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

1. Originally Posted by Swlabr
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.
So then the answers to (c) is the same as for (a) and (b)?

Originally Posted by Swlabr
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.
That's the thing, I can show it, but I would not have noticed it myself..

Again, Thanks!

2. Originally Posted by Mollier
So then the answers to (c) is the same as for (a) and (b)?
...but the answers to (a) and (b) are different...

Originally Posted by Mollier
That's the thing, I can show it, but I would not have noticed it myself..

Again, Thanks!
These types of questions (you are given a strange multiplication in this way) are relatively easy to do once you know how, and are a gift in an exam!

Firstly, look to see if the multiplication is commutative. This is usually quite easy to spot if it is.

Secondly, you need to find an identity element. So, let $\displaystyle (\alpha, \beta)$ be an arbitrary element. We want to find an $\displaystyle a$ and a $\displaystyle b$ such that $\displaystyle (\alpha, \beta)(a, b) = (\alpha, \beta)$. So, just use the multiplication given to work out what $\displaystyle (a, b)$ is.

This is your identity element. You know it is an identity on the right, but unless the multiplication is commutative you do not know if it is an identity on the left. So, knowing what $\displaystyle (a, b)$ is check to see if $\displaystyle (a, b)(\alpha, \beta) = (\alpha, \beta)$. Usually (always if you have been asked to find an identity or an inverse not a right identity or a right inverse (alternatively, left identity etc.)) this will be so. But do it all the same as it is part of the proof. Once you have shown this you have found your identity element.

Now you need to find your inverse. You want to find an element $\displaystyle (\alpha^{\prime}, \beta^{\prime})$ such that $\displaystyle (\alpha, \beta)(\alpha^{\prime}, \beta^{\prime}) = (a, b)$. Again, just use the fact that you know what the multiplication is to solve for $\displaystyle (\alpha, \beta)$.

You will usually find a generic inverse, but this is not always an inverse for all elements, as with question (a) here. Usually if this is the case you will be dividing by zero.

3. Originally Posted by Swlabr
...but the answers to (a) and (b) are different...
Aha,
$\displaystyle \alpha^2+\beta^2=(a_1+b_1i)^2+(a_2+b_2i)^2$
So for say $\displaystyle a_1=b_1=1,a_2=-1\;and\;b_2=1$ I would be dividing by zero.
In other words, for some elements there is no multiplicative inverse.

Originally Posted by Swlabr
These types of questions (you are given a strange multiplication in this way) are relatively easy to do once you know how, and are a gift in an exam!
Luckily, I am doing this just for fun so no exams are involved
I do appreciate the detailed explanation though, thanks!

4. 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. Ever."
But his point was that it is "(r, 0) if r is not 0" not having an inverse that says this is not a field!

5. Originally Posted by HallsofIvy
But his point was that it is "(r, 0) if r is not 0" not having an inverse that says this is not a field!
Yeah, but that is what was causing the confusion. Although a correct statement, it made Mollier think that this was the only time there was not an inverse.

6. It's all good now

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