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In regular math, you cannot get the square root of a negative number. Likewise, you cannot divide by zero. Both dividing by zero and taking the square of a negative have no place in real life.

However, we have systems in place to deal with the square root of a negative number (hence, imaginary numbers). Why does an option for divide by zero not exist?

yuritsuki
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    Such an option does exist. The Riemann sphere (i.e. the extended complex plane) is the set $\Bbb C\cup {\infty}$ where $\frac{z}{0}=\infty$ for any $z\neq 0$. Zero divided by zero is still a problem however. Note that the extended real numbers, $\Bbb R\cup {+\infty,-\infty}$, still have a problem with division by zero however since it is not clear whether $\frac{x}{0}$ should equal $+\infty$ or $-\infty$. – JMoravitz Jan 22 '16 at 01:40
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    Options do exist see for example https://en.wikipedia.org/wiki/Wheel_theory Some seem more useful than others. – quid Jan 22 '16 at 01:43
  • Related: https://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility – Hans Lundmark Aug 15 '18 at 12:33

4 Answers4

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I tried this myself. If I understand correctly, you intend by your question to ask "Why can we not define a number (call it $q$), such that $q \cdot 0 = 1$?". Just a bit of algebra here: we have defined 0 as the additive identity. So we have $q \cdot 0 = q \cdot (0 + 0)$. Using the distributive law (which is always required in algebras, as far as I am aware), we have: $q \cdot 0 = q \cdot 0 + q \cdot 0)$. Now subtracting $q \cdot 0$ from each side, we have $0 = q \cdot 0$, which is a contradiction. Thus, such a number cannot exist. Notice, also, that we did not assume that our number was real; in fact, we assumed nothing about the number $q$. Thus it is the properties of $0$ that decide this, not the properties of our hypothesized $q$.

Isomorphism
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  • A fantastic answer, showing how defining a q would bump into other already defined laws. It seems like i doesn't have this problem: it just requires lazy evaluation --- never actually calculating it. – Dogweather Aug 15 '19 at 18:32
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This is best understood in the framework of commutative rings.

There's a functor $$F:\mathbf{CRing} \leftarrow \mathbf{CRing}$$ given as follows: $$F(R) = R[i]/(i^2 +1).$$ We can think of $F$ as adjoining an element $i$ with the property that $i^2+1=0$. Informally, $i = \sqrt{-1}$.

There's also functor $$G:\mathbf{CRing} \leftarrow \mathbf{CRing}$$ given as follows: $$G(R) = R[j]/(j \cdot 0 - 1).$$ We can think of $F$ as adjoining an element $j$ with the property that $j \cdot 0-1=0$. Informally, $j = \frac{1}{0}$.

However, it can be seen that $G(R)$ is always the trivial ring:

$$G(R) = R[j]/(j \cdot 0 - 1) = R[j]/(0-1) = R[j]/(-1) = R[j]/R[j] \cong 1$$

So we cannot get anything useful out of $G$.

From a slightly different vantage point, main the difference between $F$ and $G$ is this: there's an obvious morphism $f_R:F(R) \leftarrow R$, and an obvious morphism $g_R:G(R) \leftarrow R$. But, whereas the morphism $f_R$ is injective for all rings $R$, on the other hand, the morphism $g_R$ is never injective, unless $R$ is the trivial ring, because $G(R)$ is always the trivial ring. So in particular, whereas $\mathbb{C} = F(\mathbb{R})$ can be viewed as an extension of $\mathbb{R}$, on the other hand, we cannot view $G(\mathbb{R})$ as an extension of $\mathbb{R}$. In fact, we cannot get anything useful like this at all.

goblin GONE
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    While this is correct, I think it's pitched a bit above the OP. – Noah Schweber Jan 22 '16 at 02:10
  • All this shows is that if you allow division by $0$, you cannot keep the structure of a commutative ring. But this doesn't mean there isn't some other natural structure you could have where division by $0$ is allowed. – Eric Wofsey Jan 22 '16 at 02:13
  • @EricWofsey, true, but commutative rings are very natural, and I don't know of any truly useful structures out there in which division by $0$ is possible. I'd be happy to be proven wrong. – goblin GONE Jan 22 '16 at 02:43
  • @NoahSchweber, true. I'll try to simplify things if I can work out how to do so. – goblin GONE Jan 22 '16 at 02:45
  • Could you point to a source that'd provide the minimal base knowledge to understand this? Is it category theory? Number theory? – Dogweather Aug 15 '19 at 18:36
  • @Dogweather, it's ring theory. I think Charles C. Pinter's "A book of abstract algebra" probably covers this kind of material. If not, just look up "intro to ring theory" or "intro to abstract algebra" or something like that and find yourself a good book. – goblin GONE Aug 16 '19 at 05:51
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We can define a quantity $i$ to have the property $i^2 = -1$, add $i$ to our set of numbers, and still have all of the rules of algebra work correctly.

Suppose we could define 1/0 to be a certain quantity, say z. Then we have this problem: If we start with the definition

$1/0 =z$, then multiply both sides by $0$, we get

$1 = z*0$, or

$1=0$, which is inconsistent.

Browning
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There is no way to extend the reals in a similar fashion while still preserving all the desired properties of the reals. For example, defining it to be $\infty$ is the most common option, but then what is $0*\infty$? How do we treat it when it come to addition?

siegehalver
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