# Thread: Do sets of all pairs of real numbers form fields?

1. ## Do sets of all pairs of real numbers form fields?

Hi,

problem:
Let $\displaystyle F$ be the set of all (ordered) pairs $\displaystyle (\alpha,\beta)$ of real numbers.

(a) If addition and multiplication are defined by
$\displaystyle (\alpha,\beta)+(\gamma,\delta)=(\alpha+\gamma,\bet a+\delta)$
and
$\displaystyle (\alpha,\beta)(\gamma,\delta)=(\alpha\gamma,\beta\ delta)$,
does $\displaystyle F$ become a field?

(b) If addition and multiplication are defined by
$\displaystyle (\alpha,\beta)+(\gamma,\delta)=(\alpha+\gamma,\bet a+\delta)$
and
$\displaystyle (\alpha,\beta)(\gamma,\delta)=(\alpha\gamma-\beta\delta,\alpha\delta+\beta\gamma)$,
is $\displaystyle F$ a field then?

(c) What happens (in both the preceding cases) if we consider ordered pairs of complex numbers instead?

attempt (at (a)):
I have gone through all the axioms for a field and am unable to find one that fails.
For example, the multiplicative inverse:
$\displaystyle (\alpha,\beta)\frac{1}{\alpha,\beta} \Rightarrow (\alpha,\beta)\left(\frac{1}{\alpha},\frac{1}{\bet a}\right)=(1,1)$
One of my friends told me that this is not a field, but I fail to see why.

Thanks!

2. for(a), (r,0) has no inverse if r is not 0!
for(b), It is a field, because the mapping: (a,b)--> a+ib, (i is the imaginary unit) is a isomorphism. Or you can varify that all axioms for field are satisfied.
for(c), (a) can't be a field for the same reason.
(b) can't be a field because the inverse of some element is not unique.

3. Originally Posted by Shanks
for(a), (r,0) has no inverse if r is not 0!
So there is an inverse for $\displaystyle (0,0)$?

Originally Posted by Shanks
for(b), It is a field, because the mapping: (a,b)--> a+ib, (i is the imaginary unit) is a isomorphism. Or you can varify that all axioms for field are satisfied.
I am not yet familiar with isomorphism, it's covered in a later chapter.
As for verifying that the axioms are satisfied, I'm having some problems with it.
Say I want to verify that there exists an multiplicative inverse:

$\displaystyle (\alpha,\beta)\cdot(\alpha',\beta')=(1,1)$
I see that it would be true for say,

$\displaystyle (1,1)\cdot(1,0)=(1-0,0+1)=(1,1)$

I am unfortunately not able to find a general expression for the inverse..

Thank you.

4. what is the inverse of (1,0)? can you answer this question?

5. If $\displaystyle (1,0)$ does not have an inverse since there is no scalar $\displaystyle \alpha$ such that $\displaystyle 0\cdot\alpha=1$. Correct?
By the same reasoning, can I not say that $\displaystyle (0,0)$ does not have an inverse?

How about my second question (b) ?

Thank you very much.

6. Originally Posted by Mollier
If $\displaystyle (1,0)$ does not have an inverse since there is no scalar $\displaystyle \alpha$ such that $\displaystyle 0\cdot\alpha=1$. Correct?
By the same reasoning, can I not say that $\displaystyle (0,0)$ does not have an inverse?

How about my second question (b) ?

Thank you very much.
What have you done to try and find this inverse so far?

You want to find an $\displaystyle a, b \in \mathbb{R}$ such that $\displaystyle (\alpha, \beta)(a, b) = 1$. This gives you 2 equations in 4 variables, but with a bit of work you should see what is going on.

7. Originally Posted by Mollier
If $\displaystyle (1,0)$ does not have an inverse since there is no scalar $\displaystyle \alpha$ such that $\displaystyle 0\cdot\alpha=1$. Correct?
By the same reasoning, can I not say that $\displaystyle (0,0)$ does not have an inverse?

How about my second question (b) ?

Thank you very much.
The 0 element (additive identity), which here is (0, 0) never has an inverse. I don't see why you keep mentioning it. The crucial point is that, in a field, every member except 0 must have an inverse. (1, 0) does not.

8. Originally Posted by HallsofIvy
The 0 element (additive identity), which here is (0, 0) never has an inverse. I don't see why you keep mentioning it. The crucial point is that, in a field, every member except 0 must have an inverse. (1, 0) does not.

His claim "for(a), (r,0) has no inverse if r is not 0!" should read "for(a), (r,0) has no inverse. Ever."

9. Originally Posted by Swlabr

His claim "for(a), (r,0) has no inverse if r is not 0!" should read "for(a), (r,0) has no inverse if r is 0."
Exactly. I am easily confused

$\displaystyle 1.\;\alpha a-\beta b = 1$
$\displaystyle 2.\;\alpha b+\beta a = 1$

$\displaystyle a_1=\frac{1+\beta b}{\alpha}$
$\displaystyle a_2=\frac{1-\alpha b}{\beta}$

Since $\displaystyle a_1=a_2$:

$\displaystyle b=\left(\frac{\alpha-\beta}{\alpha^2+\beta^2}\right)$

then

$\displaystyle a=\left(\frac{\alpha+\beta}{\alpha^2+\beta^2}\righ t)$

I think that is correct.
Thanks guys!

10. Originally Posted by Swlabr

His claim "for(a), (r,0) has no inverse if r is not 0!" should read "for(a), (r,0) has no inverse if r is 0."
your statement does not mean the same thing as mine.
what I want to clearify is that the element of the form (r,0) where r is not 0 has no inverse.
(0,0) obviously has no inverse in a field.

I appologize if I made you confused!

11. Of course, this is a justification of the "strange" multiplication law for complex numbers... clever!

Also, the multiplicative identity for (b) is (1,0), not (1,1). Ask yourself why.

12. Originally Posted by Shanks
(b) can't be a field because the inverse of some element is not unique.
So, some element $\displaystyle (\alpha,\beta)$ may have several inverses?
Would you mind explaining this in a bit more detail?

Thanks

13. Originally Posted by Kep
Also, the multiplicative identity for (b) is (1,0), not (1,1). Ask yourself why.
Well, I see that it is (1,0) but I don't know how to explain it without saying that it's caused by how multiplication of elements is defined...

14. Originally Posted by Shanks
your statement does not mean the same thing as mine.
what I want to clearify is that the element of the form (r,0) where r is not 0 has no inverse.
(0,0) obviously has no inverse in a field.

I appologize if I made you confused!
Yes, but this implies that if r=0 there is an inverse.

As to an element having more than one inverse, this is impossible for the definition of inverse here. There is a short proof of this, which I will challenge whoever believes it to not hold to do.

15. Originally Posted by Mollier
Well, I see that it is (1,0) but I don't know how to explain it without saying that it's caused by how multiplication of elements is defined...
Identities are unique. So, simply show that $\displaystyle (\alpha, \beta)(1, 0) = (\alpha, \beta)$ to get that this is your identity.

In general you would also want to show that $\displaystyle (1, 0)(\alpha, \beta) = (\alpha, \beta)$, however your multiplication here is commutative so this doesn't need to be shown.

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