## Subject questions for NOV.12TH

Forum for the GRE subject test in mathematics.
cauchy2012
Posts: 22
Joined: Sat Aug 13, 2011 9:04 am

### Subject questions for NOV.12TH

Hi,guys, i have some problems,hope you can help me out, thanks advance!

1. how many 2X2 matrices on a field of q elements are invertible?

2. Function f is a function with two linear parts. f(0)=f(2)=0, 0<x<1,max(f)=1
choose the range of the length of f?
A.(2*2^0.5, 1+5^0.5)
B.(2*2^0.5, 1+5^0.5]
C.[2*2^0.5, 1+5^0.5)
D.[2*2^0.5, 1+5^0.5]

The answer of this one is A? Since i find x could get values only between 0 and 1 and assume f is equal to 1 always between 0 and 1. f could not have a value on x=0 and x=1 for the max. so i think f can not reach 2*2^0.5 and 1+5^0.5.

3. f : X → Y is continuous bijection,
Ⅰ. If X is compact then Y is compact.
Ⅱ. If X is Hausdorff space then Y is Hausdorff space.
Ⅲ. If X is compact and Y is Hausdorff space, then f−1 exist.
Which of them are correct?

Thank you guys!!!

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Subject questions for NOV.12TH

1. was discussed not too long ago. Try the board search. For 2, I assume that the "length of f" is the length of the graph of f? Answer B is correct: You can realize 2*sqrt(2) [connect (0,0) to (1,1) to (2,0)] but not 1 + sqrt(5) [(0,0) to (epsilon,1) to (2,0) gets you arbitrarily close, but for epsilon = 0 the two lines are not the graph of a function]. But is f required to be continuous? If not, you could pick f to go from (0,0) to (1,1) and then be dead zero on the rest of the interval. This graph would have length sqrt(2) + 1 < 2*sqrt(2).

For 3, I and III are correct:

- the continuous image of a compact set is compact
- f^{-1} exists for any bijection. (The extra hypotheses give you that f^{-1} is continuous, but that's not being asked.)

II is false: Consider the "identity" map from X = the real numbers with the Euclidean topology to Y = the real numbers with the trivial topology. It's clearly continuous, 1-1 and onto, and X is Hausdorff, but Y is not Hausdorff. (Can't separate points when the only open sets are empty and the full space.)

Posts: 27
Joined: Sun Oct 17, 2010 4:57 am

### Re: Subject questions for NOV.12TH

Owl, don't you mean that C is the correct answer for 2?

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Subject questions for NOV.12TH

^ Yes, I meant that C is the best answer - though none of them is correct without an additional hypothesis on the continuity of f.

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

### Re: Subject questions for NOV.12TH

I actually believe it's a) since the maximum occurs on the open interval (0,1). Therefore f(1)=1 can't be realized

cauchy2012
Posts: 22
Joined: Sat Aug 13, 2011 9:04 am

### Re: Subject questions for NOV.12TH

blitzer6266 wrote:I actually believe it's a) since the maximum occurs on the open interval (0,1). Therefore f(1)=1 can't be realized
I agree with blitzer6266, i think both boundary value cannot be reached, so A is the answer...

cauchy2012
Posts: 22
Joined: Sat Aug 13, 2011 9:04 am

### Re: Subject questions for NOV.12TH

i didn't find the previous discussion about the first question. Would you like to explain some for me? Thank you in advance!!!

owlpride wrote:1. was discussed not too long ago. Try the board search. For 2, I assume that the "length of f" is the length of the graph of f? Answer B is correct: You can realize 2*sqrt(2) [connect (0,0) to (1,1) to (2,0)] but not 1 + sqrt(5) [(0,0) to (epsilon,1) to (2,0) gets you arbitrarily close, but for epsilon = 0 the two lines are not the graph of a function]. But is f required to be continuous? If not, you could pick f to go from (0,0) to (1,1) and then be dead zero on the rest of the interval. This graph would have length sqrt(2) + 1 < 2*sqrt(2).

For 3, I and III are correct:

- the continuous image of a compact set is compact
- f^{-1} exists for any bijection. (The extra hypotheses give you that f^{-1} is continuous, but that's not being asked.)

I did not find the previous discussion about the first one, could you explain some? Thank you advance!!!

II is false: Consider the "identity" map from X = the real numbers with the Euclidean topology to Y = the real numbers with the trivial topology. It's clearly continuous, 1-1 and onto, and X is Hausdorff, but Y is not Hausdorff. (Can't separate points when the only open sets are empty and the full space.)

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Subject questions for NOV.12TH

blitzer, cauchy: I don't understand your reasoning. Take the continuous piecewise linear function with f(0) = 0, f(1) = 1, f(2) = 0. The length of its graph is precisely 2*sqrt(2). Is this not an admissible function for f? Am I misinterpreting the question?

And for question 1, see here: http://www.mathematicsgre.com/viewtopic.php?f=1&t=671

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

### Re: Subject questions for NOV.12TH

cauchy2012 wrote:
2. Function f is a function with two linear parts. f(0)=f(2)=0, 0<x<1,max(f)=1
choose the range of the length of f?
Haha, I see what you're saying. I think this comes down to a very poor description of a function. The way I interpreted this description is:

F: [0,2] -> R is a linear piece-wise (with 2 pieces) continuous function such that

1) f(0)=f(2)=0
2) on the interval (0,1), the maximum (not just sup) of f is 1. Therefore there exists a c in (0,1) such that f(c)=1

It's a stupid question

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Subject questions for NOV.12TH

Haha, I see what you're saying. I think this comes down to a very poor description of a function. The way I interpreted this description is:

F: [0,2] -> R is a linear piece-wise (with 2 pieces) continuous function such that

1) f(0)=f(2)=0
2) on the interval (0,1), the maximum (not just sup) of f is 1. Therefore there exists a c in (0,1) such that f(c)=1
Oh, I see. I missed the part that that the maximum is supposed to be attained in the interval (0,1). My bad!

cauchy2012
Posts: 22
Joined: Sat Aug 13, 2011 9:04 am

### Re: Subject questions for NOV.12TH

Thank you buddies, i got them! And i still have some other problems.

1. If S={0,2,4,6,8} is the subset of (Z10,+), then which of the following is false
A. (S,+) is a subring of Z10
C. S has no multiplication identity
D. S is commutative

This one is an incomplete one and i think all the choices listed above are true, am i right?

2. If (Z17,*) be the multiplication group of non-zero elements of (Z17,+),then which of the following are generators of (Z17,*) :
5,8,16

A. none B. 5 only C. 8,16 D. 5,8,16

3. Which of the following is false
(1) every field is an integral domain
(2) every finite integral domain is field
(3) ring R for which x^2=x, for any x belongs to R, then R is commutative
(4) R be a ring mx=0 for any integers n, then R is finite
(5) if xy=xz and we can get y=z, commutative domain R with unity haiving above
then R is field

Thank you guys, thank you!!!

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Subject questions for NOV.12TH

1. A is probably false: a ring has two operations. It's also worth mentioning that many authors require a (sub)ring to have a multiplicative identity, which S does not have. I am not sure what the "standard" definition is. For example, the Wikipedia article on subrings assumes that rings have an identity, but the article on rings does not.

2. I believe that answer B is correct. A multiplicative group modulo a prime p is cyclic of order p-1. (For more general results, see here: http://en.wikipedia.org/wiki/Multiplica ... s_modulo_n) So check which of these elements have order 16:
5 has order 16
8 has order 8
16 has order 2

How would you compute the orders on the actual exam? Well, 16 = -1 mod 17 is easy. Then you notice that 8^4 = 16^3 and 16^3 = (-1)^3 = -1 mod 17, so the order of 8^4 is 2, so the order of 8 is 8. For 5, you suck it up and do the multiplication. Then you notice that 5^2 = 8 mod 17. Since the order of 8 is 8, the order of 5 must be 16.

Notice the following: if a number has a common divisor with p-1, then it does not have full order in the multiplicative group Z/pZ.

PNT
Posts: 37
Joined: Fri Mar 11, 2011 9:01 pm

### Re: Subject questions for NOV.12TH

For 3 i believe the answer is (5). (1) and (2) are true (4) i can't understand what it is saying (5) any commutative cancellative monoid is group embeddable which implies R is an integral domain but not necessarily a field (there are inverses but not necessarily inside R) ie Z. For (3) (x+y)=(x+y)^2=x+y+xy+yx, so xy=-yx but -yx=(-yx)^2=yxyx=yx so xy=yx

deckoff8
Posts: 23
Joined: Tue Nov 08, 2011 11:12 am

### Re: Subject questions for NOV.12TH

Quick question. Where are you getting these questions from?

cauchy2012
Posts: 22
Joined: Sat Aug 13, 2011 9:04 am

### Re: Subject questions for NOV.12TH

For 2, i think 6 is the multiplicative identity. You can check. But i still cannot figure the first one, let me think about your conclusion. Thank you. You are a good guy!

owlpride wrote:1. A is probably false: a ring has two operations. It's also worth mentioning that many authors require a (sub)ring to have a multiplicative identity, which S does not have. I am not sure what the "standard" definition is. For example, the Wikipedia article on subrings assumes that rings have an identity, but the article on rings does not.

2. I believe that answer B is correct. A multiplicative group modulo a prime p is cyclic of order p-1. (For more general results, see here: http://en.wikipedia.org/wiki/Multiplica ... s_modulo_n) So check which of these elements have order 16:
5 has order 16
8 has order 8
16 has order 2

How would you compute the orders on the actual exam? Well, 16 = -1 mod 17 is easy. Then you notice that 8^4 = 16^3 and 16^3 = (-1)^3 = -1 mod 17, so the order of 8^4 is 2, so the order of 8 is 8. For 5, you suck it up and do the multiplication. Then you notice that 5^2 = 8 mod 17. Since the order of 8 is 8, the order of 5 must be 16.

Notice the following: if a number has a common divisor with p-1, then it does not have full order in the multiplicative group Z/pZ.

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

### Re: Subject questions for NOV.12TH

For 2, i think 6 is the multiplicative identity.
Good catch! I totally missed that.

path
Posts: 16
Joined: Mon Nov 14, 2011 12:27 pm

### Re: Subject questions for NOV.12TH

Is it just me, or it seems that a handful of questions that appeared on OCT. reappeared on the Nov. test as well?

So, this means that if you took the october test, then you are almost certain to have a much higher score on the nov. one with almost no additional studying.