Division by $0$

Question

I thought it was elementary to me, but I started to do some exercises and came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions.

1. $x/0$ is Impossible ( If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both 1. and 2. are divided by zero, but only 2. has infinite solutions so as 1. has none solution, how and why ?)

2. $0/0$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?)

3. $0/x$ and $x \ne 0$, it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.

Thanks in advance.

Answer 1

The first question you need to ask is: What does "$a/b$" mean?

The answer is: "$a/b$ is the unique solution to the equation $bz = a$." (I'm using $z$ as the unknown, since you are using $x$ for other things).

Given that answer, let's discuss your points out of order:

(3) is perfectly fine: $0/x$, with $x\neq 0$, is the solution to $xz = 0$; the unique solution is $z=0$, so $0/x = z$. The reason it's unique is because $x\neq 0$, so the only way for the product to be $0$ is if $z$ is $0$.

In (1), by "impossible" we mean that the equation that defines it has no solutions: for something to be equal to $x/0$, with $x\neq 0$, we would need $0z = x$. But $0z=0$ for any $z$, so there are no solutions to the equation. Since there are no solutions to the equation, there is no such thing as "$x/0$". So $x/0$ does not represent any number.

In (2), the situation is a bit trickier; in terms of the defining equation, the problem here is that the equation $0z=0$ has any value of $z$ as a solution (that's what the "infinite solutions" means). Since the expression $a/b$ means "the unique solution to $bx= a$, then when $a=b=0$, you don't have a unique answer, so there is no "unique solution".

Generally speaking, we simply do not define "division by $0$". The issue is that, once you get to calculus, you are going to find situations where you have two variable quantities, $a$ and $b$, and you are considering $a/b$; and as $a$ and $b$ changes, you want to know what happens to $a/b$. In those situations, if $a$ is approaching $x$ and $b$ is approaching $y\neq 0$, then $a/b$ will approach $x/y$, no problem. If $a$ approaches $x\neq 0$, and $b$ approaches $0$, then $a/b$ does not approach anything (the "limits does not exist"). But if both $a$ and $b$ approach $0$, then you don't know what happens to $a/b$; it can exist, not exist, or approach pretty much any number. We say this kind of limit is "indeterminate". So there is a reason for separating out cases (1) and (2): very soon you will see an important qualitative difference between the first kind of "does not exist" and the second kind.

Answer 2

The key is to realize what a fraction $\frac ab$ really represents: $\frac ab$ is the number with the property that $\frac ab \cdot b = a$.

So, in the first case if $x \ne 0$, there is no number $\frac x0$ with the property that $\frac x0 \cdot 0 = x$ since anything times zero is zero. So $\frac x0$ is undefined. In the second, any number $y$ has the property that $y \cdot 0 = 0$, so $\frac 00$ could represent any number $y$ according to the above characterization of a fraction, so $\frac 00$ is said to be indeterminate. Finally, if $x \ne 0$ then $0$ has the property that $0 \cdot x = 0$, so $\frac 0x$ is the number $0$.

Answer 3

You will pretty much never see "0/0" or "x/0" in a situation where somebody expects you to actually interpret it as having a value. Other than simple mistakes, the times it comes up are usually of the following sort:

To solve the equation $ax=b$ for $x$, consider the expression "$b/a$".

1. If it is of the form "$0/0$", your equation has infinitely many solutions
2. If it is of the form "$x/0$" for nonzero x, your equation has no solutions
3. Otherwise, the division makes sense and the result is the unique solution

You might hear the word "form" mentioned to refer to the fact that we're talking about a formula, and not the number that might result from evaluating the formula. (especially in the context of limits)

It's important to note the difference, because the number system you use was created so that "$0/0$" and "$x/0$" are not allowed -- they are meaningless when viewed as arithmetic expressions. (the same is true for "$x/y$" if you do not already know that $y \neq 0$) There are lots of very good reasons why arithmetic was created in this way, and the other answers mention some of them.

I feel I should mention that there are other forms of arithmetic that behave differently. The projective numbers are often useful, and 1/0 makes sense in them (but $0/0$ does not). Wheel theory is somewhat more esoteric, but it provides an example where even $0/0$ makes sense. (and also clearly demonstrates the difficulties in accommodating its existence)

Answer 4

If your domain consists of a two-element set, or a set with more than two elements (where any infinite set gets classified as having more than two elements) then Arturo's answer applies.

However, if your domain consists of {0}, then things work out a bit differently, as Bill Dubuque pointed out in a comment elsewhere.

For a system with domain of {0}, you'll still define a/b as the unique solution to (bz)=a. Now, (3) still works out as perfectly fine. x/0 means the unique solution to (0z)=x. Well, x can only equal 0 in this domain, so this means the unique solution to (0z)=0. z can also only equal 0 here, so the equation does have a unique solution of 0. 0/0 means the unique solution to (0z)=0. Well, z can only equal 0 here, and if z=0, the equation does hold, so (0/0)=0 here also.

Answer 5

Let me add this quick bit in addition to what has already been said (assuming you know a little set theory). Take the following semi-random example.

Suppose you have a function $f$ with $f(5) = 14$ and $f(6) = 11$. Going backwards, what do you get? Assuming $f$ is only defined for $5$ and $6$, $f^{-1}(14) = 5$ and $f^{-1}(11) = 6$, right? Now consider a different function, which like $f$ is only defined at two points... let's call it $g$ with $g(5) = 8$ and $g(6) = 8$. Now, what is $g^{-1}(8)$? Well, it wants to be both $5$ and $6$ if it's really the inverse, but defining it either way causes a contradiction so it must be undefined.

So what's the point? In algebra many operations (like multiplying by a nonzero number) are like $f$ and and many (multiplying by zero, for example) are like $g$. Here, draw the picture with arrows... with $f$ you know where the arrows came from, with $g$ two arrows point to the same place so you don't know where you 'came from'.

Now back to multiplication. Imagine the arrows if you multiply by $\frac{1}{2}$ versus if you multiply by $0$. You can see that essentially the same thing is going on, but with infinitely many numbers instead of just a few.

BTW: There is one partial 'fix' if you know a little set theory: (new notation here, NOTE THE CURLY 'SET' BRACKETS): $f^{-1}(\{8\}) = \{5,6\}$. Another example: if $h(x) = x^2$ then $h^{-1}(\{4,25\}) = \{2,5\}$. And back to were we started from, inverting division by zero: if you have $s(x) = 0$ then $s^{-1}(\{0\}) = (-\infty,\infty)$.

Answer 6

Just to give another point of view (purely algebraic one):

Suppose you have a ring (a mathematical structure with addition and multiplication satisfying a bare minimum of laws you need to call them that), and in it there is a number $x$ such that $x \cdot 0 = 1$. Then $1 = x \cdot 0 = x \cdot (1 - 1) = x - x = 0$, which means that for any $y$ in this ring $y = y \cdot 1 = y \cdot 0 = y (1 - 1) = y - y = 0$, so this ring has just one element. Such ring is called trivial, and it is clearly not an extension of the ring of integer numbers.

If you can divide by zero, then $1/0$ is defined and it has to satisfy the property above. Thus, division by zero is possible only in the trivial ring.

Answer 7

Dividing by zero would mean multiplying by the reciprocal of 0.

But 0 has no reciprocal (because 0 times any number is 0, not 1.)

Therefore, division by 0 has no meaning in the set of real numbers.

Multiplicative property of 0

Prove:

If a is any real number, then a0 = 0 and 0a = 0

Proof:

Statement __________ Reason

1. 0 = 0 + 0 ________ 1. Identity property of addition

2. a0 = a(0 + 0) _____ 2. Multiplication property of equality

3. a0 = a0 + a0 _____ 3. Distributive property of mult. with respect to add.

4. But a0 = a0 + 0 ___ 4. Identity property of addition

5. a0 + a0 = a0 + 0 __ 5. Transitive property of equality

6. a0 = 0 _________6. Subtraction property of equality

7. 0a = 0 _________7. Commutative property of multiplication

Therefore, 0 times any number is 0, not 1.

(Source: Algebra: Structure and Method Book 1)

The two cases are:

1. Dividing a nonzero number by zero, violates the multiplicative property of zero and therefore the properties of the real numbers upon which it is proven, as shown above.

2. Dividing zero by zero, which does not violate the multiplicative property of zero, but multiplication by zero is an operation that can not be "undone."

The following argument is presented in an older edition of the above source:

If a/0 = c, then a = 0*c. But 0*c = 0. Hence, if a is not equal to 0, no value of c can make the statement a = 0*c true, while if a = 0, every value of c will make the statement true.

Thus, a/0 either has no value or is indefinite in value.

But, there must be "purely formal" steps which are assumed to be valid in the above argument; namely, if given a/0 = c is an algebraic statement, how can you then derive a = 0*c?

a/0 = c

0*(a/0) = 0*c But in transforming an equation, we should never use zero as a multiplier since we know that the final equation will be satisfied by any real number.

Furthermore, to manipulate the left hand side: 0*(a/0),

implies you are writing 0*(a/0) as 0*(1/0)*a, this must be one of the "formal" steps

and finally asserting 0*(1/0) = 1.

But 0 has no reciprocal because 0 times any number is 0, not 1, this must be another "formal" step.

1*a = 0*c

Thus, arriving at the statement: If a/0 = c, then a = 0*c.

Here is a proof by contradiction:

If (1/0) = x, and x is the multiplicative inverse of 0,

then 0 * (1/0) = 0 * x (multiply both sides by 0. (if two numbers are equal, then to multiply them by the same number, they should still be equal)

1 = 0. (since 0 * 1/0 = 1 (left hand side) and 0 * any number = 0 (right hand side))

1=0 is absurd. So the assumption that (1/0) exists and is a real number, must have been wrong (proof by contradiction).

Division is not always possible in the system of numbers consisting of the integers (6 is divisible by 2 and 3 but not by 5), but in those cases where it is, the result is always uniquely determined.

In the system of all rational numbers (that is, the integers and fractions) division is not only unique but is always realizable with one exception—division by zero. On the basis of the definition of division given above, it is apparent that it is not possible to divide a number different from zero by zero.

The result of dividing zero by zero, according to the definition, can be any number since c*0 = 0 in all cases.

It is usually preferable in algebra (in order not to violate the uniqueness of division) to consider division by zero to be impossible for all cases. Thus division by zero is defined to be undefined.

http://dynamic.uoregon.edu/~teddy/OnDivisionByZero.pdf

Answer 8

Division amounts to repeated subtraction, and subtracting 0 has no effect.