Astract Algebra Group (should be easy)

[DEMJ]

Homework Statement

Let G = \{a + b\sqrt{2} | a,b \in \mathbb{Q}\}

(a)Show that G is a group under addition

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

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 a = \frac{m}{n} and b = \frac{k}{l}. 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.

 

[Hurkyl]

G has lots of elements. Some examples are:

. 32

. -\frac{3}{8}\sqrt{2}

. \frac{1639}{65536} + \frac{4916}{257}\sqrt{2}

. \sqrt{8}

. \frac{32 - 768 \sqrt{2}}{-\frac{7}{5} + 3 \sqrt{2}}

(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)

 

[DEMJ]

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 =[.

 

[VietDao29]

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 =[.

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

x_1, x_2 \in G \Rightarrow x_1 + x_2 \in G

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

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

Note that G = \{ a + b\sqrt{2} | a, b \in \mathbb{Q} \}. That means:

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

So:

\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 (from Note 2)

\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}

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

And from Note 1, we have: x_1 + x_2 \in G.

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).

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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. :)

 

[DEMJ]

Thanks so much VietDao, that was such a clutch response in helping me get started. I understand to show associativity I must create x_3 the same way you created x_1 and x_2. Then I must show in a similar fasion that x_1 + (x_2 + x_3) = (x_1 + x_2) + x_3 for all x_1, x_2, x_3 \in G 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:

\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

\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} (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, x_1, x_2, x_3 \in G So, (x_1 + x_2) + x_3 \in G.

 

[VietDao29]

Thanks so much VietDao, that was such a clutch response in helping me get started. I understand to show associativity I must create x_3 the same way you created x_1 and x_2. Then I must show in a similar fasion that x_1 + (x_2 + x_3) = (x_1 + x_2) + x_3 for all x_1, x_2, x_3 \in G 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:

\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

\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} (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, x_1, x_2, x_3 \in G So, (x_1 + x_2) + x_3 \in G.

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.

 

[penguin007]

And what's the inverse element for "x" of (a+b*sqrt(2))??

 

[penguin007]

(following) this is 1/(a+b*sqrt(2)), but you must prove it is contained by G.

 

[DEMJ]

Ok I will try again tonight when I have time after class. Thank you so much VietDao and Penguin.