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/sci/ - Science & Math


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2171947 No.2171947 [Reply] [Original]

>> No.2171957

the absolute value of... whatever that C symbol means is less then or equal to the absolute value of all real numbers?

what does that C symbol mean? it probably is less then all real numbers, if you take all real numbers to be infinity.

>> No.2171964

>>2171957
set theory, learn it

>> No.2171966

>>2171957

Complex numbers.

And I'm pretty sure the statement is correct, but I'm not about to try constructing a bijection.

>> No.2171985

>>2171947
Mathnoob here, but I thought that the set of real numbers was a subset of the complex numbers?

>> No.2171987

Sorry, is there some way complex numbers are not equal to the set of all reals?

>> No.2171994

>>2171985

Yes.

But they're also the same size.

Welcome to set theory, leave your quaint intuitions at the door.

>> No.2171995

>>2171985
Yeah, but all sorts of crazy shit goes down when you start messing with infinite sets.

>> No.2172002

>>2171994
>Welcome to set theory, we make wrong shit up to sound 2deep4u.

fix'd

>> No.2172004

>>2171995
>>2171994
I assumed things wouldn't be that simple. I understand how the set of real numbers is larger than both the set of integers and rationals but complex numbers are still rather alien to me.

>> No.2172006

>>2172002

Go to bed, Wittgenstein.

>> No.2172008

>>2172004
just to confuse you even more, the set of rationals is the same size as the set of integers

>> No.2172012
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2172012

>>2172008
I can sort of grasp that. Would it be safe to say that for example the interval [0,1] on the real line would be larger than the set of rationals?
I remember when my mind was blown by the idea of an infinite number of transcendental numbers in any given interval. Golly gosh I love me some maths.

>> No.2172013

>>2172008
IT'S WHAT?

>> No.2172015

>>2172004

It makes more sense when you move to field theory, the Complex numbers are an extension of the Real numbers.

More specifically the field of complex numbers is isomorphic to R^2 the two dimensional vector space of real numbers

>> No.2172020

>>2172013

Rationals are countable, so you can map it to the integers, hence the same size.

Its easy to see from the definition of the rationals.

>> No.2172021

>>2172015
We haven't done a an awful lot of work using the set lot complex numbers in my Vector Spaces course at uni. We've touched extremely lightly on fields and rings and I've being told that modules are 'much less nice' to work with.

>> No.2172028

>>2172012
every interval in the reals contains uncountably many elements, so yes, bot (0,1) and [0,1] as intervals contain more elements than both the integers and rationals.

(0,1) also contains more elements than the integers + rationals. Since any countable set union-ed with a countable set is countable.

>> No.2172031

>>2171947
This is right, there's no way to prove you wrong.

>> No.2172060

>>2171947
Pretty sure they're equal, in the same way that the sizes of the rationals and integers are the same size.

If rational R = A/B, where A and B are integers, then there are |Z|^2 rationals, right? So z = x+iy where z is a complex and x,y are reals, then the number of possible z's is |Reals|^2.

>> No.2172140

>>2172020
Oh hey that makes sense. Each time you find a new rational, to know it's newness you need to know all the others you know so far. You can say "I know <integer> rationals". With rationals you explore inside, but each new ones give you also a new integer. Same size. That's so weird, because theres an infinite number of rationals between each integer.

>> No.2172150

>any numbers are real

>> No.2172279
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2172279

bijection between R and (-1,1): x -> if x = 0 then 0, x < 0 then 1/(x-1), x > 0 then 1/(x+1).

bijection between (-1,1) and (0,1): x -> (x+1)/2

bijection between the irrational numbers inside (0,1) and nonterminating binary sequences: obvious 0.0111010101111.. etc

bijection between binary sequences and the complex box {x+iy|x in (0,1),y in (0,1)}: interpret the binary sequences as base 2i numbers.

bijection between (0,1) and complex box: split both into rational (resp. complex rational) and irrational - we already have a bijection between the irrational parts, bijection between the rational parts is trivial because they're both countable (you should be able to do this).

bijection between complex box and complex plane: trivial, just use the same argument in steps 1 and 2 - twice.

Compose everything above for a bijection between R and C.