Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

walters

  • one year ago

Abstract Algebra help

  • This Question is Closed
  1. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 },f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\}\] show that T is a group or not <T,0> i've shown that 0 is associative on T can u pls help on showing the identity and the inverse

  2. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    0 is a binary operation not zero

  3. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @experimentX

  4. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    use \\ in latex for new line

  5. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[f _{6}(x)=\frac{ 1 }{ x }\]} this is the lst element

  6. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[ T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 }, \\ f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\} \]

  7. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hmm ... f1 is the identity element.

  8. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let \[f _{7}(x) \in T\] be the identity with respect to 0,then \[(f _{7}0f _{1})(x)=(f _{1}0f _{7})(x)=f _{1}(x)\]

  9. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    unfortunately, the last element ... has inverse as itself.

  10. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    from there i did this \[(f _{7}0f _{1})(x)=f _{1}(x)\] and \[(f _{1} 0f _{7})(x)=f _{1}(x)\]

  11. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    on the first part i got \[f _{7}(x)=x\] but the second part i don't know how to do it

  12. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    huh?? how did you do that??

  13. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(f _{7}0f _{1})(x)=f _{1}(x)\] then \[f _{7}(f _{1}(x))=f _{1}(x)\] \[f _{7}(x)=x\]

  14. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hmm .. what is f7, where is f7 ??

  15. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i let it to be element of T because by the definition of identity we have e*a=a*e=a

  16. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    isn't |dw:1362683941681:dw|

  17. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i got that but i am not sure what if we use f7 and f2 it seems we will have another different answer

  18. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and we know that identity element is unique

  19. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    there isn't any other identity element on the group.

  20. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if i am using f7 and f2 i have |dw:1362684306098:dw| how can i simplify it until i get x as my identity

  21. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I think you are not getting the meaning|dw:1362684677725:dw| of identity

  22. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1362684807109:dw|

  23. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hell ... the page refreshed and my work is gone.

  24. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1362685396384:dw|

  25. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    looks like i made some mistake.

  26. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so if is like this how wat am i gonna say about the inverse

  27. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    actually i mean how to write them

  28. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    could you check those stuffs?

  29. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    if inverse exist, then it must be unique.

  30. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the inverse not always is unique but the identity must always be unique

  31. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the identity is unique on this case ... f(x) = x is the only identity element.

  32. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the inverse is given by a*b=b*a=e

  33. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    where e is the identity we have found

  34. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think this is not a group

  35. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    why do you think it's not a group?

  36. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the only way to get the inverse of a certain element u must multiply two elements and this does not clearly tell us the inverse of a function.As we know that for every function wen we use the operation combining the function and its inverse we must get the identity

  37. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    here, the operation is not multiplication ... but i think there is no unique inverser. But I am not sure ... let me do this seriously on copy.

  38. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok

  39. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I am getting the same result as before have you verified associativity??

  40. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes it holds

  41. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think wat u did is similar wen trying to show that 0 is a binary operation ,because wat u have as the inverse i have them at my fisrt condition that one of checking indeed the our operation hold

  42. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    what a fool i've been ... f3, f5, f6 <-- they all are self inverse.

  43. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this is a group.

  44. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can u show them pls

  45. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    assuming associativity holds for all operations define your idenity by \( f_1(x) = x \) ... in any case you won't get \( f_x f_y = f_x \) if y \( \neq 1\)

  46. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    for every operation this is closed. earliser I showed, f2 and f4 are inverses of each other.|dw:1362690024527:dw|

  47. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1362690149777:dw| these element are inverses of themselves.

  48. walters
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok now i see it is a group

  49. experimentX
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    most likely yes ... as always I am not sure.

  50. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.