Square root of zero

Question

I'm old $35$ but starting just now with maths, so sorry if I ask non complex questions. $0$ is the only number that just has one square root. Is the explanations for this simply that 0 in arithmetic does not have a sign. If this is the explanation this seems to me like a definition matter. Thank you.

This follows more from the fundamental theorem of algebra than anything else. It is a result rather than a definition. The equation $x^2 = 0$ can be separated and factored as $(x-\color{red}{0})(x-\color{blue}{0})=0$. Compare this to $x^2 = 1$ which can be rewritten and factored as $(x-\color{red}{1})(x-\color{blue}{(-1)})=0$. The two square roots of $1$ are the $\color{red}{1}$ and $\color{blue}{-1}$ as above while for zero it so happens that the red and blue numbers coincide. — JMoravitz, Jan 13 '19 at 16:51
It's not just a definition matter, it's a fact. If $x\ne0$ then $x^2\ne0$. So the only $x$ with $x^2=0$ is $x=0$. — David C. Ullrich, Jan 13 '19 at 16:54

score 1 · Answer 1 · answered Jan 13 '19 at 16:59

For real or complex numbers, if $ab=0$, then either $a=0$ or $b=0$. This means that the complex numbers form an integral domain. It follows from this that if $a^2=0$, then $a=0$.

You could say this is the reason. If you want to use the fact that if $a$ is a root of $x^2=b$ then $-a$ is the only other root, then that is basically what you are saying : $0=-0$. But this presupposes the integral domain condition. When that fails, it could even happen that $x^2=b$ has infinitely many roots, even if $b=0$.

Every algebraic structure that can reasonably be called a set of numbers forms an integral domain. Except that this is not quite true! Suppose we are dealing with integers, but we declare that $m$ and $n$ are the same if they differ by a multiple of $4$. Then $2^2=0$, using this new notion of equality as differing by a multiple of $4$. So $2^2=0$ and $0^2=0$, but $0\neq - 2$.

score 0 · Answer 2 · answered Oct 13 '23 at 03:37

There is some staff in quantum physics that works this way, but it is funnier just to create one from scratch.

Just use same approach as when we defined imaginary number $i$ as matrix

$$ i = \left(\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right) $$

$\theta$ is what we need

First we need to learn how to treat real numbers as matrices. Lets suppose any real number is just it's real value multiplied by identity matrix, so

$$5 = \left(\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}\right)$$ This definition preserves all properties of real numbers.

Now we can try to find such matrix that $$\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)^2 = \left(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}\right) = 0$$

Now we just need to find solution to systems of equations.

I am skipping that part to results:

$$ \left\{\begin{matrix} a = -d \\ a^2 + bc=0 \end{matrix}\right. $$ From here we can choose any solution we like and it will work.

I am choosing this:

$$ a = 1, b=1,c=-1,d=-1 $$ which equals to our desired matrix:

$$ \theta = \left(\begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix}\right) $$

You can square it to see that it is indeed squares to zero

And yeah, it is linear dependent matrix, which have zero determinant.

But does it make sense?

To prove that this is legitimate definition of $\theta^2=0$ let's compute some staff that uses all real, imaginary and theta numbers. P.S remember: $i^2 = -1$, $\theta^2=0$, $i \theta \ne \theta i$ - it is noncommutative (you can just multiply them and see).

Also consider following: $$\theta i + i \theta = \left(\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}\right) = 2$$ It is property that we get just by multiplying corresponding matrices and adding them.

Here is an example: $$ 3+2i+4\theta = \left(\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}\right) + \left(\begin{matrix} 0 & -2 \\ 2 & 0 \end{matrix}\right) + \left(\begin{matrix} 4 & 4 \\ -4 & -4 \end{matrix}\right) = \left(\begin{matrix} 7 & 2 \\ -2 & -1 \end{matrix}\right) $$

So by squaring that matrix we get: $$ (3+2i+4\theta)^2= \left(\begin{matrix} 45 & 12 \\ -12 & -3 \end{matrix}\right) $$

Now let's open squaring brackets manually, add matrices and see if it works.

$$(3+2i+4\theta)^2 = 5+12i+24\theta+8(i\theta+\theta i)= \left(\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}\right) + \left(\begin{matrix} 0 & -12 \\ 12 & 0 \end{matrix}\right) + \left(\begin{matrix} 24 & 24 \\ -24 & -24 \end{matrix}\right) + \left(\begin{matrix} 16 & 0 \\ 0 & 16 \end{matrix}\right) = \left(\begin{matrix} 45 & 12 \\ -12 & -3 \end{matrix}\right) $$

As you can see this number $\theta$ is good to go and good to use mathematical object.

How to work with this super complex numbers?

You can do any operations on these super-complex numbers, such as raising a power, taking log, sin, cos, etc.

$$a+bi+c\theta$$

It generalizes to functions the same way you do with complex numbers - by eigendecomposition of resulting matrix.

See this cool dude full playlist on eigendecomposition, but this one video is most useful

https://www.youtube.com/watch?v=-1loSrioE4Y

I've got a matrix, I need these sick coefficients back from it. How?

If you done some calculations you may find that it is hard to get meaning back from these 2x2 matrix to coefficients for real, imaginary and theta numbers.

All we need to do is to solve(you know what) system of linear equations back from it.

For example we done some computing and we have a matrix

$$ A=\left(\begin{matrix} a & b \\ c & d \end{matrix}\right) $$

To map it to coefficients we need to compute

$$ k_1 + k_2 i +k_3 \theta = A $$

When you replace real, imaginary and theta with corresponding matrices, you will get yet another linear system, and I will skip all of this and give you solution:

$$ k_1 = \frac{a+d}{2} $$ $$ k_2 = \frac{a-d}{2}-b $$ $$ k_3 = \frac{a-d}{2} $$

That's it?

Yeah, you have just expanded your view on complex numbers and more powerful than previously!

I am cross-answering from https://math.stackexchange.com/posts/4785627/edit

score 0 · Answer 3 · answered Dec 07 '23 at 20:59

With $x \in (\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C})$ then yes, $x^2 = 0$ means that $x=0$.

_{Note that}

Integers $\mathbb{Z}$
Rational numbers $\mathbb{Q}$
Real numbers $\mathbb{R}$
Complex numbers $\mathbb{C}$

But there are other algebras out there. Specifically one might imagine some number ${\rm \epsilon} \neq 0$ that has the property of ${\rm \epsilon}^2 = 0$ and this defines what is called "Dual Numbers"

Considering that addition and multiplication is now defined you can build all operations from that rule, and say for example for multiplication

$$\require{cancel} (a+{\rm \epsilon} b) (c + {\rm \epsilon} d) = a\,c + {\rm \epsilon}\, b\,c + {\rm \epsilon}\, a\,d + {\rm \epsilon}^2 b\,d = \underbrace{ a\,c}_{\text{real}} + {\rm \epsilon} \underbrace{ ( b\,c + a\,d) }_{\text{dual}}$$

This has some applications, specifically in analysis and numerical methods as it can be proven than if the real part is a function then the dual part is the slope of the function, and the normal rules of calculus for derivatives are automatically enforced.

So congrats, you just invented dual numbers, or at least asked to right question that lead us to their development and the development of other so-called Clifford algebras.