# Why cant we apply the square root of a negative number to equate more imaginary numbers?

In mathematics, polynomials like $$x^2-1$$ would have a clear solution of $$x=\pm 1$$. However, without complex numbers could you solve $$x^2+1$$? No, there would be no possible solution without adding a new axis of imaginary numbers. Simply, we added this axis because it provided us with solutions within our field of mathematics, so then we assumed them to exist.

So we know that you can only assume something to exist if it produces a solution (the most important reason for the imaginary axis to exist), then my question is why don't we do the same for an additional axis called $$j$$?

For example, we created the imaginary numbers by $$\sqrt{-1}=i$$, which then we unioned this imaginary axis to our real number axis to form the complex plane $$\mathbb{C}=a+bi$$. So then, why can't we square root a negative complex number to equate $$j$$. Visually: $$\begin{gather*} \sqrt{-(a+bi)}=j\\ \textrm{We then union this new axis with the previous to form the new set of numbers:}\\ \mathbb{C}_2=a+bi+cj:a,b,c\in\mathbb{R}\\\\ \textrm{We can then further extend this to even greater numbers:}\\ \mathbb{C}_3=a+bi+cj+dk: a,b,c,d\in\mathbb{R}\\ \mathbb{C}_\infty=a+bi+cj+dk+...+z(\infty): a,b,c,d,...,z\in\mathbb{R} \end{gather*}$$ And this repeats on and on, where you take the square root of a negative number and from this method you keep on formulating a new type of axis of a number.

Why don't we already have this in mathematics? We've done it once with solely the imaginary plane, why not extend it to a three axis field? Why do we limit ourselves?

I'm more than sure there are problems out there that can be solved with a $$\mathbb{C}_2$$ number!

I must say the beauty of this is mathematically speaking we can then equate infinity or at least have a new symbol for it. $$\mathbb{C}_\infty=\infty$$

• (1) no we didn't create $\sqrt{-1}=i$. We create $i$ such that $i^2=-1$. (2) Because it turns out $\mathbb{C}$ is algebraically closed, so $a+bi$ already has squareroots in $\mathbb{C}$. – user10354138 Feb 28 at 1:38
• $\mathbb{C}$ is algebraically closed, so your $j := \sqrt{-(a+bi)} \in \mathbb{C}$. – Brandon Carter Feb 28 at 1:39
• If you seek more elaboration and insight then you should never accept an answer so quickly. This signals to readers that your question is answered. Often those with the deepest knowledge have less time to spend here, so they won't see your question until after you accepted, so they may skip it without even opening it, assuming you already have a good answer. This means you (and the site) may miss out on a great learning experience. Always wait at least a couple days before accepting so all active readers have a chance to see your question. – Bill Dubuque Feb 28 at 2:40
• You bring up a good point Bill, my curiosity has exploded into trying to think of equations that can be solved with the C_2 values, I can assume if we could create a 4th dimensional equation then it would be possible, but my brain cannot picture a 4th-dimensional equation. – Tonsofluck Feb 28 at 2:55
• @BillDubuque. I think I answered the question and also gave new input (other people agreeing with me, no downvotes/criticizing comments). I completely agree with you that one should wait for new input, but in this case the question is relatively simple to answer. I really do not care about +15 or -15 rep in this case but shouldn't people with more knowledge answer other, harder unanswered questions? – vitamin d Feb 28 at 3:32

To answer your question: the complex numbers are algebraically closed. But this should not stop you from defining new "complex" numbers, such as quaternions, octonions, sedenions etc. This is called the Cayley-Dickson construction.

• Nice answer ++1 – haidangel Mar 1 at 9:07
• Thank you for your answer, for it makes me ponder more and more of the disjoint union function such that if you disjoint union the same set of numbers they then equate to said quaternions, octonions, and sedenions, as you stated. For hopefully I will think of a better question soon in which you union the set of all polynomials and then you can use quaternions and all those fancy numbers as solutions. Thank you again, I'll be posting in the nearby future. – Tonsofluck Mar 3 at 17:54
• The usefulness of Cayley-Dickson construction construction quickly disappear due to poor algebraic properties of resulting algebraic structures. Clifford algebras would be more interesting in this prospective. – Bumblebee Mar 14 at 3:07

+1 to your query.

Generally, you only create complications (i.e. $$\mathbb{R} \to \mathbb{C}$$) to facilitate solving problems. When attacking for example $$\sqrt{x + iy}$$ there is absolutely no need to extend the Complex Numbers.

Any non-zero complex number $$z = (x + iy)$$ can be converted to polar coordinates $$re^{i\theta} : r \in \mathbb{R^+}$$.

Therefore, by DeMoivre's theorem, $$\sqrt{z} = \pm\sqrt{r} \times e^{(i\theta/2)}.$$

• I´ve noticed that the question you´ve answered has been deleted by the OP. The same thing has happened to my answer of another question of this OP. Not very nice from him. It seems that he ask a question and delete it after getting a helpful answer. – callculus Feb 28 at 3:01
• I don't understand what you're on about, I ask questions to receive answers, it doesn't matter who answers, as long as the answer guides me to better understanding @callculus. – Tonsofluck Feb 28 at 3:18
• @user2661923 no I have never deleted my own mathSE queries, instead KReiser, Paramanand Singh and Teresa Lisbon have deleted my query about disjoint unioning the same set and what it equals to – Tonsofluck Feb 28 at 3:29