behaves identically to the field of rational numbers (indeed, this is a common construction.

Further, [W] contains [Q] as a subset, so we can view [W] as an extension to the rational numbers.

We can verify that [W] contains exactly two elements not present in [Q], which I will denote:

  ⊥ = [(0,0)] = 0/0
  ∞ = [(1,0)] = 1/0 = a/0 for a!=0 and a!=⊥

The question now becomes, is "division" meaningful on [W]. If we keep the same definition of multiplication as with [Q], it is clear that we cannot define an inverse operation [0].

However, we can define a unary operation called the reciprical, given by: /[(x,y)] = [(y,x)]

And define a division operation as multiplication by the reciprical. So:

[(a,b)] div [(x,y)] = [(a,b)] * (/[(x,y)]) = [(a,b)] * [(y,x)] = [(ay,bx)]

And note that, on the subset [Q], this operation is identical to our standard notion of division.

With this, we have extended [Q] into a structure, [W], for which division by 0 is defined.

Further, by doing so, we added exactly 2 elements: 0/0 and 1/0. So we can see that 0/0 and 1/0 are distinct (in this structure) in a way that 1/0 and 2/0 are not. (as 1/0 ~ 2/0 in the same way that 1/1 ~ 2/2)

In fact, the behaviour of ⊥ ∞ should not be suprising to this forum.

⊥ behaves essentially like NaN in that it "absorbs" the other number in all operations. So:

    x + ⊥ = ⊥ and 
    x * ⊥ = ⊥.

∞ behaves mostly like you would expects:

    0 * ∞ = 0, 
    x * ∞ = ∞ | x != 0 and x != ⊥
    x + ∞ = 0 | x != ⊥

The main "weirdness" here is that ∞=-∞

In fact, this [W] I have been describing is an established structure refered to as a Wheel [1], and can be constructed over the real or complex numbers just as easily (it can also be viewed as an extension of the more well known Riemann sphere by adding ⊥)

[0] I will note that the lack of division by 0 means that we cannot define an inverse operation for multiplication in [Q] either.

[1] https://en.wikipedia.org/wiki/Wheel_theory

cygaril• 7 years ago

Assuming the definition of ~ is meant to be the same on Q and W (I think there's a typo), then it's no longer transitive and therefore no longer an equivalence relation: (0, 0) ~ (1, 2) and (0, 0) ~ (2, 3) but not (1, 2) ~ (2, 3).

gizmo6863• 7 years ago

I'm passed the window where I can edit my post, so I will correct it here.

You are correct that my original post is mistaken. I wish I could attribute that to a typo, but it was really me just working from memory without checking my work. A correct equivalence relation for a wheel over integers is as follows:

    (a,b) ~ (x,y) iff there exists s1,s2 (both non-zero integers) such that:
        (s1 * a, s1 * b) = (s2 * x, s2 * y)

When b,y are non zero, this is the same as the equivalence relationship on fractions.

cairo_x• 7 years ago

Nothing has an infinite number of nothings in it.

One has zero nothings in it...? It also has an infinite number of somethings in it. And each something has another infinity of divisible somethings within that.

Therefore anything that isn't nothing (in the universe of abstraction which doesn't exist in reality, but whatever) is infinite.

OBVIOUSLY.

NOTE: Hasn't read the article and is mathematically challenged.

Rhinobird• 7 years ago

For x⋅0=a (i.e. 1/0) no number fits. It's like a database search that finds nothing. It's a null. A point that is not in the set.

For x⋅0=0 (i.e. 0/0) any number fits. Is there such a thing as an anti-null? In the infinite set, 0/0 is infinite.

If a/0 isn't any number if you exclude 0, then a/0 should be zero. Tada!

...of course zero wasn't in the set, which makes it a null

meuk• 7 years ago

This can be explained clearly in two sentences:

Division is the inverse of multiplication (a * b / b = a / b * b = a). However, multiplication by 0 is not injective (we have x * 0 = 0 for every x), so multiplication by zero can't have an inverse operation, and both 0/0 and 1/0 are undefined.

EForEndeavour• 7 years ago

This explanation correctly explains why 1/0 and 0/0 are not numbers, but doesn't answer the question of what makes 1/0 and 0/0 different. And they aren't both undefined.

1/0 is undefined because no value of x satisfies 1/0 = x.

0/0 is indeterminate because any value of x satisfies 0/0 = x.

throwawaymath3• 7 years ago

0/0 is only indeterminate if your axioms allow or require it to be. Strictly speaking you can define 0/0 to mean something, but you will lose a lot of useful properties along the way. The only time 0/0 is not indeterminate is when you're dealing with something exotic and obscure like an algebraic wheel. It's correct to say 0/0 is undefined because in the vast majority of cases where someone doesn't explicitly call out the algebraic setting, they're working with a field.

bahhh• 7 years ago

George Boole proposed something similar in The Laws of Thought: he considered a definition of c = a/b where a = c AND b (even if he didn't write it like that).

Thus, 0/0 was an indeterminate value, and 1/0 was seen as an impossibility.

Boole's example: if m means "is mortal" and h means "is human", then saying "there is no immortal humans" is h AND NOT m = 0, and Boole explains that from it you could deduce: m = h + 0/0 and not h meaning "mortals are humans and something indeterminate which is not human"

here it's still George Boole trying to cram normal arithmetics into logic, but still.

yuriko• 7 years ago

I'm definitely joking, but

$ node

1.0/0.0

Infinity

0.0/0.0

NaN

jchook• 7 years ago

I read a relevant article about how the semantics of division by Zero ultimately rest on arbitrary decisions of the mathematics community.

Trying to dig this up I found these:

- https://www.1dividedby0.com/

- https://www.hillelwayne.com/post/divide-by-zero/

askvictor• 7 years ago

Also fun: 0^0 (not an emoji; 0 to the power of 0). Windows Calculator gives an answer of 1; Excel throws a number error, while Android hedges its bets (shows "Undefined, or 1").

murkle• 7 years ago

Extended number line is one way to formalise this https://en.m.wikipedia.org/wiki/Extended_real_number_line

What's the difference between 0/0 and 1/0?

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