Prove Complex Numbers C are isomorphic to R x R

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Prove Complex Numbers ,C are isomorphic to R x R (Real Numbers cross Real Numbers)

BUT....Before i start trying to prove this abstractly, what would be an example of a complex number isomorphic to R x R?

Shadow · **Posted:** Thu, 1 Mar 2012 04:58:01 UTC

DgrayMan wrote:

Prove Complex Numbers ,C are isomorphic to R x R (Real Numbers cross Real Numbers)

BUT....Before i start trying to prove this abstractly, what would be an example of a complex number isomorphic to R x R?

Prove they're isomorphic as what? Groups? Vector spaces? Rings? And no matter what sense you mean it in the statement "a complex number isomorphic to [tex]\mathbb{R}\times\mathbb{R}" is definitely meaningless.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

I'm doing number 3

Shadow · **Posted:** Thu, 1 Mar 2012 05:48:27 UTC

DgrayMan wrote:

I'm doing number 3

Ah, then it's easy, just recall that z=x+iy

for $x,y\in\mathbb{R}$ is how you define complex numbers, and addition works as z+w=(x+iy)+(x'+iy')=(x+x')+i(y+y')

. From the definition of addition of complex numbers, you immediately get the group isomorphism you want.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

What would be an example, like in the question i asked on arbitrary sets, daveyinaz gave an example of a function. what would be an example of a complex isomorphic to RxR?

outermeasure · **Posted:** Thu, 1 Mar 2012 09:57:29 UTC

DgrayMan wrote:

What would be an example, like in the question i asked on arbitrary sets, daveyinaz gave an example of a function. what would be an example of a complex isomorphic to RxR?

A complex number itself doesn't carry any algebraic structure, hence cannot possibly be isomorphic to $\mathbb{R}\times\mathbb{R}$ .

Shadow · **Posted:** Thu, 1 Mar 2012 16:03:13 UTC

DgrayMan wrote:

What would be an example, like in the question i asked on arbitrary sets, daveyinaz gave an example of a function. what would be an example of a complex isomorphic to RxR?

Did you never learn what the complex numbers are? Never see them as vectors in the plane? You should wikipedia them and get a better feel for them before you go trying to produce an isomorphism, as not understanding the basics of $\mathbb{C}$ will completely prevent you from producing a map. Do you know what $\mathbb{R}^2$ is? You should have seen it way back in calculus.

daveyinaz · **Joined:** Sat, 16 Aug 2008 04:47:19 UTC **Posts:** 208

Consider the mapping $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{C}$ defined by f(x, y) = x + yi

.

Assuming both $\mathbb{R} \times \mathbb{R}$ and $\mathbb{C}$ are groups under addition, where the complex addition is defined how Shadow describes and the tuples from $\mathbb{R}^2$ is done component-wise.

So there are a few things you to do here: Show that

is a homomorphism. Then show it is injective and surjective.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

homomorphism is a new word, i don't think we've talked about those in class yet. although im pretty sure you will i should have.
Ive seen complex numbers and used them in Differential equations and Calculus but i will try to wikepedia it. And RxR im assuming is something like if a,b,c,d are all real numbers numbers then RxR is f[(a b)+(c d)]= [ (a+c)+(b+d) ]

Shadow · **Posted:** Fri, 2 Mar 2012 02:05:59 UTC

DgrayMan wrote:

homomorphism is a new word, i don't think we've talked about those in class yet. although im pretty sure you will i should have.
Ive seen complex numbers and used them in Differential equations and Calculus but i will try to wikepedia it. And RxR im assuming is something like if a,b,c,d are all real numbers numbers then RxR is f[(a b)+(c d)]= [ (a+c)+(b+d) ]

Wait, an isomorphism is a special type of homomorphism. How can you be asked to show two things are isomorphic if you do not know what an isomorphism is?

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Our book has made no mention of the word homomorphism. Which is the reaction i thought i'd get here.

Shadow · **Posted:** Fri, 2 Mar 2012 09:02:11 UTC

DgrayMan wrote:

Our book has made no mention of the word homomorphism. Which is the reaction i thought i'd get here.

You should ask your professor how you're expected to do this then; it's unreasonable to ask students to do something that hasn't been taught and is not in the book under material you've learned.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

You are probably right.I thought it was just me but we as students are all having trouble on this particular chapter.

daveyinaz · **Joined:** Sat, 16 Aug 2008 04:47:19 UTC **Posts:** 208

DgrayMan wrote:

And RxR im assuming is something like if a,b,c,d are all real numbers numbers then RxR is f[(a b)+(c d)]= [ (a+c)+(b+d) ]

$\mathbb{R} \times \mathbb{R} = \{ (x, y) \mid x, y \in \mathbb{R} \}$ or in words the set containing ordered pairs where both components are elements in the real numbers.

Some examples are $(1, 1), (\sqrt{2}, \pi )$ , and $(\frac{2}{5}, 101)$ are all elements in $\mathbb{R} \times \mathbb{R}$

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Ok i looked it up and for some reason the book does not use the word 'homomorphism' to describe f(ab)=f(a)f(b), but i do know that f(ab)=f(a)f(b) is need to show isomorphisms. Gah this books makes life so difficult!

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Prove Complex Numbers C are isomorphic to R x R

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