Recent content by hopsonuk

Ok, so how do I write this out to check LHS = RHS ?

If i write this for associativity: b(x)[c(x)a] = (b x c)(x)a I cannot get the RHS = LHS..

I think my associativity is incorrect.

Is what I have written for distributivity law 2 correct?

Ok :) - Here is a re-upload:

1. Homework Statement [PLAIN]http://img87.imageshack.us/img87/9001/150674.png 2. Homework Equations 3. The Attempt at a Solution I think, I need to write: a(x)[b(+)c] Then prove: Distributivity law 1: Unsure of Distributivity law 2: Proving: a(x)[b(+)c] =...

I can't get the code to work to show my derivation, the way I thought I could is the way you said earlier is not allowed since the group is not commutative.

I haven't done it correctly as commutativity does not hold true, because I cannot times sigma by (a...an) and sigma by it's inverse.

Referring to the above in the last post.

Sorry I posted during you reply, I've had a look and it's quite easy to derive the RHS from the LHS. Do I need to prove from RHS to LHS as well?

Surely sigma multiplied by it's inverse will give the identity? Which composed with <a1...as> would give <a1...as>.

For the arbitrary element b, does this need subsituting for a or the permutation sigma?

If we plotted two functions, surely they'd look exactly the same?

Wouldn't we be able to observe if two functions are equal?

Homework Statement Homework Equations The Attempt at a Solution I honestly do not know how to start this.