# Is there such thing as an imaginary (imaginary number)?

In other words... is there such a thing that is to imaginary numbers what imaginary numbers are to real numbers? And could this be expressed as a "complex" type number? If a complex number is in the form x + yi, I guess this would be in the form of x + yi + zj? Does that exist as a concept?

• I don't see why you couldn't invent it. Just set some rules/axioms, make sure they're consistent and bam, you've invented billTavis's field of imaginary imaginary numbers. Whether it would be useful is a different matter... – tangrs May 11 '14 at 5:37
• So far none of the four answers has addressed your question of whether there is a family of numbers of the form $x+yi+zj$; the answer is no, there is not. There is some discussion of the subject in Is there a third dimension of numbers?. – MJD May 11 '14 at 6:04
• The comment above from tangrs, “I don't see why you couldn't invent it. Just set some rules/axioms, make sure they're consistent” completely elides the difficulties: there are no roles or axioms for three-dimensional numbers that are consistent, so you can't “just” do that. You run into trouble right away when you ask what $ij$ should be. In the quaternions, the answer is $k$. – MJD May 11 '14 at 6:06
• @MJD My answer agreed with you (i.e. "no"). Can you critique my answer to improve it (or is my answer completely off base). – Jared May 11 '14 at 6:28
• thanks MJD for the link to the three dimensional numbers discussion, and for confirming that there is not a concept for what I described. I accepted an answer that brought up quaternions because that is probably the closest to what I was looking for. – billTavis May 12 '14 at 17:07

Yes. Numbers akin to perhaps what you have suggested do exist. The Quaternions are a canonical example. Your basic old complex number can be written $a + ib$. Here we have numbers of the form $a + ib + jc + kd$. The rules for multiplying such numbers are discussed on the Wikipedia link above.

These have applications in applied math, especially things requiring three dimensions. Pure maths also uses them sometimes as they form a skew field.

Right. For some flavor, the complex numbers are perfectly represented by matrices of this pattern: $$\left( \begin{array}{rr} a & b \\ -b & a \end{array} \right) ,$$ with $a,b \in \mathbb R.$ In particular, $$1 \rightarrow \left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right) , \; \; \; \; i \rightarrow \left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right) .$$

Once you accept the complex numbers, the quaternions are perfectly represented by matrices of this related pattern: $$\left( \begin{array}{rr} \alpha & \beta \\ -\bar{\beta} & \bar{\alpha} \end{array} \right) ,$$ with $\alpha, \beta \in \mathbb C.$ In particular, $$1 \rightarrow \left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right) , \; \; \; \; i \rightarrow \left( \begin{array}{rr} i & 0 \\ 0 & -i \end{array} \right) , \; \; \; \; j \rightarrow \left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right) , \; \; \; \; k \rightarrow \left( \begin{array}{rr} 0 & i \\ i & 0 \end{array} \right) .$$ If you prefer, you can replace the 2 by 2 complex matrices with 4 by 4 real matrices, whereupon all the entries in the matrices for $1,i,j,k$ are $0,\pm 1.$ Just a bit harder to remember.

In a sense, yes. There is a number system called the Quaternions, which is an extension of the complex numbers. You can find the Wikipedia page here: http://en.wikipedia.org/wiki/Quaternion

In the complex numbers, we have $i^2 = -1$, but in the quaternions, which extend to $i, j, k$, we still have $i^2 = -1$, but we also have different rules such as $ij = k$ and $jk = i$. This number system plays a large role in simplifying engineering equations, just as complex numbers simplify equations describing electromagnetism and fluid dynamics.

Certainly, you can extend this system to arbitrarily many variables, so long as the basic rules (axioms) are consistent. And really, that's something mathematicians do: Create structures, give it basic rules, and see what comes out of that.

• There are just 4 basic division algebras. Real numbers, complex numbers, quaternions [which don't commute $ab \ne ba$] and octonions [which are not associative $a(bc)\ne (ab)c$]. – user121049 May 11 '14 at 6:51

No there is no such thing. To understand this you have to understand the imaginary unit: $i$ solves the equation $x^2 = -1$--that is $i^2 = -1$. Right away there are is a "problem" because $-i$ solves this equation just as well as $+i$: $\left(-i\right)^2 = (-1)^2i^2 = +1i^2 = -1$ and $\left(+i\right)^2 = (+1)^2i^2 = i^2 = -1$ so which do we choose? Do we choose $-i$ or do we choose $+i$. The answer is we choose neither and instead define $i$--we define $i$ such that $i^2 = -1$--it doesn't matter if we really mean $-i$ or $+i$--we just choose one of those two (we "choose" $+i$ because why carry around the burden of the negative when it's equivalent).

So to say there is an analogue to the complex unit $i$ would mean that there is some value of $x, y \in \mathbb{C}$ such that the following equation cannot be solved with complex numbers such that $x^2 = y$ where $x,y \in \mathbb{C}$. This is not possible, there is a value $x \in \mathbb{C}$ such that $x^2 = y$ for all values in $y\in\mathbb{C}$. We can go further and say that there exists at least one value $x \in \mathbb{C}$ such that for all values of $y \in \mathbb{C}$, $x^n = y$ where $n$ is a natural number (and you could expand to $n$ being a positive real value or a non-zero complex value).

I'm not familiar with these Qauternion values but these sound like higher dimension analogues. I am strictly talking about $1$D values. The imaginary number solves an insolvable problem from $\mathbb{R}^1$. Once we extend $\mathbb{R}^1$ to include $\mathbb{C}^1$ all such problems become solvable.

• I await the mathematicians to completely destroy what I have just said, but I think conceptually it's correct (although it may not have been stated mathematically correct). – Jared May 11 '14 at 6:02
• Actually, your intuition is right: Rephrased in technical language, you're saying that $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$, thus is algebraically closed, and so is its own algebraic closure. Source for further reading: en.wikipedia.org/wiki/Algebraically_closed_field – Neal May 11 '14 at 15:30