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

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

I want to make it.

Note by . .
1 month, 1 week ago

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aZ,pa={pa=1 if a is positive, and even pa=p if a is positive, and odd pa=1 if a is zero pa=1 if a is negative, and even pa=0 if a is negative, and odd  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

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

Agent T - 1 month, 1 week ago

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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.

Agent T - 1 month, 1 week ago

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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

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@. . Yeah sure !!

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

Agent T - 1 month, 1 week ago

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@Agent T Ohh did you mean these bigger ones? For eg {cosxtanx}\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.Pall Marton's latex guide

2.Percy Jackson's latex guide

Agent T - 1 month, 1 week ago

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@Agent T Not that.

cases.

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

. . - 1 month, 1 week ago

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@. . X(m,n)={x(n),x(n1)x(n1)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}

Agent T - 1 month, 1 week ago

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@Agent T

, or ?

. . - 1 month ago

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@. . {n/2if n0(3n+1)/2if n1(mod2)\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.

Agent T - 1 month ago

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I will prove those.

aN,p2n=1 a \in \mathbb{N}, p^{2n} = 1

aN,p2n+1=p a \in \mathbb{N}, p^{2n+1} = p

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

p0=1 p^{0} = 1 .

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

aN,p2n=1p2n=1 a \in \mathbb{N}, p^{-2n} = \frac{1}{p^{2n}} = 1

aN,p(2n+1)=p2n1=1p2n+1=1p=0 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

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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:^)

Agent T - 1 month ago

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Thank you for the explanations ,I really appreciate them!

Yo! Have a great day :)

Agent T - 1 month ago

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@Agent T You too.

. . - 1 month ago

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

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

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

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

So, ap×0=a ap \times 0 = a .

Basic calculations.

p±α p \pm \alpha cannot be calculated like an irrational number calculation like 2+elog32+3!×eπ \sqrt { 2 } + e ^ { \log _ { 3 } 2 } + \sqrt [ \pi ] { 3! \times e } .

p×α,α0 p \times \alpha, \alpha \ne 0 is equal to αp \alpha p , which is equal to multiplication of irrational numbers.

p÷0 p \div 0 is equal to 1 1 because I said 0×p=10=1÷pp×1÷p=1 0 \times p = 1 \rightarrow 0 = 1 \div p \rightarrow p \times 1 \div p = 1 .

p÷α,α0,p p \div \alpha, \alpha \ne 0, p is equal to pα \frac { p } { \alpha } , same as calculating irrational numbers.

Upgraded calculations.

p×p=p×1÷0=p÷0=1 p \times p = p \times 1 \div 0 = p \div 0 = 1 .

p×p×p=1×p=p p \times p \times p = 1 \times p = p .

pp=1 p ^ p = 1 .

nN,p2n=1 n \in \mathbb { N }, p ^ { 2n } = 1 .

nN,p2n+1=p n \in \mathbb { N }, p ^ { 2n + 1 } = p .

p=p2n=1,nN \sqrt { p } = p ^ { 2n } = 1, n \in \mathbb { N } .

Now, let us talk about mixed numbers.

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

Expansion of mixed number.

(a+bi+cp)(d+ei+fp)=ad+aei+afp+bdi+bei2+bfip+cdp+ceip+cfp2=ad+aei+afp+bdibe+bfip+cdp+ceip+cf=(adbe+cf)+(ae+bd)i+(af+cd)p+(bf+ce)ip ( 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

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

. . - 1 month, 1 week ago

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

Agent T - 1 month, 1 week ago

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The ball is in all our court, including me.

. . - 1 month, 1 week ago

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@. . 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!

Agent T - 1 month, 1 week ago

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

. . - 1 month, 1 week ago

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

Agent T - 1 month, 1 week ago

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@Agent T 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

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@. . Sounds great , thanks!

Agent T - 1 month, 1 week ago

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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 .

Agent T - 1 month ago

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

Agent T - 1 month ago

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