Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

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?

share|improve this question

marked as duplicate by egreg, Sami Ben Romdhane, Lost1, Daniel Rust, Yiorgos S. Smyrlis Jan 17 at 18:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2  
Start with $0 \times 1 = 0 \times 2$ and divide both sides by zero. –  user121926 Jan 17 at 17:17
    
@user121926, the same for $i$ and powers happens. –  JMCF125 Jan 17 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 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 at 17:28
1  
@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 at 17:35

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
Awesome! That was what I was looking for. I'll take a closer look. –  JMCF125 Jan 17 at 19:08

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.

share|improve this answer
    
Please see the other answer and the referred paper, division by zero can work. –  JMCF125 Jan 18 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 at 1:38
    
That paper does only refers the Riemann sphere. It talks about "wheels", where $0/0$ is defined. –  JMCF125 Jan 23 at 10:57
    
Take what you like. If you understand arithmetic and complex numbers properly you would know what I meant. –  user21820 Jan 23 at 14:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.