0
$\begingroup$

Assuming i does not necessarily have any other property but that it can represent (a, b) as a number in the form a + bi, how else could we define i? For instance, would it be appropriate to define a number i such that $$2^i=-1$$ instead of a number i such that $$i^2=-1$$ Are there any such numbers that are well established or some similar concept (i.e. a number which is essentially an ordered pair or, more generally, a list)?

I ask this primarily because I want to know if there is any reason not to define i in some other way, apart from the fact that there are some often used equations like Euler's formula which make use of the definition of i.

$\endgroup$
  • 2
    $\begingroup$ There is already such a complex number: Indeed if $$j=\frac{i \pi}{\ln(2)}$$ then $2^j=-1$. $\endgroup$ – Maximilian Janisch Feb 15 at 0:50
  • 5
    $\begingroup$ Does this answer your question? Why don't we define "imaginary" numbers for every "impossibility"? $\endgroup$ – Brian Feb 15 at 1:00
  • $\begingroup$ What do you want $i$ to DO? You could define $i$ as $7$ as $(a,b) $ is a number represented as $a+bi = a+7b$ for all we care. So I don't really understand what you are asking.... By the way, as Maximillian Janisch points $2^{variable} = -1$ does have a complex solution of $variable = \frac {imaginaryunit\pi}{\ln 2}$ where $imaginaryunit$ is the imaginary unit where $imaginaryunit^2 = -1$. $\endgroup$ – fleablood Feb 15 at 1:12
  • 1
    $\begingroup$ Here's an interesting but related tidbit. Did you know you can define a $j$ such that $j^2=-1$ but $j\ne i$ ? Look up quaternions, a four-dimensional number space. You can look at any complex subspace $a+bi$, $a+bj$ and $a+bk$ and observe purely complex properties within them. It isnt until you mix and match and look at the entire space $a+bi+cj+dk$ that you get different properties. So, in fact, there is no dilemma that needs resolution between the $i$ used by mathematicians and the $j$ used by engineers; they dont even have to be the same imaginary axis. $\endgroup$ – SquishyRhode Feb 15 at 1:50
2
$\begingroup$

One can come up with other "imaginary" types of numbers by demanding them to satisfy certain relations, as long as those relations would not produce a contradiction.
Interestingly enough, the example you wrote of having some number (let's call it $j$ as to not confuse it with the imaginary unit $i$) such that $2^j=-1$ introduces the idea of taking logarithms of negative numbers.
As it turns out, the study of complex numbers allows one to take logarithms of negative numbers! So in some sense, your specific relation results in us just rediscovering complex numbers. Note: Logarithms of negative numbers aren't well-defined and there are some subtle things to take care of here which should be covered in a standard first course on complex analysis.

In turns out that the complex numbers are an example of an algebraically closed field. This means that every polynomial with complex coefficients has a complex solution. Thus if you write down any sort of algebraic relation that "looks like" a polynomial, and demand some number solve it, you'll just end up with a complex number.

The complex numbers can be generalized to higher structures such as the quaternions, and the octonions, which gets at your question of a more general 'list' of numbers.
The quaternions are given by elements $a+bi+cj+dk$ where $a,b,c,d\in\mathbb{R}$ and $i^2=j^2=k^2=ijk=-1$.

To be clear, we can define all sorts of "exotic numbers", but having different definitions, mean they will behave differently. So it doesn't really make sense to ask for "a different definition of $i$", the symbol $i$ specifically refers to a solution to $x^2+1=0$. Your final question might be better phrased as why are we interested in the imaginary number $i$, as opposed to other exotic numbers? As expressed above, we do consider other kinds of numbers, and sometimes they bring us back to $i$ anyways!
It turns out that complex numbers are enormously useful in physics and endlessly fascinating for their mathematical elegance, which is why they are so much more popular than other number systems.

| cite | improve this answer | |
$\endgroup$

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