Mathematics Homework Solutions

Show a pair has a midpoint with integer coordinates

Let p = {(x1, y1), (x2, y2), (x3, y3), (x4, y4), (x5, y5)} be a set of five distinct points in the plane , each of which has integer coordinates. Show that some pair has a midpoint that has integer coordinates.

Probability: Birthdays on the Same Day

Determine the number of people needed to ensure that the probability at least two of them have the same day of the year as their birthday is at least 70 percent, at least 80 percent, at 90 percent, 95 percent, at least 98 percent, and at least 99 percent.

Application of Set Theoretic Model of Sequences

1. Using the set-theoretic model of sequences, define the following operators, giving their syntax and their semantics: (a) overwrite: given any sequence, s, over a set X, any element, e, of X and any non-zero natural number, n, return a sequence identical to s except that the element at position n is e. For example, overwri ...continues

Sets: Element of Q

Show that if a,b is an element of Q then a+b is an element of Q and ab is an element of Q.

Proof that 3^(1/2) is not rational

Prove that 3^(1/2) is not rational.

Perfect matching

Let G be a bipartite graph partitioned into vertex sets V and W. Assume all vertices have the same degree. Show G has a perfect matching.

Relationship between feasible potentials and negative dicycles.

Let G be a directed graph where c_e is the cost of arc e. If the nodes of G can be assigned feasible potentials then G has no negative dicycle.

Discrete optimization

Please see the attached file for full problem description.

Knapsack Problem : Write in Canonical Form

Let [EQUATION1] with [EQUATION2] and [EQUATION3]. The idea is to write each such set in some simple canonical form. (i) When n = 2, how many distinct knapsack sets are there? Write them out in a canonical form with integral coefficients and 1 = [EQUATION4]. (ii) Repeat for n = 3 with [EQUATION5]. *(For proper equations an ...continues

Discrete Math - Set Theory / Identity / Proof : (AUB)∩(AUB^c) = A

Prove : (AUB)∩(AUB^c) = A A union B intersection

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