Is f Isomorphic? - Defining and Proving a Function

[mathstudent79]

Homework Statement

Hi. First post here. I searched for this problem and nothing I saw helped. Sorry. Thanks in advance for help.

define a function f: C -----> S by:
f(a,b) = a+bi where a,b $$\in$$R

We leave it to reader to check that f is one-one and onto and that:

f((a,b)+(c,d)) = f((a,b)) + f((c,d))

and

f((a,b)(c,d))=f(a,b)f(c,d)

Homework Equations

a+bi = (a,b)
(a+bi)+(c+di) = (a+c) + (b+d)i
(a+bi)(c+di)=(ac-bd)+(bc+ad)i

The Attempt at a Solution

Here's what I've done so far.

f((a,b)+(c,d))=(a+c)+(b+d)i=(a+bi)+(c+di)=f(a,b)+f(c,d)

f((a,b)(c,d))=(ac-bd)+(bc+ad)i=(a+bi)(c+di)=f(a,b)f(c,d)

sp' f(a,b)=f(a',b')

then a+bi=a'+b'i
then (a,b)=(a',b')
but this is only true if a=a' and b=b' So f(a,b)=f(a',b') implies (a,b)=(a',b')

HERE'S where I'm especially not sure:

onto:

suppose we have a+bi. By definition of function, there is an a,b s.t. f(a,b)=a+bi. Therefore onto.


Thanks a lot guys.

 

[gb7nash]

define a function f: C -----> S by:
f(a,b) = a+bi where a,b $$\in$$R

I'm not sure I understand this part. Is C the set of all complex numbers? What is S? And since (a,b) is a two-tuple, shouldn't f: $$R^2$$ -> S? (or some subset of $$R^2$$ to S?)

 

[mathstudent79]

I'm sorry,

yes, C is set of Complex Nos. No argument with your observation, but I copied that directly from the book (Modern Algebra, a Natural Approach, Gardiner)

Thanks a lot.

 

[mathstudent79]

As for S, I forgot to put:

'We consider the set S of all formal expressions of the form: a+bi, where a,b$$\in$$R"

Again, with the R, direct quote.

Thanks.

 

[Sethric]

With the clarifications, the logical steps look correct here. I might flesh out more details for steps such as onto. Since they both draw from all of the reals, I feel that is an important in between step for clarification.

 

[mathstudent79]

Sethric, thank you.

As you know, the onto was the part I was especially not sure of. The book I am studying from has very informal and (to my amateur eye) intuitive proofs for onto functions. Along the lines of 'take any b in B. then f(a)=b for some a in A. Thus the function is onto.' Can you give me a couple of more details on the sort of clarifications or fleshing out you're referring to, if you don't mind? Thank you again.

 

[Sethric]

If I were proving the onto, I would go along the lines of:

Let $$a+bi \in S$$

Then $$a,b \in R$$

Therefore, since $$C = R^2, (a,b) \in C$$ and f((a,b)) = a+bi.

Essentially I would just want to show that that particular a and b are actually in the original set. Showing they were both formed from the reals is enough here, but might not be in future questions.

 

[mathstudent79]

Sethric -

Thank you very much!