# A number that jumps over the field, complex number $\mathbb { C }$.

Can I make the number field that jumps over the field, $\mathbb { C }$?

I want to make it.

Note by . .
1 month, 1 week ago

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$a \in \mathbb { Z }, p ^ { a } = \begin {cases} p ^ { a } = 1 & \quad \text { if } a \text { is positive, and even } \\ p ^ { a } = p & \quad \text { if } a \text { is positive, and odd } \\ p ^ { a } = 1 & \quad \text { if } a \text { is zero } \\ p ^ { a } = 1 & \quad \text { if } a \text { is negative, and even } \\ p ^ { a } = 0 & \quad \text { if } a \text { is negative, and odd } \end {cases}$

If you understand these, you are almost mastered this.

- 1 month, 1 week ago

Okay so are these facts about p or properties of proper numbers(p*0≠0) ?

- 1 month, 1 week ago

I think I should first look up what proper numbers are then I'll come back here to get a better understanding of what's going on.

- 1 month, 1 week ago

Of, course.

But, anyway, can you teach let me to use the simultaneous symbol? ({)

I used them in the way this Wikipedia page, but I cannot know them.

- 1 month, 1 week ago

@. . Yeah sure !!

{ This bracket can be written by adding a backslash(\) before ({).

- 1 month, 1 week ago

Ohh did you mean these bigger ones? For eg $\left\{\dfrac{cosx}{tanx}\right\}$ ?

The latex code for it is as follows:

\ (\left{ \dfrac{cosx}{tanx}\right})

I.e. adding \left before the text will work.

These are the links to a few latex guides on brilliant that I found useful:

- 1 month, 1 week ago

Not that.

cases.

$\begin {cases} 1/ 2 \end {cases}$.

- 1 month, 1 week ago

@. . $X(m,n)= \begin{cases} x(n),\\ x(n-1)\\ x(n-1) \end{cases}$

Do u wanna write something like this $\uparrow$

This is the required code:

X(m,n)= \begin{cases} x(n),\ x(n-1)\ x(n-1) \end{cases}

- 1 month, 1 week ago

, or ?

- 1 month ago

@. . $\begin{cases} n/2 &\mbox{if } n \equiv 0 \\ (3n +1)/2 & \mbox{if } n \equiv 1 \end{cases} \pmod{2}$

How about this $\uparrow$?

Reqd code:

\ begin{cases} n/2 & \mbox{if } n \equiv 0 \\ (3n +1)/2 & \mbox{if } n \equiv 1 \end{cases} \ pmod{2}

I couldn't find any other than this on stack exchange.

- 1 month ago

I will prove those.

$a \in \mathbb{N}, p^{2n} = 1$

$a \in \mathbb{N}, p^{2n+1} = p$

@Agent T, You will understand these because I mentioned them before.

$p^{0} = 1$.

I'm not sure, but I think that you will understand.

$a \in \mathbb{N}, p^{-2n} = \frac{1}{p^{2n}} = 1$

$a \in \mathbb{N}, p^{-(2n+1)} = p^{-2n-1} = \frac{1}{p^{2n+1}} = \frac{1}{p} = 0$.

I hope you will understand.

And do you understand my explanation of my problem?

- 1 month ago

Yes ,it all makes sense now. Sometimes it's just us who limit our thinking and refuse to grab a concept even if we are familiar with it:^)

- 1 month ago

Thank you for the explanations ,I really appreciate them!

Yo! Have a great day :)

- 1 month ago

You too.

- 1 month ago

It is a proper number, $\mathbb { Pr }$.

And $\mathbb { M }$ which is mixed number, $\mathbb { C } + \mathbb { Pr }$.

A proper number is a number that gets multiplied with zero, which is not equal to zero.

A proper unit is $p$.

$p \times 0 = 1$ which I want to create.

And, if $ap \times 0$, you have to calculate $a \times p \times 0$, $p \times 0$ first.

So, $ap \times 0 = a$.

Basic calculations.

$p \pm \alpha$ cannot be calculated like an irrational number calculation like $\sqrt { 2 } + e ^ { \log _ { 3 } 2 } + \sqrt [ \pi ] { 3! \times e }$.

$p \times \alpha, \alpha \ne 0$ is equal to $\alpha p$, which is equal to multiplication of irrational numbers.

$p \div 0$ is equal to $1$ because I said $0 \times p = 1 \rightarrow 0 = 1 \div p \rightarrow p \times 1 \div p = 1$.

$p \div \alpha, \alpha \ne 0, p$ is equal to $\frac { p } { \alpha }$, same as calculating irrational numbers.

$p \times p = p \times 1 \div 0 = p \div 0 = 1$.

$p \times p \times p = 1 \times p = p$.

$p ^ p = 1$.

$n \in \mathbb { N }, p ^ { 2n } = 1$.

$n \in \mathbb { N }, p ^ { 2n + 1 } = p$.

$\sqrt { p } = p ^ { 2n } = 1, n \in \mathbb { N }$.

Now, let us talk about mixed numbers.

A normal form is $a + bi + cp$, where $a, b$ are real numbers, and $c$ is complex number.

Expansion of mixed number.

$( a + bi + cp ) ( d + ei + fp ) = ad + aei + afp + bdi + bei ^ { 2 } + bfip + cdp + ceip + cfp ^ { 2 } = ad + aei + afp + bdi - be + bfip + cdp + ceip + cf = ( ad - be + cf ) + ( ae + bd )i + ( af + cd )p + ( bf + ce )ip$.

The end.

- 1 month, 1 week ago

Normally, I wanted to define $\div 0$.

- 1 month, 1 week ago

It seems interesting,what topic does it come under? I wanna explore it more!

- 1 month, 1 week ago

The ball is in all our court, including me.

- 1 month, 1 week ago

@. . Now you are gonna say things that my tiny brain can't comprehend -_-

Btw did u mean that this topic is very much related to our life or something? Eeeeeeeee I wanna knowww!

- 1 month, 1 week ago

If you solved this my problem, I will tell you more about the number.

- 1 month, 1 week ago

@. . Haha okay lemme try! Edit: As expected ,I got it wrong :/ marked the negative one.

- 1 month, 1 week ago

No problem.

If you understand my explanation, then it doesn't matter.

By the way, I promised you, so I will keep that.

As the replies are too long, I will make the new line.

- 1 month, 1 week ago

@. . Sounds great , thanks!

- 1 month, 1 week ago

I just found out that I already knew complex fields and real numbers(idk what u must be thinking XP)

But I wonder why have you used the word "proper number" rather than real numbers and "mixed" for complex??

And I guess there's an error in the part where you divided p by 0 ,kindly check that once .

- 1 month ago

@. . you might love these if u wanna go beyond complex 8)

- 1 month ago