Problems in the 4 practice tests

 Posts: 22
 Joined: Sat Aug 13, 2011 9:04 am
Problems in the 4 practice tests
Hello all guys! I am preparing for the test in Nov. And after I have finished the 4 practice test, I have some problems about these 4 tests.
1.GR0568 #40(C)
40. For which of the following rings is it possible for the product of two nonzero elements to be zero?
(A) The ring of complex numbers
(B) The ring of integers modulo 11
(C) The ring of continuous realvalued functions on [0, 1]
(D) The ring a+b^1/2 :a and b are rational numbers
(E) The ring of polynomials in x with real coefficient
2.GR0568 #44
44. A fair coin is to be tossed 100 times, with each toss resulting in a head or a tail. If H is the total number of heads
and T is the total number of tails, which of the following events has the greatest probability?
(A) H=50
(B) T >= 60
(C) 51 < H < 55
(D) H >=48 and T >= 48
(E) H <5 or H > 95
Should I deduce from the graph of normal distribution? I just cannot figure out how to calculate it.
3.GR0568 #61
61. Which of the following sets has the greatest cardinality?
(A) R
(B) The set of all functions from Z to Z
(C) The set of all functions from R to {0, 1}
(D) The set of all finite subsets of R
(E) The set of all polynomials with coefficients in R
4.GR0568 #65
65. Which of the following statements are true about the open interval (0, 1) and the closed interval [0, 1]?
I. There is a continuous function from (0, 1)onto [0, 1].
II. There is a continuous function from [0, 1]onto (0, 1).
III. There is a continuous onetoone function from (0, 1) onto [0, 1].
(A) None (B) I only (C) II only (D) I and III only (E) I, II, and III
And I also have troubles on another 2 tests: GR9367 #43, GR9367 #57,GR9367 #60
Thank you all guys! Any help will be greatly appreciated!
1.GR0568 #40(C)
40. For which of the following rings is it possible for the product of two nonzero elements to be zero?
(A) The ring of complex numbers
(B) The ring of integers modulo 11
(C) The ring of continuous realvalued functions on [0, 1]
(D) The ring a+b^1/2 :a and b are rational numbers
(E) The ring of polynomials in x with real coefficient
2.GR0568 #44
44. A fair coin is to be tossed 100 times, with each toss resulting in a head or a tail. If H is the total number of heads
and T is the total number of tails, which of the following events has the greatest probability?
(A) H=50
(B) T >= 60
(C) 51 < H < 55
(D) H >=48 and T >= 48
(E) H <5 or H > 95
Should I deduce from the graph of normal distribution? I just cannot figure out how to calculate it.
3.GR0568 #61
61. Which of the following sets has the greatest cardinality?
(A) R
(B) The set of all functions from Z to Z
(C) The set of all functions from R to {0, 1}
(D) The set of all finite subsets of R
(E) The set of all polynomials with coefficients in R
4.GR0568 #65
65. Which of the following statements are true about the open interval (0, 1) and the closed interval [0, 1]?
I. There is a continuous function from (0, 1)onto [0, 1].
II. There is a continuous function from [0, 1]onto (0, 1).
III. There is a continuous onetoone function from (0, 1) onto [0, 1].
(A) None (B) I only (C) II only (D) I and III only (E) I, II, and III
And I also have troubles on another 2 tests: GR9367 #43, GR9367 #57,GR9367 #60
Thank you all guys! Any help will be greatly appreciated!
Last edited by cauchy2012 on Wed Nov 02, 2011 4:24 am, edited 1 time in total.
Re: Problems in the 4 practice tests
1.GR0568 #40(C)
40. For which of the following rings is it possible for the product of two nonzero elements to be zero?
(A) The ring of complex numbers
(B) The ring of integers modulo 11
(C) The ring of continuous realvalued functions on [0, 1]
(D) The ring a+b^1/2 :a and b are rational numbers
(E) The ring of polynomials in x with real coefficient
This is quite tricky.
The answer is (C), and the example is
(xx) (x+x) = x^2  x^2 = 0.
You may want to check out http://sfmathgre.blogspot.com/ for some answers.
40. For which of the following rings is it possible for the product of two nonzero elements to be zero?
(A) The ring of complex numbers
(B) The ring of integers modulo 11
(C) The ring of continuous realvalued functions on [0, 1]
(D) The ring a+b^1/2 :a and b are rational numbers
(E) The ring of polynomials in x with real coefficient
This is quite tricky.
The answer is (C), and the example is
(xx) (x+x) = x^2  x^2 = 0.
You may want to check out http://sfmathgre.blogspot.com/ for some answers.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Problems in the 4 practice tests
Your answer is correct but your reasoning is invalid. xx on [0,1] IS the zero function. An easy example would be a piecewise linear function such as f= .5x on [0, .5] f= 0 on [.5,1] and g= 0 on [0,.5] g= x.5 on [.5, 1] (draw the picture). Then f*g is the zero function but f and g are not zero.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Problems in the 4 practice tests
for 3, card(S) < card(P(S)) where P(S) is the power set of S
R has "cardinality of the continuum"  card(R)
Z x Z is countable because Z is countable. Then functions from Z to Z are subsets of P(Z x Z), but card(Z x Z) = card(N),
so card(P(Z x Z)) <= card (P(N)) = card(R)
(correct)The set of functions from R to {0,1} basically is P(R), so cardinality is greater than card(R)
The set of all finite subsets of R has card(R). Basically if a countable collection of sets each has cardinality less than card(R), then the union of these also has cardinality less than card(R). Therefore, if A_n is the subsets of R which have n elements, each of these has card(R), so the union also has card(R).
The set of all polynomials with coefficients in R is almost same thing as the above oneEach polynomial of degree n is basically an ordered subset of R with size n where order matters. Thus the set of polynomials in R with degree n has card(R). Thus the union has card(R).
With these sort of tests, if you can identify which one looks like it has cardinality of P(R), that's probably the right one.. it gets complicated with larger cardinality..
For 4,
I is true.. something like .5sin(5x) + .5 should work (just draw a picture that goes up to 1 and down to 0... )
II is not. A continuous image of a compact set is compact
III is not. Continuous bijections on R are strictly monotone. When this happens, it is easy see see that f is an open map, which means the inverse is also continuous. But then the inverse would map [0,1] onto (0,1).
R has "cardinality of the continuum"  card(R)
Z x Z is countable because Z is countable. Then functions from Z to Z are subsets of P(Z x Z), but card(Z x Z) = card(N),
so card(P(Z x Z)) <= card (P(N)) = card(R)
(correct)The set of functions from R to {0,1} basically is P(R), so cardinality is greater than card(R)
The set of all finite subsets of R has card(R). Basically if a countable collection of sets each has cardinality less than card(R), then the union of these also has cardinality less than card(R). Therefore, if A_n is the subsets of R which have n elements, each of these has card(R), so the union also has card(R).
The set of all polynomials with coefficients in R is almost same thing as the above oneEach polynomial of degree n is basically an ordered subset of R with size n where order matters. Thus the set of polynomials in R with degree n has card(R). Thus the union has card(R).
With these sort of tests, if you can identify which one looks like it has cardinality of P(R), that's probably the right one.. it gets complicated with larger cardinality..
For 4,
I is true.. something like .5sin(5x) + .5 should work (just draw a picture that goes up to 1 and down to 0... )
II is not. A continuous image of a compact set is compact
III is not. Continuous bijections on R are strictly monotone. When this happens, it is easy see see that f is an open map, which means the inverse is also continuous. But then the inverse would map [0,1] onto (0,1).
Re: Problems in the 4 practice tests
ah.. yes, you are right.blitzer6266 wrote:Your answer is correct but your reasoning is invalid. xx on [0,1] IS the zero function. An easy example would be a piecewise linear function such as f= .5x on [0, .5] f= 0 on [.5,1] and g= 0 on [0,.5] g= x.5 on [.5, 1] (draw the picture). Then f*g is the zero function but f and g are not zero.
Thanks for the correction.

 Posts: 22
 Joined: Sat Aug 13, 2011 9:04 am
Re: Problems in the 4 practice tests
But in your example, f and g are both not continuous functions, which do not satisfy the requirement of the exercise...
blitzer6266 wrote:Your answer is correct but your reasoning is invalid. xx on [0,1] IS the zero function. An easy example would be a piecewise linear function such as f= .5x on [0, .5] f= 0 on [.5,1] and g= 0 on [0,.5] g= x.5 on [.5, 1] (draw the picture). Then f*g is the zero function but f and g are not zero.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Problems in the 4 practice tests
Try again
Re: Problems in the 4 practice tests
they are continuous. (drawing the graph will make it clearer)cauchy2012 wrote:But in your example, f and g are both not continuous functions, which do not satisfy the requirement of the exercise...
blitzer6266 wrote:Your answer is correct but your reasoning is invalid. xx on [0,1] IS the zero function. An easy example would be a piecewise linear function such as f= .5x on [0, .5] f= 0 on [.5,1] and g= 0 on [0,.5] g= x.5 on [.5, 1] (draw the picture). Then f*g is the zero function but f and g are not zero.
f(0.5)=0, and g(0.5)=0.

 Posts: 22
 Joined: Sat Aug 13, 2011 9:04 am
Re: Problems in the 4 practice tests
yoyobarn wrote:they are continuous. (drawing the graph will make it clearer)cauchy2012 wrote:But in your example, f and g are both not continuous functions, which do not satisfy the requirement of the exercise...
blitzer6266 wrote:Your answer is correct but your reasoning is invalid. xx on [0,1] IS the zero function. An easy example would be a piecewise linear function such as f= .5x on [0, .5] f= 0 on [.5,1] and g= 0 on [0,.5] g= x.5 on [.5, 1] (draw the picture). Then f*g is the zero function but f and g are not zero.
f(0.5)=0, and g(0.5)=0.
Yes! I got it! Thank you, but what about the others,guys?
Re: Problems in the 4 practice tests
Yes, I believe one should estimate from the normal distribution.2.GR0568 #44
44. A fair coin is to be tossed 100 times, with each toss resulting in a head or a tail. If H is the total number of heads
and T is the total number of tails, which of the following events has the greatest probability?
(A) H=50
(B) T >= 60
(C) 51 < H < 55
(D) H >=48 and T >= 48
(E) H <5 or H > 95
Should I deduce from the graph of normal distribution? I just cannot figure out how to calculate it.
Mean=np=100*0.5=50
Std Deviation=sqrt{npq}=5
Estimate it using: About 68% of values drawn from a normal distribution are within one standard deviation σ away from the mean; about 95% of the values lie within two standard deviations; and about 99.7% are within three standard deviations.
C and D should intuitively be the largest.
Furthermore, D should be larger than C, since it is closer to the centre, where the probabilities are the highest.
Not very rigorous, but gives an estimate.
Re: Problems in the 4 practice tests
Hi, I think (B) is the Card(Z^Z) instead of Card(ZxZ) and Z^Z is also countable?blitzer6266 wrote:for 3, card(S) < card(P(S)) where P(S) is the power set of S
R has "cardinality of the continuum"  card(R)
Z x Z is countable because Z is countable. Then functions from Z to Z are subsets of P(Z x Z), but card(Z x Z) = card(N),
so card(P(Z x Z)) <= card (P(N)) = card(R)
(correct)The set of functions from R to {0,1} basically is P(R), so cardinality is greater than card(R)
The set of all finite subsets of R has card(R). Basically if a countable collection of sets each has cardinality less than card(R), then the union of these also has cardinality less than card(R). Therefore, if A_n is the subsets of R which have n elements, each of these has card(R), so the union also has card(R).
The set of all polynomials with coefficients in R is almost same thing as the above oneEach polynomial of degree n is basically an ordered subset of R with size n where order matters. Thus the set of polynomials in R with degree n has card(R). Thus the union has card(R).
With these sort of tests, if you can identify which one looks like it has cardinality of P(R), that's probably the right one.. it gets complicated with larger cardinality..
For 4,
I is true.. something like .5sin(5x) + .5 should work (just draw a picture that goes up to 1 and down to 0... )
II is not. A continuous image of a compact set is compact
III is not. Continuous bijections on R are strictly monotone. When this happens, it is easy see see that f is an open map, which means the inverse is also continuous. But then the inverse would map [0,1] onto (0,1).
(B) The set of all functions from Z to Z
(C) The set of all functions from R to {0, 1}
Re: Problems in the 4 practice tests
Functions from Z to Z have card(R). Use functions from N to N instead, and write all of your natural numbers in base 2. Then given a function f: N > N, associate to it the real number in base 3 given by ternary expansion 0.f(1) 2 f(2) 2 f(3) 2 f(4) 2 ... This gives you an injection from functions N > N into the real numbers, so you know that their cardinality is at most card(P).
A similar construction also works for functions from Z to Z: 0. s(1) f(1) 2 s(1) f(1) 2 s(2) f(2) 2 s(2) f(2) 2 ... where s(k) encodes the sign of f(k), e.g. s(k) = 0 if f(k) is nonnegative and s(k) = 1 if f(k) is negative.
While not needed here, you can also get a lower bound on the cardinality. Given the decimal expansion of a real number
r = r_k r_{k1} ... r_1 . s_1 s_2 s_3 ...
assign to it the function
f(n) = r_n
f(0) = 0
f(n) = s_n
for n a natural number
This is an injection from the real numbers into functions from Z to Z.
A similar construction also works for functions from Z to Z: 0. s(1) f(1) 2 s(1) f(1) 2 s(2) f(2) 2 s(2) f(2) 2 ... where s(k) encodes the sign of f(k), e.g. s(k) = 0 if f(k) is nonnegative and s(k) = 1 if f(k) is negative.
While not needed here, you can also get a lower bound on the cardinality. Given the decimal expansion of a real number
r = r_k r_{k1} ... r_1 . s_1 s_2 s_3 ...
assign to it the function
f(n) = r_n
f(0) = 0
f(n) = s_n
for n a natural number
This is an injection from the real numbers into functions from Z to Z.