Why is Division by Zero Considered Undefined?

[heshbon]

Division by zero is undefined...but why doesn't somebody just define it?

Like they did with root (-1).

 

[NoMoreExams]

Think of what division is relative to multiplication. You can define it to be whatever you want but it has to actually make sense.

Check out some links that google spits out: http://www.google.com/search?hl=en&...tnG=Google+Search&aq=0&oq=why+is+division+by+

BTW this isn't a stupid question, people obviously wonder about things like that :)

 

[DaveC426913]

Division by zero is undefined...but why doesn't somebody just define it?

Like they did with root (-1).

Arithmetical operations have to result in unique numbers.
4 multiplied by 6 can result in one and only one number.
100 divided by 12 results in one and only one number.
1 divided by zero does not result in a unique number. What it results in is not numerically definable.

 

[MiGUi]

Division by zero is undefined...but why doesn't somebody just define it?

Like they did with root (-1).

\forall x\in \mathbb{R}, x \times 0 = 0

Then

Let be a, b \in \mathbb{R}, a \neq b

We said before that a \times 0 = 0[/tex] and also b \times 0 = 0. <br /> <br /> So, a \times 0 = b \times 0[/tex]. If we were able to divide by 0, we will have that a = b which is a contradition with our suppose.

 

[Tac-Tics]

Division by zero is undefined...but why doesn't somebody just define it?

Like they did with root (-1).

Check the archives. There are good discussions about this question.

The basic problem is there IS no way to define it in such a way that it is the full inverse of multiplication.

What does that mean? Two operators are inverses if they "undo" each other, regardless of input. Addition and subtraction are the best known example. For any x and y, x + y - y = x. You start with x, you do "+y" then you "undo" the +y with a -y.

But there is no way to do this with multiplication. You can do it for *almost* every single number. Clearly, if we have 16 * 4 / 4, that equals 16 and division has "undone" the multiplication. In fact, this works for every number except 0. Try it out.

The proof of this lies in the fact that multiplication is not one-to-one (it does not map unique inputs to unique outputs). Functions which are one-to-one all have inverse functions. Functions which are not one-to-one may only, at best, have partial inverses.

This contrasts to imaginary numbers. Square root, unlike division, IS one-to-one. No two real numbers have the same square root. However, it happens that some reals don't have square roots. The imaginary number system is a way to extend the real line to preserve the square root function's one-to-one-ness, and at the same time, upgrade square root from a partial function to a total function.

 

[wsalem]

What does that mean? Two operators are inverses if they "undo" each other, regardless of input. Addition and subtraction are the best known example. For any x and y, x + y - y = x. You start with x, you do "+y" then you "undo" the +y with a -y.

This definition is too restrictive, just say that y and it's inverse under the operation gives the identity.
Here's a counterexample. Define the binary operation on a set S = {a,b,c} as follows
<br /> \newcommand{\mc}[3]{\multicolumn{#1}{#2}{#3}}<br /> \begin{array}{|l|lll}\hline<br /> \mc{1}{l}{*} &amp; \mc{1}{l}{\bfseries a} &amp; \bfseries b &amp; \bfseries c \\ \hline<br /> a &amp; a &amp; b &amp; c \\ <br /> b &amp; b &amp; a &amp; b \\ <br /> c &amp; c &amp; b &amp; a<br /> \end{array}<br />
The operation is commutative, a is the identity, and for each x in S, the inverse is also x. Let's do b on c, cb = a let's undo b by it's inverse, then (cb)b = ab = b[/tex] but now the original c<br /> This is due to the fact that your definition would work only for associative operations!<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The proof of this lies in the fact that multiplication is not one-to-one (it does not map unique inputs to unique outputs). Functions which are one-to-one all have inverse functions. Functions which are not one-to-one may only, at best, have partial inverses. </div> </div> </blockquote>Does that mean addition is one-to-one? What is the definition of a one-to-one here?<br /> Take (7,1) \neq (5,3)[/tex] but +(7,1) = +(5,3), in other words, I can&amp;#039;t see how given an output, one can determine the unique ones which produced it (the usual one-to-one definition!)&lt;br /&gt; Actually &amp;quot;map unique inputs to unique outputs&amp;quot; doesn&amp;#039;t suggest one-to-oneness but rather that the operation is well defined, i.e given a unique x, it gets mapped to only one f(x)!

 

[Tac-Tics]

This definition is too restrictive,

I wasn't giving a full definition. Just the flavor of it.

You shouldn't really speak of an operator's inverse. You speak about sections of that operator, where only one argument is applied, such as (*2), the doubling function. The inverse function would be (/2), the halving function.

The argument is that the section (*0), which maps all real numbers to 0, has no inverse.

When speaking about addition, what I meant was that all functions (+n) have an inverse called (-n). To contrast, for all n EXCEPT 0, (*n) has an inverse (/n). That exception cannot be removed because (*0) is not one-to-one. In fact, it is quite the opposite.

 

[wsalem]

You shouldn't really speak of an operator's inverse. You speak about sections of that operator, where only one argument is applied, such as (*2), the doubling function. The inverse function would be (/2), the halving function.

Let me clarify that when I said "y and it's inverse under the operation gives the identity.", I was implicitly assuming "if it exists" both for the inverse and for the identity. Obviously an operation doesn't need to have any of these!

The argument is that the section (*0), which maps all real numbers to 0, has no inverse.

Absolutely! I never argued that!

When speaking about addition, what I meant was that all functions (+n) have an inverse called (-n). To contrast, for all n EXCEPT 0, (*n) has an inverse (/n). That exception cannot be removed because (*0) is not one-to-one. In fact, it is quite the opposite.

I'm sorry but I can't follow, what "all functions (+n)" are you talking about here, do you mean to define a function for each n, why not just a single function that sends each element to it's inverse?

To contrast, for all n EXCEPT 0, (*n) has an inverse (/n). That exception cannot be removed because (*0) is not one-to-one.

Surely the exception cannot be removed, but then if we define such function, actually *0 wouldn't be "not one-to-one" but rather undefined!
This is what my argument is about, I can't understand what do you mean by one-to-one!

 

[wsalem]

Tac-Tics, I should have assumed the definition meant to be informal, so I hope my reply didn't offend you. I simply couldn't understand what you meant by one-to-one! But this is off the main argument. I think you already made your point which I definitely agree with. Cheers.

 

[Tac-Tics]

Tac-Tics, I should have assumed the definition meant to be informal, so I hope my reply didn't offend you.

You did not offend.

I might as well add one more idea about division. The standard accepted definition is that division isn't even defined at 0. If we're talking about the function f(x) = 1/x (or using "section" notation (1/)), 0 isn't even part of the domain. The advantage of this is that f is actually a continuous function (by the epsilon-delta definition) even though its graph clearly makes a large "jump" from negative infinity to positive infinity.

 

[wsalem]

I'm glad to hear that.

Surely, there's an algebraic explanation why it is undefined, as was elaborated by you and others, but here's an interesting example.
The same operations of R (+ and *) defined on a set with just one element. We can actually freely divide by 0 since we already know that the multiplicative identity equals the additive identity.

But I realize also that the original poster didn't question why it's undefined but rather why we can't find a way to come around it, which I believe was put nicely in the fourth post by MiGUi, which if such inverse exists, it would result in equating all the elements (in other words, reduces the set R to a singleton set, which is contradictory).

 

[symbolipoint]

Simple-minded thought: a number, n is a specific (although unknown, variable) number, and zero is a specific number. n divided by 0 is infinity? But infinity is NOT a number. Division by zero is impossible. Good? Not Good?

 

[NoMoreExams]

I wouldn't go with that argument, infinity is a "number" in the extended reals isn't it?

 

[heshbon]

Is it possible to construct an algebraic system where zero is not unique, prehaps have a unique zero for every element which would than permit division by zero?

Or maybe think of the zero as an operater rather than an identity element where the inverse would represent division by zero.

sorry for the general questions but any response would be greatly appreciated.

 

[wsalem]

Algebraicly speaking, there's a structure called wheels, which does something similar to that. Not sure how useful it is though! See http://dx.doi.org/10.1017/S0960129503004110

Division by zero is still not defined in the extended real line, since you would get both +\infty and -\infty. However, it can be defined on the extended complex plane C \cup \{\infty\} since there's only one infinity, and you would get \frac{1}{0} = \infty. But the structure you get is NOT a field, like R or C. So you won't be treating it as an algebraic structure and get the usual contradictions!