2

Good afternoon,

The square root of $-1$, AKA $i$, seemed a crazy number allowing contradictions as $1=-1$ by the usual rules of the real numbers. However, it proved to be useful and non-self-contradicting, after removing identities like $\sqrt{ab}=\sqrt a\sqrt b$.

Could dividing by $0$ be subject to removing stuff like $a\times\frac1a= 1$?

A way I can think of defining this is to have $\varepsilon=\frac10$. Remeber it may be that $\varepsilon\neq\varepsilon^2$, we don't know whether $\left(\frac ab\right)^n=\frac{a^n}{b^n}$, and other things alike; so they can't disprove anything.

This seems fun! What about defining a set $\mathbb E$ of numbers of form $a + b\varepsilon$? There would be interesting results as $(a+b\varepsilon)\times0 =b$.

Are these rules self consistent? Can such a number exist? If not, what rules can we change to make it exist? If yes, what can we find about it? How can it help us?

If there is no utter way of dividing by zero without getting a contradiction no matter the rules used, how can that be proven?

JMCF125
  • 2,738
  • 1
  • 20
  • 34
  • 3
    Start with $0 \times 1 = 0 \times 2$ and divide both sides by zero. – user121926 Jan 17 '14 at 17:17
  • @user121926, the same for $i$ and powers happens. – JMCF125 Jan 17 '14 at 17:23
  • @JMCF125: No, for complex numbers we define the complex exponential and logarithm and use that to define $a^b = e^{b\ln(a)}$. If the logarithm is multi-valued, power rules actually work out. If the logarithm uses a branch-cut, then power rules work 'within' the branch-cut. – user21820 Jan 17 '14 at 17:26
  • @JMCF125 Multiply both sides by $\frac{1}{0}$, if you prefer. Can you give an example of what you're talking about with $i$? – user121926 Jan 17 '14 at 17:28
  • @user21820, maybe as one needs $\pm$ to make complexes work, an analogous operator is required to divide by zero. – JMCF125 Jan 17 '14 at 17:32
  • @JMCF125: No, that is wrong too. There are 3 complex cube-roots of unity, and $\pm$ is not enough to 'solve' the problem. I suggest you learn a bit about complex numbers to see what I meant about multi-valued functions and branch cuts. See http://en.wikipedia.org/wiki/Complex_logarithm. – user21820 Jan 17 '14 at 17:35
  • 2
    @JMCF125 If you want that the usual rules continue to apply, you have $\varepsilon^2=\frac{1}{0}\frac{1}{0}=\frac{1^2}{0^2}=\varepsilon$. So indeed $\varepsilon^2=\varepsilon$. Divide by $\varepsilon$ (which is non zero), and you get $\varepsilon=1$. Oops. So, we must throw away the usual rules; no, thanks. – egreg Jan 17 '14 at 17:35
  • @egreg, who said $\frac ab\frac cd=\frac{ac}{bd}$? – JMCF125 Jan 17 '14 at 17:43
  • @JMCF125 And why should I throw away the rules just for having a useless thing? Humanity has made without an inverse of zero for thousands of years; algebra has been invented about 1000 years ago and it works quite well. – egreg Jan 17 '14 at 18:13
  • @egreg, I didn't say you should throw it away. I said it may be (and turns out it is) possible to divide by zero if you do. Whether you think this should or not be used is another question. – JMCF125 Jan 23 '14 at 11:00
  • $1/0=\infty$ on projective real line or complex plane. It is used more rare than affine real line. – Anixx Sep 17 '16 at 00:17

2 Answers2

5

If $\,0\,$ has an inverse then $\ \color{#c00}1 = 0\cdot 0^{-1} = \color{#c00}0\,\Rightarrow\ a = a\cdot \color{#c00}1 = a\cdot \color{#c00}0 = 0\,$ so every element $= 0,\,$ i.e. the ring is the trivial one element ring.

So you need to drop some ring axiom(s) if you wish to divide by zero with nontrivial consequences. For one way to do so see Jesper Carlström's theory of wheels, which includes, e.g. the Riemann sphere.

Bill Dubuque
  • 272,048
1

In short, there is no way to define $\frac{1}{0}$ without getting nothing interesting, as user121926 has already mentioned in his comment.

However, in measure theory we can define $0 \times \infty = 0$ and it keeps certain theorems simple. Nevertheless division by zero still cannot be allowed.

user21820
  • 57,693
  • 9
  • 98
  • 256
  • Please see the other answer and the referred paper, division by zero can work. – JMCF125 Jan 18 '14 at 11:46
  • The Riemann sphere still doesn't allow every element to be divided by zero. For one, $0/0$ is still undefined or in some cases need to be evaluated by taking limits. So what is the point? By the way, there are always ways to define all sorts of funny systems that are still consistent but not necessarily useful. – user21820 Jan 23 '14 at 01:38
  • That paper does only refers the Riemann sphere. It talks about "wheels", where $0/0$ is defined. – JMCF125 Jan 23 '14 at 10:57
  • Take what you like. If you understand arithmetic and complex numbers properly you would know what I meant. – user21820 Jan 23 '14 at 14:51