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
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.)
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
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.
For 2, i think 6 is the multiplicative identity.
Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”
Users browsing this forum: Alicelxd, Andreas, caterpillar, heshan666, thegamer and 17 guests