Can Division by Zero Be Defined in a More General Number Group?

[zetafunction]

the sum would be the usual

$$a+b.(1/0)+c+d.(1/0)= (a+b)+(c+d).(1/0)$$

the problem with the product is that you make it worse , in the product a term $$(1/0).(1/0)$$ would appear

 

[Hurkyl]

Well, then, with those operations, the set of {a + b (1/0)} is not a ring.

Of course, the set of all polynomials in one variable is a ring, and we can name that variable 1/0 if we choose. But does that have anything to do with division by zero?

 

[jcender]

Now, since complex numbers require the special definition of √(-1), could you just define division by zero (arbitrarily, say, as "1/0 = m") and make an even more general number group?

Consider defining a number m and also redefining zero with it as a way of extending the number system and thus defining division by zero. By defining both zero and "m" through division instead of subtraction, they can be multiplicative inverses. How might this be carried out?

Let's follow your example of the special definition of the √(-1) where i indicates an imaginary number when written together with a Real number, and choose a symbol that will indicate nothingness when written with a Real number. In other words, instead of the single number 0, there will be a set of absent numbers represented by Real numbers together with an absence symbol.

So let m be n$$\infty$$ where n is a Real number, and let the absence symbol be $$\overline{\infty}$$. These numbers of nothing would then be n$$\overline{\infty}$$. In general, division by zero numbers would be $$\frac{x}{n\overline{\infty}} = (x/n) \infty$$. As with a reciprocal under a fraction bar, the inverse inverts. A zero number divided by another zero number would equal the quotient of their Real parts.

Note that the example of the imaginary numbers can only be followed so far. Normally at least, the additive identity property would preclude forming numbers made up of Real numbers plus absent numbers in a way analogous to the complex numbers.

The bar in the absence symbol is an inverse bar. It indicates absence and functions somewhat like a fraction bar except there is never a numerator. (This prohibition entails an exception for the division rule of equality although one much less onerous than the lack of definition of division by 0.) The infinity symbol here represents an array of the Real numbers. (Other definitions are possible.) To see some of the advantages of this notation, look to pp. 379-382, 580-2, 897-907, 916-25 in the book The Road to Reality by the physicist and mathematician Roger Penrose. He uses it for n-real dimensional space, without, however, a multiplicative inverse. Also note the compatibility of the new numbers with the complex extended plane or Riemann sphere referred to on this thread by g_edgar on Jul30-09. Indeed they may be seen as a generalization of it.

A number of issues not addressed here such as implications for physics, defining the absence symbol more fully, defining subtraction, an unsigned zero, arithmetic with the inverses of zero numbers, multiplication of zero numbers, a geometric interpretation, the empty sets unsuitability as a rigorous basis for the absent numbers, as well as other issues, are dealt with in the attached paper. Also not addressed here is the question "Does this number system form a field or not?"

 

[yuiop]

We don't imagine -- we construct a new number system (e.g. the projective numbers) in which 1/0 is defined, and use that one instead of the real numbers.

By the way, both of these limits are of the form 0/0:

$$\lim_{x \rightarrow +\infty} \frac{x}{x^2}$$

$$\lim_{x \rightarrow +\infty} \frac{x^2}{x}$$​

Assuming you meant in the limit as x goes to zero (as Mentallic mentioned) then can 0/0 is undefined even in the projective system, so can we not apply L'Hopital's rule and get:

$$\lim_{x \rightarrow 0} \frac{x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{2x}=\infty \qquad ?$$

For the case that the limit of x goes to infinity (in the projective system):

$$\lim_{x \rightarrow \infty} \frac{x}{x^2}=\lim_{x \rightarrow \infty} \frac{1}{2x}=0$$

In fact it seems we can note that x/x² is equivalent to 1/x and obtain the same limits as above even without applying L'Hopital's rule in the projective system.

 

[Max™]

Be careful using up arrows for stuff like this.

http://en.wikipedia.org/wiki/Knuth's_up-arrow_notation