Absolute value of complex number

[gotjrgkr]

Homework Statement

Let R be a set of real numbers derived from rational numbers.
And, let R* be a set consisting of all ordered pairs of the form (x,0) where x is an element of R.

For a complex number z = (a,b) = a+ib, I've learned that the absolute value of z is the number (a^2+b^2)^1/2.

I wonder which one of R and R* contains the absolute value ((a^2+b^2)^1/2) of z

Homework Equations

The Attempt at a Solution

 

[micromass]

Hi gotjrgkr,

Strictly mathematically, $$\sqrt{a^2+b^2}$$ will be an element of $$\mathbb{R}$$. Since you seem to say that $$\mathbb{R}^*$$ (which is a notation I've not encountered yet) is the set of all couples (x,0). This would mean that $$\sqrt{a^2+b^2}$$ is not in $$\mathbb{R}^*$$, since it is not of the form (x,0).

That said, there is a canonical isomorphism between $$\mathbb{R}$$ and $$\mathbb{R}^*$$. So, I wouldn't be surprised if people identify the two sets and would actually say that $$\sqrt{a^2+b^2}$$ is in $$\mathbb{R}^*$$. But strictly mathematical, this is not correct.

 

[Antiphon]

Yes, and I'm thinking also that it's not in either since sqrt(2) isn't in R or R*.

 

[micromass]

Yes, and I'm thinking also that it's not in either since sqrt(2) isn't in R or R*.

And since when is sqrt(2) not in R?

 

[gotjrgkr]

Hi gotjrgkr,

Strictly mathematically, $$\sqrt{a^2+b^2}$$ will be an element of $$\mathbb{R}$$. Since you seem to say that $$\mathbb{R}^*$$ (which is a notation I've not encountered yet) is the set of all couples (x,0). This would mean that $$\sqrt{a^2+b^2}$$ is not in $$\mathbb{R}^*$$, since it is not of the form (x,0).

That said, there is a canonical isomorphism between $$\mathbb{R}$$ and $$\mathbb{R}^*$$. So, I wouldn't be surprised if people identify the two sets and would actually say that $$\sqrt{a^2+b^2}$$ is in $$\mathbb{R}^*$$. But strictly mathematical, this is not correct.

yeah, I also thought about that case like you first.
I actually study P.M.A by rudin nowadays.
And in the first chapter ,as you know, the complex number system has been introduced.
There is a phrase arguing that we wright x=(x,0) where numbers x,0 in the right side of the equation are contained in R and the absolute value of any complex number z is sqrt of z*(conjugate of z). When I saw this phrase, because z*(conjugate of z) = (a^2+b^2,0) where z=(a,b) and a,b are contained in R by using the definition of absolute value of z as I've written it above, if I follow the convention x=(x,0), then I would get the result such that (a^2+b^2)^1/2=((a^2+b^2)^1/2,0) is contained in R*.
But if I accept this result then i couldn't explain an another result related with metric space; if i take it , then because the value(real number) of metric on system of complex numbers would be of the form(x,0) for some x contained in R.So, the value is contained in R*.
But the metric of euclidean space has its value (real number) in R. And i think the value(real number) of a metric space must be in the same space (R or R*).
I'm sorry for my poor english ability in conveying my idea.
Please try to understand some errors...

 

[micromass]

Ah, well, in that case Rudin simply identified the two spaces. This might look tricky and incorrect, but mathematicians do this very often.
In this case,

$$\sqrt{a^2+b^2}=(\sqrt{a^2+b^2},0)$$

have been identified. So both are in R and in R*. But, what does identification mean? Well, what Rudin doesn't say is that he assumes an isomorphism $$\phi:\mathbb{R}\rightarrow \mathbb{R}^*$$ and that he actually writes

$$\phi(x)=(x,0)$$.

So, whenever he talks about x and (x,0), he assumes those two to be the same under the isomorphism phi!

 

[gotjrgkr]

Ah, well, in that case Rudin simply identified the two spaces. This might look tricky and incorrect, but mathematicians do this very often.
In this case,

$$\sqrt{a^2+b^2}=(\sqrt{a^2+b^2},0)$$

have been identified. So both are in R and in R*. But, what does identification mean? Well, what Rudin doesn't say is that he assumes an isomorphism $$\phi:\mathbb{R}\rightarrow \mathbb{R}^*$$ and that he actually writes

$$\phi(x)=(x,0)$$.

So, whenever he talks about x and (x,0), he assumes those two to be the same under the isomorphism phi!

1.I haven't studied about isomorphism yet. Could you tell me where I can find that notion?
And if you don't mind, i'd like to ask you one more question about euclidean space related with this topic.
2.As I said above, if i accept that the value of metric function on a system of complex numbers is contained in R*, isn't it contradiction to the fact that the value of metric function on any k-dimensional euclidean space is contained in R??
Therefore, i tried another way to solve this problem; Because R* also satisfies all axioms which are needed to be R, I thought R* could be redefined as a system of real numbers.And this procedure is similar with construction how rational numbers are defined from R( at first, R is derived from a set of rational numbers and the set of rationals has been redefined as a subset of R). Then by using this newly defined real numbers x= (x,0), i could get the result such that the value of metric function on the euclidean space is also contained in R*. Therefore, their set theorytic forms coincide each other; i mean the values of metric functions of C(a set of complex numbers) and k-dimensional euclidean space are contained in the same space R*.
Do you agree with my argument?
If there's any wrong part, please tell me about that.
I really appreciate you about reading my long questions and sharing my problems.

 

[micromass]

1) Don't worry about it then. With isomorphism, I just mean a suitable "identification"
2) This is a very good solution to the problem! Just redefine R to mean R*, this looks fine to me!

 

[gotjrgkr]

1) Don't worry about it then. With isomorphism, I just mean a suitable "identification"
2) This is a very good solution to the problem! Just redefine R to mean R*, this looks fine to me!

Oh, really?
But it seems to me that there's another problem related with the definition of integration of complex valued function of real variable if i assume that...
Anyway, thank you very much!