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walters

  • 2 years ago

Abstract Algebra help

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  1. walters
    • 2 years ago
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    \[T=\left\{ f _{1} (x)=x,f _{2}(x)=\frac{ 1 }{ 1-x },f _{3}(x)\frac{ x }{ x-1 },f _{4}(x)=\frac{ x-1 }{ x },f _{5}(x)=1-x,f _{6}(x)=\frac{ 1 }{ x }\right\}\] 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
    • 2 years ago
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    0 is a binary operation not zero

  3. walters
    • 2 years ago
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    @experimentX

  4. experimentX
    • 2 years ago
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    use \\ in latex for new line

  5. walters
    • 2 years ago
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    \[f _{6}(x)=\frac{ 1 }{ x }\]} this is the lst element

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

  7. experimentX
    • 2 years ago
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    hmm ... f1 is the identity element.

  8. walters
    • 2 years ago
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    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
    • 2 years ago
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    unfortunately, the last element ... has inverse as itself.

  10. walters
    • 2 years ago
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    from there i did this \[(f _{7}0f _{1})(x)=f _{1}(x)\] and \[(f _{1} 0f _{7})(x)=f _{1}(x)\]

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

  12. experimentX
    • 2 years ago
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    huh?? how did you do that??

  13. walters
    • 2 years ago
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    \[(f _{7}0f _{1})(x)=f _{1}(x)\] then \[f _{7}(f _{1}(x))=f _{1}(x)\] \[f _{7}(x)=x\]

  14. experimentX
    • 2 years ago
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    hmm .. what is f7, where is f7 ??

  15. walters
    • 2 years ago
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    i let it to be element of T because by the definition of identity we have e*a=a*e=a

  16. experimentX
    • 2 years ago
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    isn't |dw:1362683941681:dw|

  17. walters
    • 2 years ago
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    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
    • 2 years ago
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    and we know that identity element is unique

  19. experimentX
    • 2 years ago
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    there isn't any other identity element on the group.

  20. walters
    • 2 years ago
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    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
    • 2 years ago
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    I think you are not getting the meaning|dw:1362684677725:dw| of identity

  22. experimentX
    • 2 years ago
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    |dw:1362684807109:dw|

  23. experimentX
    • 2 years ago
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    hell ... the page refreshed and my work is gone.

  24. experimentX
    • 2 years ago
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    |dw:1362685396384:dw|

  25. experimentX
    • 2 years ago
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    looks like i made some mistake.

  26. walters
    • 2 years ago
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    so if is like this how wat am i gonna say about the inverse

  27. walters
    • 2 years ago
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    actually i mean how to write them

  28. experimentX
    • 2 years ago
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    could you check those stuffs?

  29. experimentX
    • 2 years ago
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    if inverse exist, then it must be unique.

  30. walters
    • 2 years ago
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    the inverse not always is unique but the identity must always be unique

  31. experimentX
    • 2 years ago
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    the identity is unique on this case ... f(x) = x is the only identity element.

  32. walters
    • 2 years ago
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    the inverse is given by a*b=b*a=e

  33. walters
    • 2 years ago
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    where e is the identity we have found

  34. walters
    • 2 years ago
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    i think this is not a group

  35. experimentX
    • 2 years ago
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    why do you think it's not a group?

  36. walters
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    ok

  39. experimentX
    • 2 years ago
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    I am getting the same result as before have you verified associativity??

  40. walters
    • 2 years ago
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    yes it holds

  41. walters
    • 2 years ago
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    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
    • 2 years ago
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    what a fool i've been ... f3, f5, f6 <-- they all are self inverse.

  43. experimentX
    • 2 years ago
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    this is a group.

  44. walters
    • 2 years ago
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    can u show them pls

  45. experimentX
    • 2 years ago
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    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
    • 2 years ago
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    for every operation this is closed. earliser I showed, f2 and f4 are inverses of each other.|dw:1362690024527:dw|

  47. experimentX
    • 2 years ago
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    |dw:1362690149777:dw| these element are inverses of themselves.

  48. walters
    • 2 years ago
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    ok now i see it is a group

  49. experimentX
    • 2 years ago
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    most likely yes ... as always I am not sure.

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