# Q: Can you fix the “1/0 problem” by defining 1/0 as a new number?

The original question was: I’ve been relearning some things about imaginary numbers and the concept behind the number $i$ got me thinking about something else. Is there a reason that the quantity $1/0$ could not be defined in a similar way to $i$, so that functions could a real component, imaginary component, and an undefined component? If so, what implications would this have to mathematics? I can’t see it being very useful but I was curious to see what you think.

Physicist: If that worked it would be extremely useful. However, it just doesn’t jive with the axioms of arithmetic.
People in the math biz are always making up place holders for things that aren’t known or, in some cases, can’t exist. Euler (pronounced “Oiler”, as in “one who oils”) wanted to come up with a number system that included a solution to x2+1=0. He called that solution “$i$ “, for “Imaginary number” or possibly “Incredibly awesome number” (I think it’s the latter).

As it happens, there are no problems involved with defining $i$.  In fact, it really cleans up a lot of math.  1/0 on the other hand is kind of a train wreck.

Define $Q$ (for “Quite a bit more awesome than $i$“) as the solution to $0x=1$. That is, define $Q$ as “1/0”, so it’s the “multiplicative inverse” of 0.  Right off the bat there’s a problem:
$\begin{array}{ll}1=Q\cdot 0\\\Leftrightarrow 1=Q\cdot (1-1)\\\Leftrightarrow 1=Q-Q\\\Leftrightarrow 1=0 \end{array}$

This is because zero is defined as “x-x” for any number or variable x.  So by assuming $1=Q\cdot 0$ (and the laws of arithmetic) you reach an impossible conclusion!

You could try to patch this problem, for example by declaring that $Q-Q\ne 0$.  Even so:

$\begin{array}{ll}1=Q\cdot 0\\\Rightarrow 1\cdot 0=Q\cdot 0^2\\\Rightarrow 0=Q\cdot 0\\\Rightarrow 0=1 \end{array}$

Again, by defining $1=Q\cdot 0$ to be true, you’re led to a contradiction.  Mo’ logic, mo’ problems.  You could fix this problem by declaring that associativity doesn’t apply to $Q$.  That is, $(Q\cdot 0)\cdot 0 \ne Q\cdot (0\cdot 0)$.  Losing associativity is a big deal though.  Without it you can barely do anything.

You can keep going, finding more problems and declaring more “fixes”, but in short order you’ll find that by the time you’re done patching things you’ll have more problems, exceptions, and caveats than just “1/0 is undefined”.

Best to just leave 1/0 undefined.

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### 26 Responses to Q: Can you fix the “1/0 problem” by defining 1/0 as a new number?

1. Boreeas says:

1 = Q*0
=> 1*0 = Q*0²
=> 0 = Q*0
=> 0 = 1

How did you get from step 3 to step 4? If Q*0 = 1, then the first step is already true, and the other steps unnecessary, or am I wrong?

2. The Physicist says:

This is a “proof by contradiction”.
The first step is defined to be “true”, but it can’t be because it leads to a false statement (0=1).
I’ll go back and clear that up.

3. Neal says:

Another way of phrasing this problem: for all (rational) numbers but 0, the map “multiplication by *number*” is bijective, so it can be inverted. For 0, the map “multiplication by 0” is very horrendously not even injective. So it can’t be undone.

4. Alex says:

Actually it’s been done

http://en.wikipedia.org/wiki/Wheel_theory

“Wheels are a kind of algebra where division is always defined. In particular, division by zero is meaningful.”

In essence the reciprocal of zero is defined, but the “normal” rules of arithmetic such as “x-x=0” may not necessarily hold.

5. The Physicist says:

Awesome!
Still, given the choice between that set of rules and “don’t divide by zero”, I’d go with the second choice.

6. Neal says:

Modification of earlier comment: In any ring R, (right) multiplication maps R into its self-maps. Non-zero-divisors go to non-invertible maps. In particular, 0 is not a zero divisor, so (right) multiplication by 0 is not invertible: “x/0” makes no sense.

7. Paul says:

The wheels construction looks cool. Though we can still say that one can’t “fix” the 1/0 problem — and stay within the framework of (commutative) rings / fields (i.e. the usual “axioms of arithmetic”). In contrast with the inclusion of i to go from reals to complex, which is the simplest of field extensions.

Neal I believe in your description, “units” should take the place of “zero divisors.”

8. Neal says:

Yes, thank you.

9. marshal says:

here you try to manipulate Q=1/0 to Q*0=1 this would be done by multiplying both sides by zero which means Q*0=1 or 1*0/0=1 or simply 1=0/0, which is a contradiction . in order for your proof by contradiction to work you have to accept your unintentional corollary, 1=0/0 which would produce a very bizarre set of numbers. now accepting this new invention there is the conclusion that 1=0 or 0=1=0/0 and Q=1/0.

10. The Physicist says:

I should have been more clear.
“Q*0=1” is the definition, and “Q=1/0” is merely a convenient way of writing Q, since that’s the way every other multiplicative inverse is written. But be careful not to put any weight on it (it doesn’t hold up).

11. tonyf says:

As Boreeas pointed out, for your second example, even in ordinary arithmetics 0=Q*0 does not imply 0=1, 0=Q*0 does not imply anything.

For your first example, as you noticed yourself, the problem is defining Q-Q=0. Actually, the question of giving 0/1 a value has already been solved, in two somewhat related ways: so-called non-standard analysis, and many computer implementations of arithmetics. In the latter we introduce two numbers in addition to the standard ones: NaN and Inf, informally we can think of them as “not a number” and “infinity”. Here 1/0=Inf, Inf-Inf=NaN. Your first example would work because Inf is not the multiplicative inverse of 0, instead any positive number divided by zero is directly defined as Inf (and as -Inf for negative), and 0*Inf=NaN. So the trick to get all arithmetic operations well defined (i.e., also division by zero when only ordinary numbers are concerned) is that it is not enough to introduce only one additional non-conventional number (“Q”), you need two new numbers (“Q=Inf” _and_ “NaN”) to get both ordinary number divided by zero well defined, and all the operations concerning the new non-conventional numbers well defined in a consistent way.

12. Tom says:

Well im just throwing this out there… 1/0 doesnt work in general as zero times any number will not equal one, even if you define an arbitrary symbol to equal this impossible number, that symbol really represents an either negative or positive infinite limit and cannot be defined as a single value, and as for defining Q as NaN and Inf is counterproductive for the original question, as operations using two non-existant numbers cannot be arithmetically calculated not to mention you just defined one symbol as two different values, which cannot exist in real functions, so then when does this become useful?

13. The Physicist says:

You’re certainly right, there is no number that when multiplied by zero is one. Which is why “Q” in the post isn’t an actual number, it’s just some mathematical object with the correct property, in very much the same way that “i” is just an “impossible” mathematical object.
I think what tonyf wanted to do was abandon Q altogether and replace it with two new symbols. But, there are plenty of examples of systems involving more than one made-up mathematical objects. For example, “quaternions” include the usual “i”, but also include “j” and “k”, which have identical properties to i, but interact weirdly with each other.

x > 0
x/0 = y
x = y*0
x > AND = 0, which is a contradiction.

A better one:
x ≠ 0
x/0 = y
x = y*0
x ≠ AND = 0, which is a contradiction.

16. John Morgan says:

This thread on 1/0 puts me in mind of something I used to tell my kids when they were very young and fascinated by the idea of infinity (this was before the public school system managed to beat all interest whatever in mathematics out of one of them-to my instant, white-hot, and abiding fury.) I’m no great shakes at explaining mathematics to the very young, but the Riemann construction showed me what I think is a good way to think about infinity and numbers:

“Infinity is NOT a number, but you can sometimes think of it as a place”

Having said that, the person who posed the original question was close to the idea of an extension to the reals (AKA the “point at infinity”) which Riemann so memorably visualized.

17. pi says:

If I define a function called f(x) = x^2 – 6x + 9 / x^2 – 9
then, if x = 3, then f(3) = 0/0, which is undefined.
But, if I simplify the fraction by factorization, it would equal, f(x) = (x-3) / (x+3). If you evaluate for f(3) now, it will now equal to 0/6 = 0.
Another example is g(x)= x-1/x-1
If I put in g(1)=0/0, which is undefined.
But g(x) can be simplified to 1/1 =1 . And g(1) is 1.
How is this possible? Doesn’t it violate mathematics? I mean if one expression can give you an answer, then simplifying it should also give you the answer because you have only simplified the approach to the solution. How can it have two solutions?

18. The Physicist says:

Simplifying doesn’t quite leave the equation alone. In the examples you gave the new simplified functions keep their undefined points. So, (x-1)/(x-1) = 1, for x≠1. The “for x≠1” is important there, but it’s subtle.

19. Gavin Grubb says:

If you divide any number that’s not zero by zero you will get negative or positive infinity because zero tries to turn the number into zero by turning it into X groups of zero. (Hence why division makes a number into something smaller.) Now if you’re going to say that the answer should be zero because it makes it into zero groups, you’re wrong!

Dividing any number by zero is basically breaking it down by any means possible until you reach nothing. Now multiplying by zero is like dividing by any other number; it gets smaller. When dividing, it’s the exact opposite; it gets [U]Value[/U].

What about 0/0? It suppose to be every number that can be multiplied by zero, or 1, -500, 0, or infinity… This means the theory of something coming from nothing can be true, but it’s not exactly clear…

What if, 0/0 was suppose to happen, that would mean the equation was suppose to give value to anything. This could be the base of our number line…

20. The Wonderer says:

I have often thought of the same thing as this, but you have proved me wrong that this cant be done unless you make some special rules to go along with it. Alephs have their own special rules why not this?

21. I think the problem deals with SO MUCH imaginary numbers, which in essence isn’t real. To mean something; a number MUST be in front of a [unit], but zero cancels all units.!!! so if you reach definition for 1/0 then i can turn my lead into gold… a*[lead]=b*[gold],

In essence 1/0 is the philosopher’s stone which in essence is again OUT OF THIS universe [units]… because one apple IS EVEN AN ESSENTIALLY DIFFERENT to another apple, just in math world can be treated as equal….

22. Physics Pi says:

Could 1/0 be infinity and 2/0 be 2*infinity and so on? That would mean that 0*infinity would be a range that includes every real number since 0/0 could equal any real number.

23. Angel says:

@tonyf:

NaN and inf are not actual numbers. They don’t have definitions appliable to mathematics. They merely eliminate the floating point of n/0.

24. Angel says:

Seeing that quaternions disobey multiplicative commutativity and octonions disobey some forms of multiplicative associativity, I see no problem with ignoring multiplicative properties in a system in which 1/0=Q is defined. In fact, the imaginary unit i itself disobeys distributive exponentiation. For instance, i^2 = sqrt(-1)^2 = sqrt[(-1)^2] = sqrt(1) = 1, which is a contradiction by definition, because that would mean i=-1 or 1, which is false.

If we define Q to be the number that satisfies the equation x^(-1)=0, then, in order to apply coherent arithmetic to our new extension to the number system, we need to define the indeterminate values. In particular, we ought to define 0/0 and therefore 0^0 both to equal 1, such that the definition of Q is satisfied. Then, it must follow that Q – Q = 1, and furthermore, Q + 1 = Q.

Another property we need to assign in order to create a coherent system is that L < Q < M, where L is an infinitely large positive number (a.k.a non-finite natural) and M is an non-finite negative. The geometric implication for this is that the real number line is only a line at the locality of finite numbers, but the entirety of it is actually a circle. Extending this construction to the complex numbers creates not a flat plane, but a torus. We now have that 0*x=0, x unequals 1/0, where n unequals 0, and 0*x=1, x=1/0, so now multiplication by 0 is a piecewise function, and the property of multiplicative nullification is restricted to values of non-Q nature. Moreover, there now exists an additive counterpart to the property of nullification; namely, Q + x = Q: this property is also restricted and untrue for all x, in order to maintain coherence. For example, the aforementioned is false if x = Q: Q does not equal 2Q or -Q. Multiplicative associativity is no longer true for all numbers, and the property of additive inverses is no longer always true either, but the property of multiplicative inverses is now always true, even for 0. This system is very similar to the wheel algebra theory, but it is still different in some respects.

Other indeterminates are now to be defined and fixed for the sake of coherent arithmetic. 1^Q=Q, and in general, x^Q=Q except for x=0. This implies that for nonzero x, x^(-Q) = 0. 0^Q = 0, so that 0^(-Q) = Q. Q/Q=1, Q^0=1. We have actually eliminated problems and restrictions with many exponentiation rules, though we have needed to ignore multiplicative associativity and perhaps some distributive properties, as well as restricting the additive inverse.

It seems impractical and inapplicable to have a number system in which there is a number Q for which Q + x = Q for an infinite range of values of x. However, if there exists a number 0 for which 0*x=0, then I see no problem with the earlier formula.

One final identity that would need to be acknowledged: Q/0=Q^2.