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

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• Jan 5th 2010, 09:09 PM
Mollier
Do sets of all pairs of real numbers form fields?
Hi,

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

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

(b) If addition and multiplication are defined by
$(\alpha,\beta)+(\gamma,\delta)=(\alpha+\gamma,\bet a+\delta)$
and
$(\alpha,\beta)(\gamma,\delta)=(\alpha\gamma-\beta\delta,\alpha\delta+\beta\gamma)$,
is $F$ 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:
$(\alpha,\beta)\frac{1}{\alpha,\beta} \Rightarrow (\alpha,\beta)\left(\frac{1}{\alpha},\frac{1}{\bet a}\right)=(1,1)$
One of my friends told me that this is not a field, but I fail to see why.

Thanks!
• Jan 5th 2010, 10:19 PM
Shanks
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.
• Jan 6th 2010, 12:15 AM
Mollier
Quote:

Originally Posted by Shanks
for(a), (r,0) has no inverse if r is not 0!

So there is an inverse for $(0,0)$?

Quote:

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:

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

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

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

Thank you.
• Jan 6th 2010, 01:42 AM
Shanks
what is the inverse of (1,0)? can you answer this question?
• Jan 6th 2010, 01:56 AM
Mollier
If $(1,0)$ does not have an inverse since there is no scalar $\alpha$ such that $0\cdot\alpha=1$. Correct?
By the same reasoning, can I not say that $(0,0)$ does not have an inverse?

How about my second question (b) ?

Thank you very much.
• Jan 6th 2010, 02:03 AM
Swlabr
Quote:

Originally Posted by Mollier
If $(1,0)$ does not have an inverse since there is no scalar $\alpha$ such that $0\cdot\alpha=1$. Correct?
By the same reasoning, can I not say that $(0,0)$ 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 $a, b \in \mathbb{R}$ such that $(\alpha, \beta)(a, b) = 1$. This gives you 2 equations in 4 variables, but with a bit of work you should see what is going on.
• Jan 6th 2010, 03:18 AM
HallsofIvy
Quote:

Originally Posted by Mollier
If $(1,0)$ does not have an inverse since there is no scalar $\alpha$ such that $0\cdot\alpha=1$. Correct?
By the same reasoning, can I not say that $(0,0)$ 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.
• Jan 6th 2010, 03:22 AM
Swlabr
Quote:

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."
• Jan 6th 2010, 03:33 AM
Mollier
Quote:

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 :)

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

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

Since $a_1=a_2$:

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

then

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

I think that is correct.
Thanks guys!
• Jan 6th 2010, 03:46 AM
Shanks
Quote:

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!
• Jan 6th 2010, 03:59 AM
Kep
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.
• Jan 6th 2010, 04:00 AM
Mollier
Quote:

Originally Posted by Shanks
(b) can't be a field because the inverse of some element is not unique.

So, some element $(\alpha,\beta)$ may have several inverses?
Would you mind explaining this in a bit more detail?

Thanks
• Jan 6th 2010, 04:08 AM
Mollier
Quote:

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...
• Jan 6th 2010, 04:20 AM
Swlabr
Quote:

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.
• Jan 6th 2010, 04:24 AM
Swlabr
Quote:

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 $(\alpha, \beta)(1, 0) = (\alpha, \beta)$ to get that this is your identity.

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