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Astract Algebra Group (should be easy)

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]G = \{a + b\sqrt{2} | a,b \in \mathbb{Q}\}[/tex]

    (a)Show that G is a group under addition

    (b)Show that G - {0} is a group undermultiplication.

    3. The attempt at a solution
    I think this problem should be easy if I could get on the right start.

    (a) All I know is to prove that G is a group is to show it is associative, has an identity, and has inverses under addition. The problem is I have no clue where to start since I have never seen this before. I am guessing I should let [tex]a = \frac{m}{n}[/tex] and [tex]b = \frac{k}{l}[/tex]. Now how do I go about showing this is associative? Don't I need an extra element, say c, to show that it is associative? If so, can I just arbitrarily create one?
    Thank you and any help is greatly appreciated.
     
  2. jcsd
  3. Sep 24, 2009 #2

    Hurkyl

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    :confused:

    G has lots of elements. Some examples are:

    . [tex]32[/tex]

    . [tex]-\frac{3}{8}\sqrt{2}[/tex]

    . [tex]\frac{1639}{65536} + \frac{4916}{257}\sqrt{2}[/tex]

    . [tex]\sqrt{8}[/tex]

    . [tex]\frac{32 - 768 \sqrt{2}}{-\frac{7}{5} + 3 \sqrt{2}}[/tex]

    (These last two take some extra work to show they are in G. All of the other examples are essentially nothing more than pattern matching)
     
  4. Sep 24, 2009 #3
    I understand that those are all elements of G (except maybe the 4th one). But how do I even begin to show G is a group under addition? I thought the first thing to show is associativity in G, but maybe I have no clue what I am talking about because I am really confused right now =[.
     
  5. Sep 24, 2009 #4

    VietDao29

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    You must first show that + is a binary operation, i.e, to show that:

    [tex]x_1, x_2 \in G \Rightarrow x_1 + x_2 \in G[/tex]

    (or in other word, G is closed under addition)

    ----------------------------------

    I'll do this as a simple example for you:

    Note that [tex]G = \{ a + b\sqrt{2} | a, b \in \mathbb{Q} \}[/tex]. That means:
    1. If there's some element x that can be expressed as [tex]x = a + b\sqrt{2}[/tex], where a, b are rational numbers, then [tex]x \in G[/tex]. (Note 1)
    2. For every element [tex]x \in G[/tex], there exists 2 rational numbers, namely, a and b, such that: [tex]x = a + b\sqrt{2}[/tex]. (Note 2)

    So:

    [tex]\forall x_1, x_2 \in G, \exists a_1, b_1, a_2, b_2 \in \mathbb{Q} : \left\{ \begin{array}{l} x_1 = a_1 + b_1 \sqrt{2} \\ x_2 = a_2 + b_2 \sqrt{2} \end{array} \right[/tex] (from Note 2)

    [tex]\Rightarrow x_1 + x_2 = (a_1 + b_1 \sqrt{2}) + (a_2 + b_2 \sqrt{2}) = (a_1 + a_2) + (b_1 + b_2) \sqrt{2}[/tex]

    Since [tex]a_1, b_1, a_2, b_2 \in \mathbb{Q}[/tex], we also have: [tex]a_1 + a_2, b_1 + b_2 \in \mathbb{Q}[/tex]

    And from Note 1, we have: [tex]x_1 + x_2 \in G[/tex].

    So, G is closed under addition.

    (You can skip the 'from (Note 1)' and 'from (Note 2)' parts, I just put it there to help you see things more clearly).

    ----------------------

    And, to prove associativity property, you'll need 3 elements.

    Other requirements can be proven in somewhat the same way. Let's see if you can handle it on your own. :)
     
  6. Sep 24, 2009 #5
    Thanks so much VietDao, that was such a clutch response in helping me get started. I understand to show associativity I must create [tex]x_3[/tex] the same way you created [tex]x_1[/tex] and [tex]x_2[/tex]. Then I must show in a similar fasion that [tex]x_1 + (x_2 + x_3) = (x_1 + x_2) + x_3[/tex] for all [tex] x_1, x_2, x_3 \in G[/tex] So here is what I just did on a sheet of scratch paper which is basically a carbon copy of what you did with the extra x_3 in there so I am not sure if I showed it correctly:

    [tex]\forall x_1, x_2, x_3 \in G, \exists a_1, b_1, a_2, b_2, a_3, b_3 \in \mathbb{Q} : \left\{ \begin{array}{l} x_1 = a_1 + b_1 \sqrt{2} \\ x_2 = a_2 + b_2 \sqrt{2} \\ x_3 = a_3 + b_3 \sqrt{2} \end{array}\right[/tex]

    [tex]\Rightarrow x_1 + (x_2 + x_3) = a_1 + b_1 \sqrt{2} + (a_2 + b_2 \sqrt{2} + a_3 + b_3\sqrt{2}) = a_1 + b_1 \sqrt{2} + (a_2 + a_3 + (b_2 + b_3)\sqrt{2}) = a_1 + a_2 + a_3 + (b_1 + b_2 + b_3)\sqrt{2}[/tex] (Starting here I do not know if I am doing this right I think I skipped some logic here at the end but I do not understand how to show associativity from my work up to this point) Thus, [tex]x_1, x_2, x_3 \in G[/tex] So, [tex](x_1 + x_2) + x_3 \in G[/tex].
     
  7. Sep 24, 2009 #6

    VietDao29

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    When proving something, you cannot just do it without any aims. What is the goal of your proof? This goal will make one proof diferes so much from the other. The method used may be the same, but the flow, the steps may be completely different.

    In my example, I am trying to show that: x1 + x2 is in G, so I'll group it like that. But your goal is not showing that x1 + x2 + x3 is in G!! So grouping like what you've done is of no use.

    You are trying to show that:

    (x1 + x2) + x3 = x1 + (x2 + x3) right?

    Can you expressed (x1 + x2) + x3, and x1 + (x2 + x3) in terms of a1, a2, a3; and b1, b2, b3? Then let's try to do some manipulations to show that they are equal.
     
  8. Sep 24, 2009 #7
    And what's the inverse element for "x" of (a+b*sqrt(2))??
     
  9. Sep 24, 2009 #8
    (following) this is 1/(a+b*sqrt(2)), but you must prove it is contained by G.
     
  10. Sep 24, 2009 #9
    Ok I will try again tonight when I have time after class. Thank you so much VietDao and Penguin.
     
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