s is odd, so we can write s = 2k + 1, where k is an integer.
We note that ##\alpha^n## is a cycle of length s for all integers n.
Then ##\alpha = \alpha^{s+1} = \alpha^{2(k+1)} = \alpha^{k+1}\alpha^{k+1}##.
so ##\alpha## is the square of ##\alpha^{k+1}##.
Thanks for the help and correction, I agree that (12345)12345) = (13524)
So we can write ##\alpha^na_i = a_j## where ##j = (i + n) \mod s##
Therefore ##\alpha a_i = a_{((i + 1) \text{mod s})}##
and ##\alpha^{s+1}a_i = a_{((s+1+i) \text{mod s})}##
Since (i + 1) = (s + 1 + i) (mod s)
we have...
Homework Statement
In the following problems let ##\alpha## be a cycle of length ##s##, and say
##\alpha = (a_1a_2 . . . a_s)##.
5) If ##s## is odd, ##\alpha## is the square of some cycle of length s. (Find it. Hint: Show ##\alpha = \alpha^{s+1}##)
Homework Equations
The Attempt at a...
OK i made a mistake reading the question, sorry about that, but think I can use post 3's hint..
We observed that raising a permutation, ##\alpha## of length s, to the ##s^{th}## power, we have ##\alpha^s = e##.
Therefore ##\alpha^{-1} = \alpha^{-1}e = \alpha^{-1}\alpha^s = \alpha^{s-1}## []
Homework Statement
In the following problems, let ##\alpha## be a cycle of length s, say ##\alpha = (a_1a_2 ... a_s)##
3)Find the inverse of ##\alpha## and show that ##\alpha^{-1} = \alpha^{s-1}##
Homework Equations
I've observed in the previous problem that there are ##s## distinct powers of...
Thanks for the reply.
Since we chose the "loosest" constraint's upper bounds, ##U_1 \ge x_1, U_2 \ge x_2## in constraint 3, that means they are also upper bounds of the first constraint's variables and the third constraint's variables, since an upper bound is just some number that is greater...
Homework Statement
Formulate as a mixed integer programming problem but do not solve. Maximize ##x_1 + x_2## subject to ##2x_1 + 3x_2 \le 12## or {##3x_1 + 4x_2 \le 24## and ##-x_1 + x_2 \ge 1##} ##x_1, x_2 \ge 0##
Homework Equations
The Attempt at a Solution
if the first constraint is met...
This all makes sense, ill keep it in mind on the rest of this chapter. Also i'll probably make a note next to the constraint to clarify that I think we can bring at most one snack. Thank you.
Ok so the constraint would be ##x_1 + x_2 + x_3 \le 3x_4## This means if we don't bring the can opener, then ##x_4 = 0##. Therefore ##x_1, x_2, x_3 = 0##. I was hesitant to use this because I thought there would be cases where we brought the unopennable items without the can opener, so their...
I think if say.. I brought 2 identical toothbrushes, then they would just count as 2 separate items with the same weight/volume/value. In total we get ##n## items, but I don't think they have to be unique. The values/weights/volumes aren't specified.
Homework Statement
8 (a) (The Knapsack Problem) A backpacker's knapsack has a volume of V in.^3 and can hold up to W lb of gear. The backpacker has a choice of ##n## items to carry in it, with the ##i##th item requiring ##a_i## in.^3 of space, weighing ##w_i## lb, and providing ##c_i## units of...
That is my mistake, I see that −∞ is not in (−∞, ∞). What I meant was "If an interval is approaching −∞, can it also be bounded below?". I'm sorry for the carelessness.
Homework Statement
Consider the sets below. For each one, decide whether the set is bounded above. If it is, give the supremum in ##\mathbb{R}##. Then decide whether or not the set is bounded below. If it is, give the infimum. Finally, decide whether or not the supremum is a maximum, and...
So basically, if a = -|m| - 200 ##\epsilon## A and a < m, it follows that there is no lower bound, because for all elements m ##\epsilon## A, there exists an a ##\epsilon## A, such that a < m.
edit: had some questions.. but rereading your post a few times answered them...
Homework Statement
Prove that {##x \epsilon \mathbb{R} : x^2 \ge 1##} is "not" bounded below.
EDIT: I Looked closely and realized there is a "not" that we all had to write in...sorry if you lost some time..
Homework Equations
Defintion: We say a nonempty subset ##A## of ##\mathbb{R}## is...
Ok, and to show minimum we would do this:
Suppose ##h## is a minimum of ##(0,2)##. Then ##0 < h < 2## by definition of minimum. But ##0 < \frac {h}{2} < h < 2##. Thus h is not a minimum, a contradiction. We conclude that ##(0,2)## does not have a minimum. []
note: for the minimum, we could...
Homework Statement
Give an example of a bounded set that has neither a maximum nor a minimum. (The proof below is given by the book).
We claim that the set ##(0,2)## is bounded and has neither a maximum nor a minimum.
Proof: For each ##x \epsilon (0,2)##, we know that ##0 < x < 2##. Therefore...
Thank you for the suggestion.
So using this and the next post, my transitivity step would look like..
(Transitivity) Suppose a ~ b and b ~ c. Then ##\frac {a^k} {b^k} = 1## and ## \frac {b^n} {c^n} = 1## for some ##k, n \epsilon \mathbb{Z^{+}}##. Raising the first equation by n and the second...
Yea I see your point that the expansion into a + bi is unnecessary.. Here is the updated proof,
Proof:
(Reflexivity) Suppose ##a \epsilon \mathbb{C}##. Then ##a^k = a^k##, so a ~ a. So ~ is reflexive.
(Symmetry) Suppose a ~ b. Then ##a^k = b^k## so ##b^k = a^k##. So b ~ a. So ~ is symmetric...
Homework Statement
If a, b ##\epsilon \mathbb{C}##, we say that ##a## ~ ##b## if and only if ##a^k = b^k## for some positive integer ##k##. Prove that this is an equivalence relation.
Homework Equations
The Attempt at a Solution
Proof:
(Reflexivity): Suppose ##a \epsilon \mathbb{C}##...
Thanks for the response
The z in the antecedent was supposed to be an x.
I understand your second point, I made a(a bunch) of mistakes typing what I had.
To your third point, this was another typo.. here is the corrected version:
original statement to negate: ##\forall x ((x \in \mathbb{Z}...
Homework Statement
Consider the following statement:
##\forall x, ((x \in \mathbb{Z} \wedge \neg(\exists y, (y \in \mathbb{Z} \wedge z = 7y))) \rightarrow (\exists z, (z \in \mathbb{Z} \wedge x = 2z)))##
a)Negate this statement.
b)Write the original statement in English.
c) Which statement is...
I hadn't thought about that,
if we let x-2y = -5 and x-y = -2 then x = 1 and y = 3 which is a solution to the problem. (1-6)(1-3) = (-5)(-2) = 10
if we let x-2y = -10 and x-y = -1, then x = 8 and y = 9 which is a solution. (8-18)(8-9) = (-10)(-1) = 10
I forgot to add my actual answer.. since...
then x-y is larger.. so from my above post, I would automatically know x-y =5 or x-y = 10
Ok so if x-y = 10 then x-2y = 1. Solving these equations gives x = 19 and y = 9. We confirm this by (19-9)(19-2(9)) = 10*1 = 10. So this is one solution.
If we let x-y = 5 then x-2y = 5. Solving these...
Thanks for the response, I think you did the proving when made x=8 and y=3.
I made a mistake.. i'm not sure what the x intercepts represented when I graphed it..
I think since I couldn't just see it, I could have set ##x-2y = 1,2,5,## or ##10## and then set ##x-y = 1,2,5## or ##10## depending...