Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

walters

Let G be a group such that (a*b)^2=a^2*b^2 can pls show the associativity

  • one year ago
  • one year ago

  • This Question is Closed
  1. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1362323673416:dw|

    • one year ago
  2. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    i am not sure

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

    Isn't the associativity of the binary operation one of the requirements fore being a group?

    • one year ago
  4. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes it is

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

    Then why do you have to prove it when G is already a group? :/

    • one year ago
  6. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    actually my question was about showing that the group is abelian so i am not sure whether should i show commutativity only or i have to show them all

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

    :) Given that G is a group, to show that it's abelian, all that's necessary is to show commutativity. Is it given in the question that G is a group?

    • one year ago
  8. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

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

    Then it's easy to show commutativity :) I'm certain you've already done it XD

    • one year ago
  10. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes ok now i understand the question

    • one year ago
  11. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    wait..

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

    NO rush :) I'm going to savour this question because I love group theory :D

    • one year ago
  13. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    let the mapping \[\tau _{ab}\] for a,b element R ,maps the real into the reals by the rule \[\tau _{ab}:x rightarrowa x+b\] and g={\[\tau _{ab}\]:a\[\neq\]=0} be a group under the composition of mappings Prove that N={\[\tau _{1b}\] element G} is a subgroup

    • one year ago
  14. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1362324922660:dw|

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

    Hang on, I'm going to need a while to digest that :D

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

    Okay, so, you know the subgroup test?

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

    A subset H of a group G is a subgroup of G if for any elements a and b in H, ab⁻¹ is in H.

    • one year ago
  18. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    A subset H Group is a subgroup iff 1 the identity element in G is also in H 2 if h1 ,h2 element in G is also in G 3 if h element H,then h^-1 element H

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

    Oh. No shortcuts, eh? :D Okay, first of all, is the identity element in N? The identity element is the one that would send x to x, IE the identity function :D

    • one year ago
  20. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    i am not sure

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

    Okay. Maybe we should translate the definition of N from gibberish to English :D sloowly :) \[\large N= \left\{ \tau_{1b}\ \right\}\] It means all the mappings that send x to ax + b where a = 1 So, in English :D (not really English, but in simpler terms) \[\large N = \left\{\tau|\tau \in G, \tau(x)=x+b, \ \ b\in \mathbb{R} \right\}\]

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

    So, is the identity function... \[\large id(x)=x\] in N?

    • one year ago
  23. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes,

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

    So that's the identity element down :D Now, let's try the second necessity, in that given any two elements of the set N, if we compose them, the composition is still in N. Any idea how to proceed?

    • one year ago
  25. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    i think we have to show that it is the group under the binary operation |dw:1362326767652:dw|

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

    Fair enough :) Let's take the set G, first., shall we?

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

    Let there be two functions f and g such that f and g are in G. In other words f(x) = ax + b g(x) = cx + d Where a,b,c, and d are real numbers Catch me so far?

    • one year ago
  28. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

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

    Well, consider fog(x) fog(x) = a(cx + d) + b = acx + ad + b and is this still in G?

    • one year ago
  30. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

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

    So we've shown closure. The rest is easy, right? :D

    • one year ago
  32. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    are u not suppose to use |dw:1362327123975:dw|

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

    yeah, but that's just notation :D I'll leave that to you :> \[\huge f(x) = \tau_{ab} \ \ \ \ g(x) = \tau_{cd}\]

    • one year ago
  34. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

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

    So, back to the question about N being a subgroup?

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

    We've shown the identity, and now we need to show closure.

    • one year ago
  37. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i see it

    • one year ago
  38. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    @experimentX

    • one year ago
  39. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    for that one of subgroup

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

    I am not so good with abstract algebra either ... but let's see it.

    • one year ago
  41. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    this is what i did since G is a group ,let 1,b element of real number then |dw:1362353281934:dw| then i don't know wat to do next

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

    hmm ... if G is a group then why do you have to show associativity? isn't it associative by hypothesis?

    • one year ago
  43. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i see that since it is the group it is not neccessary one thing that i can't do it is that of showing N is a subgroup

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

    when is N, normalizer?

    • one year ago
  45. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    is not N for natural numbers?

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

    sorry, is N normalizer?

    • one year ago
  47. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes it is

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

    given that a, b in G, then (a*b)^2=a^2*b^2 looks like we could use this http://en.wikipedia.org/wiki/Normal_subgroup

    • one year ago
  49. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    there are 2 different questions (given that a, b in G, then (a*b)^2=a^2*b^2) and that onee of subgroup

    • one year ago
  50. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    let me retype my questions

    • one year ago
  51. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Let G be a group such that (a*b)^2=a^2*b^2 for all a,b element G show that <G,*> is abelian

    • one year ago
  52. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    2 let the mapping \[\tau _{ab}\] for a,b element of R ,maps the reals into the reals by the rule \[\tau _{ab}: x \rightarrow ax +b\] and G ={\[\tau _{ab}:a \neq0\]} be a group under the composition of mappings Proe that N={\[\tau _{1b}\ G]} s a subgroup of G

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

    |dw:1362355116918:dw| use associativity.

    • one year ago
  54. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    is already a group i only have to show that is abelian so to show comutativity only

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

    for any two element, the above picture shows commutativity.

    • one year ago
  56. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1362355288828:dw|

    • one year ago
  57. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    but this is confusing because the * operation does not mean is multiplication wat if this operation is under (-) LHS will not be equal to RHS

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

    yes it's not multiplication ... but associativity and inverse are properties of group, we are only allowed to exploit that.

    • one year ago
  59. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    SO IS IT GOING TO BE|dw:1362355812287:dw|

    • one year ago
  60. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    SO IF I CAN PUT THE SQUARE ROOT both side Lhs is it going to be equal to RHS

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

    No, let assume \(a\) , \(b \in G\) \[ (a*b)^2 = (a*b)*(a*b)\\ (a*b)*(a*b) = a^2*b^2 = (a*a)*(b*b)\\ a^{-1}*(a*b)*(a*b) = a^2*b^2 = a^{-1}*(a*a)*(b*b)\\ (a^{-1}*a)*b*(a*b) = a^2*b^2 = (a^{-1}*a)*a*(b*b)\\ (e)*b*(a*b) = a^2*b^2 = (e)*a*(b*b) \\ b*(a*b) = a*(b*b)\\ b*(a*b)*b^{-1} = a*(b*b)*b^{-1}\\ b*a*(b*b^{-1}) = a*b*(b*b^{-1})\\ b*a = a*b\] Since two arbitrary elements of G are commutative, the group G is albelian.

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

    http://planetmath.org/Normalizer.html http://planetmath.org/NormalSubgroup.html Let N be normalizer of G i,e. \( H \subseteq G \) Since it's abelian \( aNa^{-1} = aa^{-1}N = eN = N\; \forall a \in G \) Hence N is normal subgroup. However I am not sure ,,, I am a physicist, abstract algebra is one of my weakest point.

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

    Oh... I missed it :( Why did I have to fall asleep?! Hey, @experimentX I like abstract algebra and all, but does this group stuff have any applications, say, in Physics?

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

    yeah ... in QM it is used to check symmetry of operators. And there is Lie algebra supposedly used in string theories ... but i haven't advanced to that point.

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

    LOL I thought abstract algebra was just invented when Mathematicians got bored from Calculus XD But yeah... Abstract Algebra = My Favourite branch of Maths :)

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

    yeah ... i admit, it's elegant (although when i first encountered, i couldn't comprehend) ... there is a free lecture series on Harvard Extension ... but i hardly get time.

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

    and of course ... just viewing lecture pretty useless without doing exercise (perhaps pratice is the best way of understanding)

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

    I hope this guy sticks around :) I'm itchin' for more abstract algebra :D

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

    I hope i'll get enough time to do this http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra

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

    Linear Algebra comes before Abstract Algebra?

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

    also @terenzreignz have you check this site? http://math.stackexchange.com/ http://math.stackexchange.com/questions/tagged/abstract-algebra yeah, it's easy that way ... since you are oriented with matrices since early school. http://math.stackexchange.com/questions/tagged/abstract-algebra

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

    Don't vector spaces fall under Linear Algebra? I don't think I'd have understood it nearly as well had I not read up on Groups first...

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

    i think linear algebra is subset of Abstract algebra ... yeah vector spaces fall under linear algebra ... (but linear algebra is a bit broad) ... it's both applied and abstract.

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

    i was reading this book ... http://www.amazon.com/Linear-Algebra-Undergraduate-Texts-Mathematics/dp/0387964126 it had (little part) some group theory inside it. I think linear algebra falls under Abstract algebra ... since both they are linear algebra ... just add the condition of linearity, and vector spaces ... they are linear by their definition. And matrices are vector spaces of which has max application on applied sciences.

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

    *since both they are algebra

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

    Vector spaces seem to be between Groups and Rings, and yet, above rings, since they involve Fields... It's a complicated relationship :D The Vector Space is an abelian group, it doesn't have multiplication defined on it (unless it be an inner product space, and even then, the products aren't vectors), but it does have scalar multiplication, which distributes in the same manner as in rings, but vector spaces involve fields... Best not to think about this stuff :D

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

    I haven't seen that way since I am still novice on abstract algebra ... looks like youare right... and i think inner product is generalization of scalar product.

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

    Just appreciate the beauty of algebra... it's an awesome break from the rigors of calculus :D

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

    yeah ... but somehow i enjoy and work more on calculus. Still i would like to learn abstract algebra and Real Analysis (in dept).

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

    Ahh... Real Analysis... Just when you thought Calculus couldn't possibly get worse XD

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

    When calculus gets worse ... it's Real Analysis, When real analysis goes worse ... I am hopeless ... i guess set theory then.

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

    I thought Complex Analysis comes after Real Analysis?

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

    The complex analysis I am doing is mostly applied. I am well oriented with Real Analysis (nearly up to Lebesgue integral)... but the methods of real analysis are very hard.

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

    you can easily think of solution on calculus ... you begin at something and use these particular theorems ... and arrive at result. this is pretty straight forward. But Real Analysis is different thing ... nearly all theorem requires a long paragraph of explanation. and it has to be locally correct. Real Analysis lies somewhere between set theory and calculus.

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

    Now I'm scared :D

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

    From mathematical perspective elementary calculus is a child play.

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

    to sum up ... in real analysis, it might be easy to understand a theorem ... but hard to solve problem.

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

    and also note that ,, when calculus get's worse ... we often use techniques of Real Analysis to solve problem.

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

    Hmmm... that makes sense... but does little to console me :D Now I'm rethinking my purpose in life XD

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

    haha ... that is the ultimate question ... if you trust WolframAlpha then it calculates as 42

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

    http://www.wolframalpha.com/input/?i=the+ultimate+meaning+of+life%2C+universe+and+everything

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

    Well, if they say so :D

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

    well ... cya around :)

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

    Yeap :) Thanks for... uhh.... stuff :D

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

    yw!!

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

    We kind of used this thread as a chat-medium.... sorry about that. What do you need help with?

    • one year ago
  97. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    by subgroup i know that i must show that Let H <= G(group) H=G iff H is not empty ie e elment H 2 gh^-1 element of H for every g,h element of H

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

    That's the subgroup test :D

    • one year ago
  99. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    so this means i must first check the identity of mappings and the identity should be in N

    • one year ago
  100. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    to conclude that N is not enpty

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

    You've thus concluded that N is not empty, right? To recap, N is the set of all functions that send x to x+b, for some real number b, right?

    • one year ago
  102. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    but should i not suppose to get the exact value as that one in mappings

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

    Exact value of what?

    • one year ago
  104. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    identity element in G

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

    I guess so. Let's let e(x) be the identity element in G, shall we?

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

    Now, by definition, for any tau in G \[\huge \tau\circ e(x)=\tau(x)=e\circ\tau(x) \]

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

    Are you following me so far? :)

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

    Now, tau always takes the form \[\large \tau(x)=ax+b \ \ \ \ a,b\in\mathbb{R}\] Following the first equation, we have \[\huge \tau(e(x))=a\cdot e(x)+b=ax+b=\tau(x)\]

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

    And it just goes to show that \[\huge e(x)=x\] So the identity element, is, after all, the identity function :D

    • one year ago
  110. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    i've let \[\tau _{pg}\] element of G be identity with respect to 0 opereration yes i also got x on my rough work

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

    So, with that done, clearly e(x) is in N, meaning N is not empty :) If you're still not convinced, let's recap N the 'proper Mathematical way' \[\huge N=\left\{\tau(x) \ | \ \tau(x)=x+b , \ \ b\in\mathbb{R}\right\}\]

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

    And clearly, the identity map is in N, just take b=0 Convinced now that the identity is in N? :)

    • one year ago
  113. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    ok now ths become more interesting how am i gonna answer this question should i start by showing that it is a group "especially by showing the identity of G" or should i start by answering the subgroup 's question

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

    Use the subgroup test. Show it's not empty (we've already done that) and show that for any two elements in N \[\huge \alpha,\beta\in N\]\[\huge \alpha\circ\beta^{-1}\in N\]

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

    This was what you told me earlier, right?

    • one year ago
  116. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    am i going to use \[\tau _{cd}\] and \[\tau _{ef}\] because of my condition is of type |dw:1362487393261:dw|

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

    You can, but we're dealing with elements of N, which means the coefficient of x is 1.

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

    Remember, again, the original definition of N \[\huge N=\left\{ \tau_{1b} \ \ | \ \ b\in\mathbb{R} \right\}\]

    • one year ago
  119. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    so it will look like this let |dw:1362487871485:dw| but thre is something i don't understand on the subscripts how am i going to writeit because i have cd and ef (am i gonna have c and e equal to 1 or what)?

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

    Just write \[\huge \tau_{1c} \ \ \ and \ \ \ \tau_{1d}\]

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

    Are you still there? Let's get back to the subgroup test... Let \[\huge \alpha,\beta\in N\]

    • one year ago
  122. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i am still trying to work it aside

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

    Then \[\huge \alpha(x)=x+a\]\[\huge\beta(x)=x+b\]for some\[\huge a,b\in \mathbb{R} \]

    • one year ago
  124. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    so wat will be this |dw:1362488547392:dw|

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

    No... Okay, let's look for the inverse of beta... \[\huge \beta^{-1}(x)\]

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

    By definition of the inverse... \[\huge \beta\circ\beta^{-1}(x)=e(x)\]

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

    So... \[\huge \beta(\beta^{-1}(x))=\beta^{-1}(x)+b=x\]

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

    Which can only mean \[\huge \beta^{-1}(x)=x-b\] So far so good?

    • one year ago
  129. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    what ?

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

    We've found the value of the inverse of beta.

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

    So, let's calculate \[\huge \alpha\circ\beta^{-1}(x)=\alpha(\beta^{-1}(x))\]

    • one year ago
  132. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    but emenyana(wait) is |dw:1362489338312:dw|

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

    \[\huge =\beta^{-1}(x)+a\]\[\huge =x-b+a\]

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

    No, see, we're not taking the multiplicative inverse, we're taking the inverse of the function. Remember that the binary operation is composition, not muliplication

    • one year ago
  135. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i see

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

    so, \[\huge \alpha\circ\beta^{-1}(x)=x+(a-b)\]And since a and b are both real, so is a-b And therefore \[\huge \alpha\circ\beta^{-1}\in N\] And further implying \[\huge N\le G\] ∎ That's it, and thanks, it was fun D

    • one year ago
  137. walters
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks very much ,see u on the other question i am still busy with it i will consult wen i get stuck it is the proof

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

    Ok :)

    • one year ago
    • Attachments:

See more questions >>>

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.