GR9768 #58 #59

Forum for the GRE subject test in mathematics.
Post Reply
Posts: 10
Joined: Sat Sep 19, 2009 5:55 pm

GR9768 #58 #59

Post by sjin » Thu Oct 01, 2009 5:36 am

58 Let f be a real-valued function defined and continuous on the set of real number R. Which of the following must be true of the set S={f(c):0<c<1}?
1. S is a connected subset of R
2. S is an open subset of R
3. S is a bounded subset of R

I don't know how 1 and 2 are right.

59. A cyclic group 5 has an element x such that the set {x^3, x^5, x^9} has exactly two elements. The number of elements in the set {x^(13n) : n is a positive integer} is

I am not sure my way of thinking is correct. Correct me if I am wrong.

We know that {x^3, x^5, x^9} has exactly two elements, which means {x^3,x^5} are the only two elements and we know the order of x^3 is 3.

Therefore, {x^(13n):n is a positive integer} consists of three elements. They are x^13,x^26, and x^39, correct?

Posts: 9
Joined: Tue Sep 29, 2009 1:07 pm

Re: GR9768 #58 #59

Post by conrado.math » Thu Oct 01, 2009 8:19 am

#58: 1 is correct because $$S$$ is the image of a connected set (an interval) by a continuous function, and therefore must be connected. Besides, 2 is not correct. If we take $$f(x) = \left|x-\frac{1}{2}\right|$$, then $$S = \left[0,\frac{1}{2}\right)$$ is not open.

#59: What do you mean by "A cyclic group 5"?

Posts: 9
Joined: Thu Jun 25, 2009 10:57 am

Re: GR9768 #58 #59

Post by toughluck » Thu Oct 01, 2009 3:36 pm

So for 59 it should be a cyclic group of order 15. So think of the following group {x^0,x^1,x^2,.....x^14} under multiplication which is isomorphic to the group Z 15. Then you have to find which element is used so that {x^3, x^5, x^9} has exactly two elements and you have the choices of x^0, x^1, x^3 and x^5 which are the generators of cyclic subgroups.
If you use x^0 that group that has to have three elements will only have one x^0 so that is out.

If you use x^1 the group will have three elements x^3, x^5 and x^9 so that is out

If you use x^3 the group will have three elements x^9, x^0, and x^12 so that is out.

If you use x^5 the group will have two elements x^0 and x^10 so we'll use that generator.

Then we put x^5 into x^(13n) so if n is 1 we get x^65 which goes to x^5.
If n=2 we get x^130 which goes to x^10
If n= 3 we get x^195 whih goes to x^0

Therefore we have three elements in that subgroup. Hope that helps.

Posts: 29
Joined: Sun Jun 28, 2009 3:42 pm

Re: GR9768 #58 #59

Post by aas56 » Tue Oct 06, 2009 2:45 pm

My way of doing it:
SInce The cyclic group has order 15 then the order of x is either 1 or 3 or 5 or 15.
but order of x can't be 1 sine in this case the set would containe one element
and can't be 5 neither 15 since the set would contain 3 elemeents
verifying for 3 : we get {x^3,x^5,x^9}={e,x^2}
the x^3=e, x^(13n)=x^n.x^(12n)=x^n.e=x^n={e,x,x^2}.
then the answer is A

Posts: 4
Joined: Sat Sep 05, 2009 11:36 pm

Re: GR9768 #58 #59

Post by isih » Mon Oct 12, 2009 9:22 pm

Another way of looking at it is the folowing:

Since the set has 2 elements on of the following is true: either x^3=x^5 or x^5=x^9 or x^3=x^9. The first 2 gives us x^2=1=x^4 which is a contradiction since 2 doesn't divide 15. Thus x^6=1. Since 6 doesn't divide 15 we obtain that x^3=1. Now <x^{13}>=order{x^13}. But ord(x^13)=3/(13,3)=3/1=3. (In general if ord(x)=d then ord(x^k)=k/gcd(d,k))

Posts: 6
Joined: Wed Nov 04, 2009 1:31 am

Re: GR9768 #58 #59

Post by chival » Fri Nov 06, 2009 7:59 am

anyone can please tell me in #58 why S is bounded ?

can {0<c<1} be mapped on to (0, +infinite)?

Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

Re: GR9768 #58 #59

Post by mtey » Fri Nov 06, 2009 8:23 am

Notice that f is continuous on all $$\mathbb{R}$$ and thus on $$[0,1]$$. So $$f([0,1])$$ is compact, i.e. closed and bounded and $$f((0,1)) \subset f([0,1])$$

If f was continuous only on $$(0,1)$$ this won't be true take $$f(x) = \frac{1}{x}$$

Post Reply