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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?

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marked as duplicate by Jendrik Stelzner, quid, JMoravitz, rschwieb, MJD Jan 22 '16 at 2:13

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    $\begingroup$ 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 $-\infty$. $\endgroup$ – JMoravitz Jan 22 '16 at 1:40
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    $\begingroup$ Options do exist see for example en.wikipedia.org/wiki/Wheel_theory Some seem more useful than others. $\endgroup$ – quid Jan 22 '16 at 1:43
  • $\begingroup$ Related: math.stackexchange.com/questions/259584/… $\endgroup$ – Hans Lundmark Aug 15 '18 at 12:33
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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 $q$.

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  • $\begingroup$ 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. $\endgroup$ – Dogweather Aug 15 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.

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    $\begingroup$ While this is correct, I think it's pitched a bit above the OP. $\endgroup$ – Noah Schweber Jan 22 '16 at 2:10
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jan 22 '16 at 2:13
  • $\begingroup$ @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. $\endgroup$ – goblin Jan 22 '16 at 2:43
  • $\begingroup$ @NoahSchweber, true. I'll try to simplify things if I can work out how to do so. $\endgroup$ – goblin Jan 22 '16 at 2:45
  • $\begingroup$ Could you point to a source that'd provide the minimal base knowledge to understand this? Is it category theory? Number theory? $\endgroup$ – Dogweather Aug 15 at 18:36
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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.

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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?

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