Question

let S denote the set of all ordered pairs of real numbers. For any pair alpha=(asub1,asub2),beta=(bsub1,bsub2) of element of S define

alpha+beta=(asub1+bsub1,asub2+bsub2)
alpha*beta=(asub1bsub1+3asub2bsub2,asub2bsub1+asub1bsub2)
determine whether or not S is a field together with these operations for addition and multiplication.If so, prove that all the conditions in the definition of a field are satisfied,If not, show that at least one condition does not hold.

Answer 1

It's been a loooooooooong time since I did this kind of stuff :) but from a quick look, my first thought was that it wasn't clear that there was a multiplicative identity.....

... but then I realized, .... I think that (1,0) will work as a multiplicative identity?

Like I said, it's been a while. lol :)

Additive inverse should be easy enough.

Just need to figure out multiplicative inverse... and then check associativity, commutativity, and distribution, right? I didn't check those, but that should be straight-forward just churning through the formulas.

Answer 2

Multiplicative inverse is not looking good...... or maybe I'm just tooooo rusty at this, lol.

Answer 3

it's really too hard isn't it? that's why i need someones help =)

Answer 4

Just to be clear what we have here:

\(\Large \alpha=(a_{1},a_{2}),~\beta=(b_{1},b_{2})\)
\(\Large \alpha+\beta=(a_{1}+b_{1},~a_{2}+b_{2})\)
\(\Large \alpha\cdot\beta=(a_{1}b_{1}+3a_{2}b_{2},~a_{2}b_{1}+a_{1}b_{2})\)

Right?

Answer 5

yup

Answer 6

Have you checked multiplicative associativity, distribution, and commutativity?

I ask only because, if one of those falls apart, then you know it isn't a field and can just come up with the counter-example.

I was trying to figure out the inverse by using a couple of points, to see if there was some intuitively obvious inverse.... but it was UG.LY. lol so maybe it isn't a field at all?

Answer 7

Thank you very much for helping me for this =))))

Answer 8

Sorry I'm not more help. I just tried multiplicative commutativity with the points:
(1,2) (0,3) (-1,2)

and unfortunately, it works... LOL! Now, that does NOT prove commutativity, of course, but I was hoping it would NOT work and then we would know that it's NOT a field.

I feel like the multiplication is where it would have to fall apart, if it isn't a field, because the product rule seems more random than the addition rule. But I guess if all the properties can be proved in general, then it all comes down to that pesky multiplicative identity......... <scratching head>

Answer 9

that's what i also felt the first time i look on that problem

Answer 10

SORRY, I mean I tried DISTRIBUTION, not commutativity.

Answer 11

ok, IF I did it right, when I tried multiplicative associativity with:
a=(1,2), b=(0,3) and c=(-1,4)

it did NOT work. If that's true, that will blow this baby's chances of getting to be a field. But double check, maybe with 3 other points. I didn't use anything simple like (0,0) or (1,1) in case that would work but was an anomaly.

Answer 12

Easier points:
a=(0,-1) b=(1,1) c=(2,0)

\(\Large c(ab)\neq(ca)b\)

Again.... if I did it correctly..... lol. so not a field.

Answer 13

you're so persevered on answereing my questions and i thank you for that =]