MihirKotbagi

Why we can't define 1/0


a month ago | 3 minute read

Recently, I was thinking about imaginary numbers, and more specifically about all the ways that imaginary numbers help us with understanding the real world.

They show up in many equations relating to various aspects of science (especially in quantum mechanics), from the time-dependent Schrödinger equation

$$i \hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},t) = \hat H \Psi(\mathbf{r},t)$$

and complex Minkowski spacetime

$$x^2+y^2+x^2+(ict)^2=constant$$

to Fourier transforms

$$\hat f(\xi)=\int^\infty_{-\infty}f(x)e^{-i2\pi\xi}dx$$

and even momentum-space wave functions, where

$$\Psi_p(x)=e^{ipx/\hbar}$$

is a solution to the time-independent Schrödinger Equation.

Nowadays, it seems absurd that these were once considered useless, but they initially had little relevance in math, let alone the physical sciences.

Imaginary numbers initially arose as a way to deal with equations that produced square roots of negative numbers instead of dismissing the underlying math as incorrect. For example, the quadratic polynomial

$$2x^2 - 2x + 1$$

has a negative discriminant (\(b^2 - 4ac = 2^2-4(2)(1) = 4-8 = -4\)) so its roots are imaginary. Before the advent of imaginary numbers, the equation

$$2x^2 - 2x + 1 = 0$$

would have been regarded as pointless, much like how we view the equation

$$x\cdot0=1$$

today.

However, we now would say that the solutions of that polynomial are \(\frac{1+i}{2}\) and \(\frac{1-i}{2}\), so why does the other equation seem so ridiculous?

Let's say we define a constant \(u\) such that

$$u\cdot0=1$$

This neatly solves the previous problem; we can just say \(x = u\).

Expanding on this fledgling number system, we can define the following operations:

$$x/0 = x\cdot{u}$$

$$x\cdot{u}\cdot0 = x$$

$$(x/0+y/0)\cdot0 = x + y$$

However, this introduces a major issue that might not immediately be apparent. For example, let's begin with the simple statement

$$0 = 1 - 1$$

Then, using the definition \(u\cdot0=1\) this can be rewritten as

$$0 = u\cdot0-u\cdot0$$

From here, factoring yields

$$0 = u(0-0) = u(0) = 1$$

Following the rules we established, we managed to "prove" that \(0 = 1\). Clearly, something has gone awry, and finding what it is unravels our entire premise.

The initial statement

$$u\cdot0=1$$

is uniquely pernicious; it violates the multiplicative property of zero (any number multiplied by zero is zero) under the guise of defining our "new constant" \(u\).

This means that we would have to give up the distributive property of multiplication (i.e. \(u(0-0)\not ={u\cdot0-u\cdot0}\)) just to avoid coming to the conclusion that \(0 = 1\).

Despite this setback, it turns out there is a system that uses the notion of infinites (denoted by \(\varepsilon)\) and infinitesimals (denoted by \(\omega\)) to achieve a similar result. Infinities are just values larger than any natural number, while infinitesimals are values that are closer to zero than any real number while not being equal to zero. It is crucial that \(\omega\not ={0}\) in order to avoid violating any fundamental properties of multiplication like before with \(u\).

In a similar manner to before, we define these terms with the equation

$$\varepsilon=1/\omega$$

The terms \(\varepsilon\) and \(\omega\) obey the commutative,

$$(x\cdot\varepsilon) + (y\cdot\omega) = (y\cdot\omega) + (x\cdot\varepsilon)$$

$$(x\cdot\varepsilon)\cdot (y\cdot\omega) = (y\cdot\omega)\cdot (x\cdot\varepsilon)$$

associative,

$$x\cdot\varepsilon + y\cdot\varepsilon = (x+y)\varepsilon$$

$$x\cdot\omega + y\cdot\omega = (x+y)\omega$$

and the distributive properties.

$$z(x\cdot\varepsilon + y\cdot\varepsilon) = x\cdot z\cdot\varepsilon + y\cdot z\cdot\varepsilon$$

Finally, we can re-examine the sequence of statements that invalidated our previous premise to confirm that \(\varepsilon\) and \(\omega\) are internally consistent.

$$0 = 1 - 1$$

$$0 = \varepsilon\cdot \omega - \varepsilon\cdot \omega$$

$$0 = \omega(\varepsilon - \varepsilon)$$

$$0 = \omega(0)$$

$$0 = 0$$

Although we annoyingly still have to say that \(1/0\) is undefined, at least there is a system that allows us to think about operations involving infinitely large or small quantities.