How do you show that two functions are equal?

[hopsonuk]

Homework Statement

View attachment 32368

Homework Equations

The Attempt at a Solution

I honestly do not know how to start this.

 

[micromass]

How do you show that two functions are equal?

 

[hopsonuk]

How do you show that two functions are equal?

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

 

[micromass]

How would you "observe" such a thing??

Functions are very rigourously defined, and so is their equality. I suggest you look up somewhere how equality of functions is defined...

 

[hopsonuk]

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

 

[micromass]

Yes, of course. But the problem is that not every function can be plotted. You will often meet functions that have very abstract domain and codomain, or that have a very abstract definition. These functions can not be plotted. This is the situation right now.

So you'll need to come up with another method of seeing whether two functions are equal...
Did you see the definition of a function? Did you define equality? I suggest you look those things up...

 

[hopsonuk]

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

 

[micromass]

Well, you'll need to show for every b that

$$\sigma((a_1~...~a_n)(\sigma^{-1}(b))=(\sigma(a_1)~...~\sigma(a_n))(b)$$

 

[hopsonuk]

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

 

[micromass]

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

But we don't multiply sigma by it's inverse! We "multiply" sigma by $$(a_1~...~a_n)$$. And then we "multiply" it by $$\sigma^{-1}$$.

What we can NOT do is:

$$\sigma\circ (a_1~...~a_n)\circ \sigma^{-1}=\sigma\circ \sigma^{-1}\circ (a_1~...~a_n)= (a_1~...~a_n)$$

We can not do this because our group is not commutative!

 

[hopsonuk]

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?

 

[hopsonuk]

Well, you'll need to show for every b that

$$\sigma((a_1~...~a_n)(\sigma^{-1}(b))=(\sigma(a_1)~...~\sigma(a_n))(b)$$

Referring to the above in the last post.

 

[micromass]

It depends on how you made the derivation. In general, if you derive the LHS from the RHS, then you don't need to make the other derivation.
But maybe you could post your derivation, so I can make sure?

 

[hopsonuk]

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.

 

[hopsonuk]

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.