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Showing a set is a Group

  • #1

Homework Statement



Show that [tex] G = \left\{x | -1 < x < 1 \right\} [/tex] with the relation

[tex] a\ast b = \frac {a+b}{ab +1}[/tex]

Homework Equations



1) Definition of a group.

A group is a set G equipped with an operation * and a special element e in G, callled the identity, such that

i) for every a,b,c in G
[tex]a\ast( b \ast c) = (a \ast b})\ast c [/tex];
ii) [tex]e\ast a =a [/tex] for all a in G
iii) for all a in G, there is a' in G with [tex] a' \ast a =e[/tex]


2) Definition of a group number 2

A group is a nonempty set G equipped with an operation that satisfies the following axioms:

1. if a,b are in , then [tex] a \ast b[/tex] is in G
2.for every a,b,c in G
[tex]a\ast( b \ast c) = (a \ast b})\ast c [/tex];
3. [tex]e\ast a =a [/tex] for all a in G
4. for all a in G, there is a' in G with [tex] a' \ast a =e[/tex]


The Attempt at a Solution




I already showed that G is a group using the first definition of a group but the first axiom of the second definition is difficult to show.

The second definition is a stronger definition which leads me to think that the first definition is not correct.

Which of these definitions is the correct one ?

Also does anyone know any tricks for showing that
[tex] -1 < \frac {a+b}{ab +1} < 1 [/tex] for a,b in G ?

Basically, how to satisfy axiom 1 of the second definition.
 
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Answers and Replies

  • #2
Dick
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You could take a direct approach using calculus. Define f(b)=(a+b)/(1+ab). Now look at f'(b). And I think the first definition just assumes ab is an element of G without stating it explicitly as an axiom. Of course, you need closure.
 
  • #3
You could take a direct approach using calculus. Define f(b)=(a+b)/(1+ab). Now look at f'(b).
I thought of that but I figured this was an algebra course.
Besides, all I would get from the derievative of f'(b) is the intervals of increase and decrease, no ?

And I think the first definition just assumes ab is an element of G without stating it explicitly as an axiom. Of course, you need closure.
The author did a bad job with that one. It certainly threw me off!
 
  • #4
Dick
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I thought of that but I figured this was an algebra course.
Besides, all I would get from the derievative of f'(b) is the intervals of increase and decrease, no ?
Yes, you get intervals of increase and decrease. You want to use that to find the extrema.
 
  • #5
Yes, you get intervals of increase and decrease. You want to use that to find the extrema.

Hahaha...I forgot all my calculus. I guess another issue would be to show that a*b doesn't take on it's end points also, i.e -1 and 1.

If I find and a minima and maxima on the interval it won't tell me whether the function attains the value; which it obviously does not.

Do you know any other elegant solutions using only algebra ?
 
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  • #6
Dick
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Hahaha...I forgot all my calculus. I guess another issue would be to show that a*b doesn't take on it's end points also, i.e -1 and 1.

If I find and a minima and maxima on the interval it won't tell me whether the function attains the value; which is obviously does not.

Do you know any other elegant solutions using only algebra ?
I'm sure there is some trick but nothing immediately comes to mind, though there certainly is one. The argument using calculus is pretty simple. I'd use a straightforward mechanical way of answering before spending huge amounts of time looking for a trick.
 
  • #7
I'm sure there is some trick but nothing immediately comes to mind, though there certainly is one. The argument using calculus is pretty simple. I'd use a straightforward mechanical way of answering before spending huge amounts of time looking for a trick.
I already handed in the problem set. This question was the only one I couldn't do completely.

I didn't want to use calculus at all since the course does not require, or suppose any knowledge of calculus. I didn't think my work would be credited.

Also, I cunningly used the first definition(that's the acutally one from the course textbook) to show it was a group since I couldn't find a pure algebraic solution.

However, my conscience bothers me.:rolleyes:
 
  • #8
epenguin
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Also does anyone know any tricks for showing that
[tex] -1 < \frac {a+b}{ab +1} < 1 [/tex] for a,b in G ?
I wouldn't call it a trick, but since a, b both > -1, the denominator is positive. (Just to feel safer though actually it would make no difference if it wasn't). So you can multiply by it, and get

- (ab + 1) < a + b < ab + 1

and conclude your inequalities will be true if those above are true.

The second inequality will be true if

a + b - ab < 1 ineq. (1)

To prove that it is, start with

a + (1 - a) = 1

The bracket is positive by postulates about a.

If you multiply this bracket by b to form

a + b(1 - a)

then you have diminished the whole if b is negative , and you've also diminished if b is positive because b < 1 , so you have

a + b(1 - a) < a + (1 - a) = 1

so have shown ineq. 1.


You may need to satisfy yourself this works whatever the sign of a, b, and you could even write it better than I have.

You can also do it sketching rectangles inside squares of side 1 which would be useful reinforcement and convincing. That maybe is the trick.

Did you do the other parts of this? I wondered briefly about the identity element. It doesn't seem to me there is one that fits the <1 condition. ≤ 1 I think would do.

Anyone?
 
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  • #9
That makes a lot of sense. I will try out this method.

I don't think I will need to draw anything.

Your method works but I will have to prove every inequality I use.

I was hoping for trick like AM-GM or cauchy schwartz but perhaps I am been to picky XD.
 
  • #10
Dick
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Did you do the other parts of this? I wondered briefly about the identity element. It doesn't seem to me there is one that fits the <1 condition. ≤ 1 I think would do.

Anyone?
0 is the identity, isn't it?
 
  • #11
Yeah it is.

Also the inverse of a is -a.
 
  • #12
epenguin
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0 is the identity, isn't it?
Yes. I came back to correct but you got here first!

If I am more nearly right now, if 1 was also included it wouldn't be a group any more; on the other hand if 0 was then excluded it would be again?
 
  • #13
Yes. I came back to correct but you got here first!

If I am more nearly right now, if 1 was also included it wouldn't be a group any more; on the other hand if 0 was then excluded it would be again?
No. If one was included then you have division by zero problem, no?
If zero was excluded then we don't have an identity hence, no inverse.
EDit

Actually i am wrong adding 1 changes nothing but obviously adding -1 does.
 

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