Are Perplex Numbers on R2 a Field According to the Axioms?

[evad1089]

Homework Statement

On R², define the binary operators
(x,y)+(u,v)=(x+u,y+v)
(x,y)+(u,v)=(xu+yv,xv+yu)
The set R², along with these definitions of addition and multiplication, for the perplex numbers.
(a) Show that \cdot: R² \rightarrow R is associative.
(b) Either prove or disprove: The triple (R²,+,\cdot) is a field.

Homework Equations

Definition of a field.

The Attempt at a Solution

Well, I really have no idea where to start. I have been reading the definition a field over and over and cannot seem to get a grip on the beginning of the proof. I believe on part b I would need to prove/disprove the described field fits the 6 (according to my notes) field axioms. Part a, I am not even sure what it is asking. Just a little help starting is all I should need. My book has nothing on this particular area and my usual resource Wikipedia, has just confused me more (http://en.wikipedia.org/wiki/Split-complex_number).

Thank you in advance,
Dave

 

[VeeEight]

For (a), take 3 arbitrary elements from your set R^2 and show that the operation defined is associative. That is, a * (b * c) = (a * b) * c, where a,b,c are members of your set (R^2, and * is the operation defined.)

For (b), to disprove it, you should find one field axiom it does not obey. If you can't find one, then go through all of the axioms.

 

[evad1089]

So, I have figured out part a. Turns out a was (dot) (colon) R² -> R² not (therefore) R² -> R.

For anyone in the same boat as me here is the solution.

Suppose (a,b), (c,d), (e,f) \in R², we will show (a,b) \bullet ((c,d)\bullet(e,f)) = ((a,b) \bullet (c,d))\bullet(e,f).
1. (a,b) \bullet ((c,d)\bullet(e,f)) = (a,b) \bullet (ce+df,cf+d3)
2. =(ace+adf+bcf+bde,acf+ade+bce+bdf)
3. ((a,b) \bullet (c,d))\bullet(e,f) = (ac+bd,ad+bc)\bullet(e,f)
4. =(ace+bdf+adf+bcf,acf+bdf+ade+bce)
5. =(ace+adf+bcf+bde,acf+ade+bce+bdf)
Since line 2 equals line 5, with a little substitution, (a,b) \bullet ((c,d)\bullet(e,f)) = ((a,b) \bullet (c,d))\bullet(e,f). Therefore, \bullet : R² -> R² is associative.

 

[evad1089]

Now I am still seriously puzzled about b. I am pretty sure that it is not a field, but it seems to fit every axiom. It is associative and commutative for addition and multiplication. And seems to be distributive for multiplication over addition. There appears to be an additive identity of (0,0), a multiplicative identity of (1,0). I see no problem with the field having a additive inverse.

Well this leaves the multiplicative inverse; I believe this might be the key to proving (R²,+,\bullet) is not a field, but as to proving this idea, I am at a loss.

 

[Mark44]

Homework Statement

On R², define the binary operators
(x,y)+(u,v)=(x+u,y+v)
(x,y)+(u,v)=(xu+yv,xv+yu)

Do you have a typo in the second line above? Otherwise you have defined addition in two ways.

The set R², along with these definitions of addition and multiplication, for the perplex numbers.
(a) Show that \cdot: R² \rightarrow R is associative.
(b) Either prove or disprove: The triple (R²,+,\cdot) is a field.

Homework Equations

Definition of a field.

The Attempt at a Solution

Well, I really have no idea where to start. I have been reading the definition a field over and over and cannot seem to get a grip on the beginning of the proof. I believe on part b I would need to prove/disprove the described field fits the 6 (according to my notes) field axioms. Part a, I am not even sure what it is asking. Just a little help starting is all I should need. My book has nothing on this particular area and my usual resource Wikipedia, has just confused me more (http://en.wikipedia.org/wiki/Split-complex_number).

Thank you in advance,
Dave

 

[evad1089]

Yes. The second one is supposed to be (x,y)\bullet(u,v)=(xu+yv,xv+yu).

 

[evad1089]

Well I figured out b.

Since one of the requirements of a field is that the multiplicative inverse is defined for all elements of the field. We can prove the triple (R²,+,\bullet) is not a field. As a side note the multiplicative identity of this non field is (1,0).

Suppose (3,3)\inR² has a multiplicative inverse, (a,b). Then (3,3)\bullet(a,b)=(1,0). Evaluating the left side we obtain, (3a+3b,3b+3a)=(1,0). Since 3a+3b=3b+3a, but 1\neq0, this is a contradiction. Therefore (3,3)\inR² does not have a multiplicative inverse. Since one of the reuirements of a field is the multiplicative inverse is defined for all members of the field, the triple (R²,+,\bullet) is not a field.

Uhhhh... well I guess good luck to the next person. Hope this helps.