GRE Pratice Test Problems 50, 56, 61 GR:9768

[moo5003]

50. How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]

A) One
B) Two
C) Three
D) Four
E) Infinite
(Correct Answer D)

Since f(x)^2 = x^2 we know f(x) = +/-x for every x in [-1,1]

My first guess was that there were two continuous functions namely f(x) = x and f(x) = -x, I'm unsure how they constructed two more and was wondering if someone could explain the answer to me in more detail. My only conclusion was that they used |x| and -|x| but I was under the impression that these are not continuous, I may just be confusing differentiable with continuity however.

56. For every set S and every metric d on S, which of the following is a metric on S?

A) 4 + d
B) e^d - 1
C) d - |d|
D) d^2
E) Root(d)
(Correct Answer E)

4 + d is incorrect since 4 + d(x,x) != 0

e^d - 1 is incorrect since it fails the triangle inequality. EX: S=Z d(x,y) = |y-x|

d(0,1) + d(1,2) >/= d(0,2) but under e^d-1 this is the ienquality~
2e - 2 >/= e^2 -1 which is inconsisent

C) Is 0 for everything which is not a metric on any set with more then 1 element.

D/E) I thought both of these were metrics could someone please clarify why d^2 fails to be one?

61. What is the greatest integer that divides p^4-1 for every prime number p greater then 5?

A) 12
B) 30
C) 48
D) 120
E) 240
(Correct Answer E)

I had no idea how to start this question other then plugging in 6 finding a prime factorization and then comparing it with the other integers to see if I could eliminate possibilities (I highly doubt this is the correct way to go about this problem).

 

[Mark44]

50. How many continuous real-valued functions f are there with domain [-1,1] such that (f(x))^2 = x^2 for each x in [-1,1]

A) One
B) Two
C) Three
D) Four
E) Infinite
(Correct Answer D)

Since f(x)^2 = x^2 we know f(x) = +/-x for every x in [-1,1]

My first guess was that there were two continuous functions namely f(x) = x and f(x) = -x, I'm unsure how they constructed two more and was wondering if someone could explain the answer to me in more detail. My only conclusion was that they used |x| and -|x| but I was under the impression that these are not continuous, I may just be confusing differentiable with continuity however.

The two absolute value functions are continuous everywhere, so yes, you are confusing differentiability with continuity.

61. What is the greatest integer that divides p^4-1 for every prime number p greater then 5?

A) 12
B) 30
C) 48
D) 120
E) 240
(Correct Answer E)

I had no idea how to start this question other then plugging in 6 finding a prime factorization and then comparing it with the other integers to see if I could eliminate possibilities (I highly doubt this is the correct way to go about this problem).

It's a start. You chose 6 to start with, and factored the resulting expression. The trouble is, p has to be prime. What about factoring p^4 - 1? Can it be factored?

 

[Dick]

For the first one, |x| and -|x| are continuous, they aren't differentiable. You are confusing the two. Why didn't you look them up? For the second one try d^2 on the same example you used before. It fails the triangle inequality. For the third, look at the gcd of 7^4-1 and 11^4-1 to see what the largest number could be. Now try and prove that it is the one by factorizing p^4-1.

 

[moo5003]

So I factored p^4-1 and realized that they are just the roots of unity in the complex numbers namely:

e^2ipi/4, e^ipi, e^6ipi/4, 1 ~ p^4 -1 = (p+1)(p-1)(p - e^2ipi/4)(p - e^6ipi/4)

Now I just need to figure out where to go from here.



Dick: I'm unsure how to go about finding the GCD (in a timely fashion) for numbers such as 7^4-1 and 11^4-1 ~ I'll keep looking but the closest I found was finding the GCD of 2^p and 2^q. p,q prime.

 

[Dick]

You want to factor p^4-1 over the real numbers, not the complex numbers. I'll get you started p^4-1=(p^2-1)(p^2+1). One of those factors will factor some more. For the GCD just look up Euclid's algorithm.

 

[moo5003]

So, 7^4-1 = 2400
11^4 - 1 = 14640

Using Euclids algorythm
14,640 | 2400
2400 | 240
GCD: 240

Note: Is there a way to do this without finding 7^4 or 11^4? On the test I will not have access to a calculator and while its possible to find numbers like this its usually time consuming.

p^4-1 = (p+1)(p-1)(p^2+1)

My best guess to prove if factors all integers of this form is by putting 240 into prime factors and some how extracting those factors from the above expression.

240 = 2^4 * 3 * 5

p+1 is divisible by 2 for any prime above 5 thus (p+1)/2 is a integer, this argument is valid for (p-1) and (p^2+1) as well.

Thus we know 2^3 divides p^4-1.

I'm lead to believe that either p+1 or p-1 must be divisible by 2^2 since if for example p+1 = 2 * k for some prime k then
p -1 = 2 * k - 2 = 2 * (k-1) = 2 * 2 * m for some prime m.

Thus without a rigorous proof in full let's assume 2^4 divides p^4-1.

Now we must show 3 and 5 divide p^4-1.

 

[Dick]

I'm not sure there is a shortcut. Let's just finish it and then you tell me. 240=2^4*3*5. That means that the 'greatest integer' are looking for must divide 240. Let's start with the 2 factors. How many factors of two divide (p-1)(p+1)(p^2+1) for a prime p>5?

 

[moo5003]

3 factors of 2 divide (p+1)(p-1)
1 factor of 2 divides (p^2 + 1)
still trying to show 3 and 5 factor in this somehow.

 

[adriank]

It might help to know that every prime greater than 3 is either 6k - 1 or 6k + 1 for some integer k.

 

[Dick]

3 factors of 2 divide (p+1)(p-1)
1 factor of 2 divides (p^2 + 1) still trying to show 3 and 5 factor in this somehow.

Ok, can you show there must be a factor of 3? Of 5?

 

[Dick]

3 factors of 2 divide (p+1)(p-1)
1 factor of 2 divides (p^2 + 1) still trying to show 3 and 5 factor in this somehow.

p-1, p and p+1 are three consecutive integers...

 

[moo5003]

So 3 is apparent from p-1, p , p+1 but my proof for one part of 5 seemed to be a little long.

Let p = 6k + 1 (the -1 case is pretty much the same save a few things)

If 5|(p+1) or (p-1) we are done.

Suppose 5!|p or (p+1) or (p-1) then

5!|6k, 6k+1, 6k-1, or 6k+2

Thus 5|6k+3, and 6k-2.

Claim: 5|P^2 +1

P^2 + 1 = 36k^2 + 12k + 2
= 36k^2 + 12k + 6 - 4
= 36k^2 - 4 + 5m for some integer m (Note: 5|6k+3)
= 4(9k^2 - 1) + 5m

If we can show 5|9k^2 -1 we are done

9k^2 -1 = (3k+1)(3k-1) but 5|6k-2 thus 5|3k-1

While I obviously would not have gotten this question correct had I been taking it on the exam, it seems even now it would take far to long to do it efficiently. I was wondering in what way I could see the answer much sooner.

 

[Dick]

I wouldn't use the p=6k+1 or 6k-1 thing. It's not something I make it a point to remember. For 5, we know 5 doesn't divide p so if it doesn't divide either p-1 or p+1 then p mod 5 must be either 2 or 3. If that's the case then what is p^2+1 mod 5?

 

[moo5003]

p^2+1 mod 5 would be 5 or 10 ie: 0, if p mod 5 = 2 or 3 respectivley. Thanks for the help Dick and everyone else!

 

[Dick]

If you put it all together, it's not THAT hard. The only reason for finding the GCD of 7^4-1 and 11^4-1 was have some idea of what to look for. Like, do I need to check for divisibility by 7? In a test situation you could also figure that out by looking at the suggested answers (unless they were evil and put 7*240 in the list).