Imaginary numbers

[Anonymous No.6615122]

Hello /sci/ low IQ amateur here.

I was reading about imaginary numbers and saw often example ? * ? = -1 and as you all here know answer is i * i = -1. Nothing surprising here but friend of mine came up with solution (-1) * 1 = -1 which at least in my calculator turns out to be totally correct so anyone here could explain what's whats wrong with this alternative solution, if anything?

[Anonymous No.6615135]

>>6615122
>? * ?
> (-1) * 1
srsly?

[Anonymous No.6615138]

>>6615122
protip: -1 != 1

[Anonymous No.6615167]

I was reading about natural numbers and saw often example ? + ? = ? and as you all here know answer is 0 + 0 = 0. Nothing surprising here but friend of mine came up with solution 7 + 0 = 7 which at least in my calculator turns out to be totally correct so anyone here could explain what's whats wrong with this alternative solution, if anything?

[Anonymous No.6615176]

>>6615167

ummh.. okay thank you for the reply. Thought i was the low IQ here, guess you can get much lower.

[Anonymous No.6615180]

>>6615176
wow

no you really can't.

[Anonymous No.6615194]

>>6615180

Well, if so could you kindly explain to me why (-1) * 1 = -1 is correct or wrong? You reply didn't really bring anything on the table except really bad trolling attempt where you fell on your own shoe laces because you obviously have no clue about imaginary numbers.

[Anonymous No.6615197]

The only winning move is not to play.

[Anonymous No.6615198]

>>6615194

It's not wrong, it's just missing the entire point of I.

[Anonymous No.6615211]

>>6615194
When the problem says ?*?=-1, it is implying that both ?'s are the same...

[Anonymous No.6615217]

>>6615194
>>6615211
yeah, it should say ? ^ 2 not ? * ?, fuck

[Anonymous No.6615250]

>>6615122
>>6615217
?^2 = -1.
i.e. both the ? have to be the same. i*i =-1 is fine because i is the same as i.

-1*1 = -1 is not the same because -1 and 1 are different.

[Anonymous No.6615398]

>>6615122
The whole of complex analysis is symmetric under √-1 <-> -√-1

[Anonymous No.6616106]

>>6615122
>x * x = -1

>(-1) * (1) = -1

>x = (-1) = 1

>-1 = (1)

OP, your friend is a mathematician.

/sarcasm

Your friend is an idiot.

[Anonymous No.6616127]

>>6616106
I think OP's friend really <span class="math">is[/spoiler] a mathematician. He has quite cleverly exploited OP's low intelligence.

[Anonymous No.6616130]

Nothing. Imaginary numbers are a crock of shit.

[Anonymous No.6616131]

>>6615122
however there is also (-i)*(-i) = -1

[Anonymous No.6616300]

>>6616131
and also
(i*i*i)*(i*i*i)= -1

woah guys, I think we are close to a breakthrough, maybe imaginary numbers have something to do with circles! o.O

[Anonymous No.6616307]

A*B=-1
A=-1/B

There are infinitely many of these number pairs even in real numbers. Actually only 0 has not such pair.

[Anonymous No.6616310]

>>6616300
>circles
You're aware of the polar form of complex numbers, yes?

[Anonymous No.6616398]

>>6616310

[Anonymous No.6616628]

[Anonymous No.6616764]

>>6615138
wrong, -1! = -1

[Anonymous No.6617095]

>>6616310
e^(i*pi/2) have no idea what you're talking about.

[Anonymous No.6617189]

>>6615194
If you can't understand something so simple without an explanation you can't understand it with an explanation

[Anonymous No.6617193]

>>6616764
wrong!
Factorial is not defined for negative numbers!
Not even gamma function.

[Anonymous No.6617203]

>>6615122
OP, do you not understand the image you linked?

[Anonymous No.6617311]

>>6617193
<span class="math">\Gamma [-n]= \tilde { \infty }~~n=0,1, \cdots[/spoiler]

[Anonymous No.6617313]

>>6616130
>this is what normalfags actually believe

[Anonymous No.6617315]

>>6617193
You mean negative integers. It's defined for negatives that aren't integers

[Anonymous No.6617354]

>>6615122
>>6615122
>>6615122
Are you mentally retarded?

Serious question

[Anonymous No.6617367]

>>6617315
i'm actually curious about the reasoning behind this.

[Anonymous No.6617390]

[Anonymous No.6618157]

>>6617367
Without bitching about analytic continuation of the gamma function

n!=n*(n-1)! is basically the main point of the factorial

0!=0*(-1)!

1=0*(-1)!

So the factorial isn't defined at -1 and you can do the same thing with -2 etc

[Anonymous No.6618172]

Is zero a purely imaginary number seeing as its real part is 0?

Or is zero a real number seeing as its imaginary part is 0?

Or is zero just a special case in C just like it's considered neither positve or negative.

[Anonymous No.6618174]

>>6618172
0 is a real number
0 is also an imaginary number
kindof like clopen sets
going by most definitions anyway but this shit is just semantics it doesn't matter

[Anonymous No.6618175]

>>6618174
k thanks anon

[Anonymous No.6618176]

>>6618175
oh forgot to add, can we think of this in the terms of the complex plane?

if all imaginary numbers lie on the imaginary axis, and all real numbers lie on the real axis, and if 0 is the intersection of those axes then 0 is both real and imaginary

[Anonymous No.6618215]

>>6617193
>Factorial is not defined for negative numbers!

Yes it is: -½!=√π

[Anonymous No.6618219]

>>6618172
I∩R={0}

[Anonymous No.6619514]

I always thought imaginary numbers were considered the ones which have non-zero Im, which leaves 0 as real. At least that's how I usually see it treated in functional analysis and self-adjoint operators in qm.

[Anonymous No.6619575]

>>6619514
Well yea because 0i is still zero.

[Anonymous No.6619885]

i^2=-1
i^4=1
i^6=-1
i^8=1

[Anonymous No.6619897]

>>6618174
It's not an imaginary number or a real number, just a complex number.

[Anonymous No.6619945]

>>6615122
On the off-chance this isn't just a troll thread and it's an honest question:

The thing about i (and -i) is that when you multiply it by ITSELF it gives you 1, or, in symbolic language, (i)^2 = (-i)^2 = 1. i was needed to solve polynomials (things like x^2+x, x^4-2x^3-5, etc) where the roots (the values of x for which the polynomial equals 0) were roots of negative numbers.

Multiplying (-1) by 1 and getting -1 back isn't "as special", because we're not talking about multiplying the same number by itself (-1 is not the same as 1). You can write this in the more general form:

y*x=-1
y=-1/x
If you plot the graph of this equation, you'll see that you can find a y for every x without ever needing to leave the reals (as long as the x's you input are real themselves), except for 0 (because division by 0 is a very complicated thing which can't really happen in the reals, but which you can "approximate" by using things called limits or by leaving the reals).

[Anonymous No.6621250]

>>6619945
>The thing about i (and -i) is that when you multiply it by ITSELF it gives you 1, or, in symbolic language, (i)^2 = (-i)^2 = 1.

>definition of i
>sqrt(-1) = i

>i^2 = sqrt(-1)^2 = 1
>how to math

[Anonymous No.6622193]

>>6619945
in op's pic it shows i^2 = -1 i^ 3= -i and i^4 = 1. Does that mean that i^5 = i ?

[Anonymous No.6624224]

>>6622193
yes

i^2 = -1

i^4 = 1

i^5 = i^4 * i = 1* i = i

[Anonymous No.6624262]

>>6622193
Yes
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i

The pattern repeats itself thereafter.

[Anonymous No.6624806]

>>6615194
Those are different numbers. That is why it is wrong.

[None !EmFv0xqPZs No.6626119]

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