Proving a set under an operation is associative. (Abstract Algebra)

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  • #1
jdinatale
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Homework Statement



godel.jpg


I'm trying to prove that this is a group. I already established elsewhere that it is a binary operation, so now I am onto proving associativity. I've tried many examples and so I'm confident it is associative, but now I just have to prove that.


The Attempt at a Solution



Let [itex]x, y, z \in G[/itex]. Then [itex]x*(y*z) = x*(y + z - [y + z]) = x + y + z - [y + z] - [x + y + z - [y + z]][/itex]

Then [itex](x*y)*z = (x + y - [x + y])*z = x + y + z - [x + y] - [x + y + z - [x + y]] [/itex]

Now the problem is coming up with an equality to show this. We would have to show that [y + z] = [x + y]. I guess you could do a ton of cases where you show what happens when [itex]y + z \geq 1[/itex] and [itex]y + z < 1[/itex], same goes for [itex]x + y[/itex], but I'm now sure that even that would work.
 
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Answers and Replies

  • #2
micromass
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Did you see quotient groups?? Perhaps you can prove your structure to be isomorphic to

[tex]\mathbb{R}/\mathbb{Z}[/tex]

Since the above is a group, it follows that your G is also a group.
 
  • #3
jdinatale
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Did you see quotient groups?? Perhaps you can prove your structure to be isomorphic to

[tex]\mathbb{R}/\mathbb{Z}[/tex]

Since the above is a group, it follows that your G is also a group.

No, I've never seen a quotient group or the term isomorphic (although I know what isomorphic means). We haven't even defined a subgroup. This is only the third day of abstract algebra class and he hasn't gotten to either of those terms. We are suppose to be able to prove this just using definitions of a group, definitions of associativity, etc.

I would normally be up for the challenge of proving it using an alternate method like the one you suggested, but the assignment is due tomorrow and I have other assignments to do, so I won't have time to read up on the quotient group stuff yet.
 
  • #4
Dick
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No, I've never seen a quotient group or the term isomorphic (although I know what isomorphic means). We haven't even defined a subgroup. This is only the third day of abstract algebra class and he hasn't gotten to either of those terms. We are suppose to be able to prove this just using definitions of a group, definitions of associativity, etc.

I would normally be up for the challenge of proving it using an alternate method like the one you suggested, but the assignment is due tomorrow and I have other assignments to do, so I won't have time to read up on the quotient group stuff yet.

You can't show [x+y]=[x+z]. It's not true as you can see by making examples. What you can show is things like [x+[y]]=[x]+[y] when x and y are positive.
 
  • #5
jdinatale
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You can't show [x+y]=[x+z]. It's not true as you can see by making examples. What you can show is things like [x+[y]]=[x]+[y] when x and y are positive.

Thank you for responding in this thread and in my last one. I'm not sure how your example helps. I know you can't show that [x + y] = [x + z]...however, we would have to do this to show that these two equations are equal, since [x +y] and [x + z] are the only difference between the two equations.

Let [itex]x, y, z \in G[/itex]. Then [itex]x*(y*z) = x*(y + z - [y + z]) = x + y + z - [y + z] - [x + y + z - [y + z]][/itex]

Then [itex](x*y)*z = (x + y - [x + y])*z = x + y + z - [x + y] - [x + y + z - [x + y]] [/itex]


I've tried a ton of examples, and I am almost certain that this is associative, but I'm not sure how to show the above equations are equal.
 
  • #6
vela
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Look at [x+y+z-[y+z]] as [a-] where a=x+y+z and b=y+z.
 

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