anonymous
  • anonymous
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.
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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DebbieG
  • DebbieG
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.
DebbieG
  • DebbieG
Multiplicative inverse is not looking good...... or maybe I'm just tooooo rusty at this, lol.
anonymous
  • anonymous
it's really too hard isn't it? that's why i need someones help =)

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DebbieG
  • DebbieG
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?
anonymous
  • anonymous
yup
DebbieG
  • DebbieG
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?
anonymous
  • anonymous
Thank you very much for helping me for this =))))
DebbieG
  • DebbieG
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.........
anonymous
  • anonymous
that's what i also felt the first time i look on that problem
DebbieG
  • DebbieG
SORRY, I mean I tried DISTRIBUTION, not commutativity.
DebbieG
  • DebbieG
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.
DebbieG
  • DebbieG
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.
anonymous
  • anonymous
you're so persevered on answereing my questions and i thank you for that =]

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