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Is it possible in mathematics to use a third number line based on division by zero; in addition to the real and imaginary number lines?

This is because some solutions blow up when there is a division by zero. Would it be possible to solve them with this new number line?

$\therefore$ on the z axis we would have $\frac{1}{0}$ , $\frac{2}{0}$ , $\frac{3}{0}$ , etc. where, $p = \frac{1}{0}$ and $p \cdot 0 = 1$ .

division by zero graph

Is this a viable number system?

A similar question to this one: Is there a third dimension of numbers?

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In order to append numbers to the number system, you must explain how they interact with the standard number system.

You need to tell us how to avoid the following: $$p-p=0$$

$$\frac{1}{0}-\frac{1}{0}=\frac{0}{0}$$

So $$\frac{0}{0} = 0$$

Now $$\frac{2}{3}+\frac{0}{0} = \frac{2}{3} +0 = \frac{2}{3}$$

And, assuming that we still add fractions by finding a common denominator:

$$\frac{2}{3}+\frac{0}{0}=\frac{2\cdot0}{3\cdot0}+\frac{3\cdot0}{3\cdot0}=\frac{0+0}{0} = \frac{0}{0}=0$$

We conclude that $$\frac{2}{3} = 0$$

As others have pointed out, there are other things that need to be avoided as well. You need to explain this new number system more than simply saying I'm going to make division by zero possible.

Added: You also need to explain associativity (and probably commutativity) in you new system to avoid:

$$2\cdot(0\cdot p) = 2\cdot 1 = 2$$ and $$(2\cdot 0)\cdot p = 0\cdot p = 1$$

So $$2 = 1$$

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If you define $p\cdot0:=1$ then you get problems with the distributive property: $$1=p\cdot 0= p\cdot(0+0) = p\cdot0+p\cdot0=1+1=2$$ Since the distributive property defines multiplication as we know it, you can not drop it without totally changing its meaning and thus not solving the dividing by zero problem.

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  • $\begingroup$ If $p \cdot 0 = 1$ then $p \cdot (0 + 0) = p \cdot 0 = 1$ $\endgroup$ – Brendan Darrer Sep 11 '17 at 21:05
  • $\begingroup$ @BrendanDarrer Apparently you propose to abolish the distributive property, at least for your new numbers. Or do we keep it for $p \cdot (1+2)$? $\endgroup$ – Jim H Sep 11 '17 at 22:35
  • $\begingroup$ Yes, in this number system $p \cdot (1 + 2) = 3p$ $\endgroup$ – Brendan Darrer Sep 11 '17 at 23:14
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One way to approach this problem could be to imagine zero as a very small number.

So to start with, let us define,

$$small \to 0$$

$$\therefore$$

instead of $$ (2 \cdot 0) \cdot p = 1 $$

as above @Jim H. Use:

$$2 \cdot small \cdot \frac{1}{small} = 2$$

As

$$(2 \cdot small) \cdot \frac{1}{small} \neq small \cdot \frac{1}{small} = 1$$

$$= 2 \cdot \frac{small}{small} \cdot 1 = 2$$

Do not use $(2 \cdot small) \to 0$ during multiplication and division, otherwise it will produce nonsense such as 2 = 1.

$\underline {The\, p\, number\, system}$

If a very small number, $$s \to 0$$ $$p = \frac{1}{s} \to \infty$$

then, $$\frac{p}{p} = \frac{1}{s} \cdot \frac{s}{1} = 1$$

$$\therefore p \cdot p = \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s^2} = p^2$$

Also, we could invent a p number,

$$4 + 3p$$

multiply it by s,

$$(4 + 3p) \cdot s = 4 \cdot s + 3 \cdot \frac{1}{s} \cdot s$$

$$4 \cdot s \to 0$$

$$\therefore (4 + 3p) \cdot s = 3$$

$\underline {Adding\, and\, multiplying\, fractions\, in\, p\, number\,system}$

$$p + p = \frac{1}{s} + \frac{1}{s} = \frac{1 + 1}{s} = 2p$$

$$p - p = \frac{1}{s} - \frac{1}{s} = \frac{1 - 1}{s} = \frac{0}{s} = 0$$

$$\frac{2}{3} + \frac{p}{3} = \frac{2 + p}{3}$$

Instead of $$\frac{0}{0} = 0$$ use $$\frac{s}{s} = 1$$

$$\frac{2}{3} + \frac{s}{s} = \frac{2}{3} + 1 = \frac{5}{3}$$

$\underline {Adding\, two\, p\, numbers}$

$$(3 + 2p) + (5 + p) = 8 + 3p$$

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$\textbf {Rules of the p number system using 1/0 and $\infty$ }$

If $p = \frac{1}{0}$ and $p \cdot 0 = 1$. Then I suggest the following rules.

$$p - p = \frac{1}{0} - \frac{1}{0} = \frac{0 \cdot 1 - 0 \cdot 1}{0 \cdot 0} = \frac{0 \cdot (1 - 1)}{0 \cdot 0} = \frac{0}{0} = 1$$

problem lies here (if it is a problem?) $$\frac{0}{0} \neq \frac{0}{\delta x} = 0$$

where $\delta x \to 0$ (see previous answer), going by the rule of $\frac{0}{0} = 1$ .

$\underline{ RULES \, that \,appear\, to \,work}$

$$\frac{2}{3} + \frac{0}{0} = \frac{2 \cdot 0 + 3 \cdot 0}{3 \cdot 0} = \frac{5 \cdot 0}{3 \cdot 0} = \frac{5}{3}$$

$$p \cdot p = \frac{1}{0} \cdot \frac{1}{0} = \frac{1}{0^2} = p^2 = \infty \cdot \infty = \infty^2$$

$$p + p = \frac{1}{0} + \frac{1}{0} = \frac{0 \cdot 1 + 0 \cdot 1}{0 \cdot 0} = \frac{0 \cdot (1 + 1)}{0 \cdot 0} = \frac{2}{0} = 2p$$

$$p \cdot 0 = 1 \neq p \cdot (0 + 0)$$

$$p \cdot (0 + 0) = p \cdot 2 \cdot 0 = \frac{1}{0} \cdot 2 \cdot 0 = 2$$

$$0 + 0 + 0 \,+ ... = 3 \cdot 0 \,+ ...$$

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I don't think you can divide 1 by 0.

Because,

$$\infty \cdot 0 = 0$$

$$\therefore \frac{0}{0} = \infty$$

There are infinite 0's in 0 .

$$\therefore \frac{1}{0} = error$$

This is because 0 is not a number. It is a nothing number. There is nothing to divide with! Unless the number actually exits, e.g. a small number such as $\delta x \to 0$.

$$\therefore \sum_{1}^{\infty} 0 = 0 + 0 + 0 \,+... = 0$$

So far in the above discussion we have: $$\frac{0}{0} = 0\,,\, 1 \, \,or \,\, \infty $$

depending on how it's defined.

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  • $\begingroup$ In the same way $\sqrt {-1}$ = error, until a complex number was invented $ i = \sqrt {-1}$ $\endgroup$ – Brendan Darrer Sep 16 '17 at 12:23

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