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## #1 Help Me ! » Find the order of the cyclic subgroup of D2n generated by r » 2011-02-03 14:56:14

xsw001
Replies: 0

Find the order of the cyclic subgroup of D2n generated by r.

The order of an element r is the smallest positive integer n such that r^n = 1.
Here is the representation of Dihedral group D2n = <r, s|r^n=s^2=1, rs=s^-1>
The elements that are in D2n = {1, r, r^2, ... , r^n-1, s, sr, sr^2, ... , s(r^n-1)}
Dihedral group is non-abelian (cannot commute), but cyclic group is abelian (can commute)

Okay it basically ask us to use the generator r from dihedral group to form a subgroup of D2n that is cyclic group.
So obviously we can't choose the term that has s(r^i) for i=1, ... , n-1 since it's not power of r.
that leaves us the set of choices {1, r, r^2, ... , r^n-1}, identity 1 has to be there and it commutes with all the elements in the group.

Since the order of D2n=2n, now the subgroup has half of its entries, and the property of D2n such that r^n=1, therefore the order of r is n.  Does it implies that the order of cyclic subgroup {1, r, r^2, ... , r^n-1} of D2n is n then?

## #2 Re: Exercises » Prove Lower Integral <= 0 <= Upper Integral » 2010-10-29 15:27:08

Never mind, I got it.
Every interval of nonzero size contains a rational number.
So the min of f(x) on the interval MUST be <=0
and the max of f(x) on the interval is >=0.

## #3 Exercises » Prove Lower Integral <= 0 <= Upper Integral » 2010-10-29 15:06:28

xsw001
Replies: 3

Suppose f:[a, b]-> R is bounded function
f(x)=0 for each rational number x in [a, b]
Prove Lower Integral <= 0 <= Upper Integral

Proof:
f(x) = 0 when x is rational
both L(f, p) = U(f, P) = 0
and L(f, p) <= Lower Integral <= Upper Integral <= U(f, p)

This function seems like discontinous even though there aren't any information of functional value when x is NOT rational.  It looks like that the Intermediate Value Theorem needs to be appplied.

So I have to prove that the Lower Integral <=0, and the Upper Integral >=0.  So the function itself has to cross f(x)=0 isn't it?

Any suggestions would be greatly appreciated.

## #4 Help Me ! » Show set has no limit points » 2010-10-24 10:04:13

xsw001
Replies: 0

D={set of real numbers consisting of single numbers}
Show set D has no limit points, and show the set of Natural numbers has no limits points.

I know it's a very simple question.  I dont know my way of approaching this is appropriate or not.  Let me know.  Thanks.

A finite set of real numbers consisting of single numbers is not a sequence and doesnt converge to a specific number.  Therefore cant have limit points.

I know the fact that the set of Natural numbers are denumerable (infinite countable), and it diverges, therefore natural numbers have no limit point.

## #5 Re: Help Me ! » prove that f(a) < f(c) < f(b) » 2010-10-24 05:01:36

I know, I agree with you Bob.  It is so obvious for any given continuous injective function, it is strictly monotone, either strictly increasing or decreasing, with the given domain [a,b], then it has to be strictly increasing instead of strictly decreasing, or else the domain would have stated [b,a], isn't it?

Here is the sketch of the proof.
Assume by contradiction that a<c<b, and case 1) f(a)<f(b)<f(c) or case 2) f(c)<f(a)<f(b)
Since the function is one-to-one, therefore the graph of the continuous function can't oscillate, so it is strictly monotone, either strictly increasing or decresing.
case 1) if a<c<b but f(a)<f(b)<f(c)
then f(a)<f(b)<f(c) => a<b<c if it is strictly increasing which is a contradiction.
also f(a)<f(b)<f(c) => c<b<a if it is striclty decreasing which is also a contradiction.
case 2) if if a<c<b but f(c)<f(a)<f(b)
then f(c)<f(a)<f(b) => c<a<b if it is strictly increasing which is a contradiction.
also f(c)<f(a)<f(b) => b<a<c if it is striclty decreasing which is also a contradiction.
Hence if a<c<b, then f(a)<f(c)<f(b), and the function is strictly increasing.

## #6 Help Me ! » prove that f(a) < f(c) < f(b) » 2010-10-23 12:29:12

xsw001
Replies: 3

Let f:[a,b]->R be continuous and one-to-one such that f(a)<f(b).
Let a<c<b.  Prove that f(a)<f(c)<f(b)

My first instinct is to apply intermediate value theorem. Let me know whether my proof makes sense or not.

Proof:
Since f: f:[a,b]->R is continuous and one-to-one.
Therefore f is strictly increasing function.
Suppose a<c<b
According to Intermediate Value Theorem
There exists f(c) such that f(a)<f(c)<f(b)

## #7 Help Me ! » Find functions with given domain and range » 2010-10-23 05:28:22

xsw001
Replies: 1

1) Find a continuous function f: (0,1)->R with f[(0,1)]=R
I couldn't think of any function except tangent, but its domain is NOT (0,1) though? Any suggestions?

2) Find a continous function f: (0,1)->R with f[(0,1)]=[0,1]
I couldn't think of any function that I know. Any suggestions?

3) Find a continuous function f: R->R this is strictly increasing and f(R)= (-1,1)
The graph is somewhat look like f(x)=x^(1/3), but not exactly though since the domain doesn't fall into (-1,1). Any suggestions?