Why there is no solution to $\frac1x=0$

Question

We all have learnt that the extension from $\mathbb R$ to $\mathbb C$ was to enable us to solve the equation $x^2=-1$, and we added the letter '$i$' as the solution of this equation. If we didn't, $x=\pm \sqrt{-1}$ would have made no sense at all.
But what about $$\frac1x=0$$ Why there is no unit like the imaginary unit for this? Suppose we use '$i_2$' for this, then there would be the solution to these types of equations also, and probably a new area of math may develop.

Answer 1

It depends on what you want to do with $x.$

There are no useful algebraic structures where the neutral element for addition divides the neutral element for multiplication, because this leads to the conclusion that all numbers are identical. To make this claim more precise, suppose that we have at least a ring structure with a unit element for multiplication. Then

$$1 \cdot 0=1 \cdot (1+(-1))= 1+(-1)=0=1+(-1)=(1+(-1)) \cdot 1=0 \cdot 1$$

so the neutral element for addition is an absorbing element for multiplication. Now if there exists an $x$ such that $0.x=1,$ by multiplying both sides with an arbitrary element of the ring we get that all elements are equal to $0.$

There is a useful topological extension of the real numbers, and for that purpose we can use the symbol $\infty$ (without a sign). It can be used to simplify some statements about limits, and it can be safely multiplied by any number different from $0.$

Answer 2

There are several extensions in this direction. However, none of them is treating your situation completely. We can introduce new symbols like $+\infty$ $-\infty$ but these cannot belong to any set that would have all properties of real numbers transparent like associativity, commutative property, uniqueness and so on.

Division is defined from multiplication, we say that $\frac{a}{b}=c$ if $a=bc$. Imagine that we have extended real numbers with T that would have $T \cdot 0 = 1$ from where you would have $\frac{1}{T}=0$. This new element would have to obey the same rules as any other real number. (In case of complex numbers you can do with $i$ the same thing you can do with any other real number.)

If so we would have for any real number $r \neq 0, r \neq 1$, $r \cdot T \cdot 0 = r \cdot 1$ or by simple exchange $T \cdot r \cdot 0 = r \cdot 1$ Unfortunately this last is $T \cdot (r \cdot 0) = r \cdot 1 = r$ which means $T \cdot 0 = r$ and this is failing the restrictions imposed over $r$.

So there is no extension of real numbers in this direction that would keep the structure of real numbers and define a new element $T$ with $\frac{1}{T}=0$.

The approaches that successfully extend real numbers in this direction are replacing 0 with some sort of infinitesimal value, a value closer to 0 than any other real number but still larger than 0, and only then eventually map all such infinitesimally small numbers to 0. If you do this you need two new elements: infinitely large $\omega$ and infinitely small $\epsilon$, $0<\epsilon<|r|, r \neq 0$ and $\omega>|r|$ where r is any real number.

Now you have $\frac{1}{\omega}=\epsilon$. There is a function $st$ known as standard part that is converting infinitesimals to real numbers and we know that $st(\epsilon)=0$ but $\omega$ unfortunately cannot have a standard part for the reason we have explained above, since this fictive element $st(\omega)$ cannot belong to any extension of R and keep all its properties. Still you can write

$$st(\frac{1}{\omega})=0$$

If you take that equality is comparing standard parts of both sides, since $st(0)=0$, you have an even nicer expression that almost solves your question as

$$\frac{1}{\omega}\stackrel{st}{=}0$$

This last is probably the closest you can have, something that still answers your question.

Answer 3

You hunch that this leads to a new area of math is correct provided one interprets your division as is done in the context of the so-called "indeterminate forms" in calculus, such as $1^\infty$ or $\frac{0}{0}$. Namely, one can interpret division by $0$ as division by an infinitely small number $o$ much as Newton did it. Then $\frac{1}{o}$ can be well-defined in which case it is an infinite number. Such extended number systems were developed already at the end of the 19th century by Paul du Bois-Reymond, Hahn, and others.

Answer 4

As with the complex numbers you will have to drop properties. One common way you do it in complex analysis is to introduce $\infty$, simply stated this represents all types of infinities (without regard of the argument).

The property that is dropped is that you can't no longer always multiply, add or subtract (or even divide). The construct nevertheless has it's advantage.

There are numerous examples using normally valid formulas that will break if you introduce any such element. For example $x (1/x)=1$ is normally true for $x\ne 0$, but for this solution we have $1/x = 0$ so $x (1/x)=x\cdot 0=0$. So you have to drop either the identity $x (1/x)=1$ for nonzero $x$ or $x\cdot 0=0$ for all $x$. In the complex analysis case one gives up both as one leaves them undefined if $x=\infty$.

Answer 5

The use of $\mathbb{C}$ is algebraic in that it is the promised algebraic completion of $\mathbb{R}$, where $\mathbb{R}$ completes $\mathbb{Q}$ in the sense of the metric, in that $\mathbb{C}$ is a field containing $\mathbb{R}$ on which every nonconstant polynomial has roots. What you're proposing would be something like a completion on rational functions. But we cannot consistently take a field (for our purpose, $\mathbb{R}$) and introduce an element, call it as you might $i'$, such that $0 \cdot i' = 1$, as $0 \cdot x = 0$ for all $x$. To see this, consider $$0 \cdot x = (1 - 1) \cdot x = (1 \cdot x) \cdot ( (-1) \cdot x) = (1 \cdot x) - (1 \cdot x) = 0 . $$ Note the things I used to get that, in particular the way addition and multiplication distribute. To get $i'$ to work, we would have to lose that. That's not to say you couldn't make an algebraic object with two operations on it with analogous properties, but they would not interact as they do in good old $\mathbb{R}$ (or $\mathbb{C}$), for if they did, then as I have shown $i'$ wouldn't work.

Answer 6

From

$$\frac1x=0$$ we deduce

$$-\frac1x=\frac1{-x}=0.$$

Then $i_2=-i_2$.

Mh, pretty bad start for a new unit.