INFS 501: H/W 5.3.16 - Complete Claim: A B= A B . Proof 1: First we show P A B P A P B by taking x P A B . Now, x P A B means x A B . Thus, x A B A , so x A , and x P A . Likewise, x P B . Thus, x P A P B . Since x P A B was arbitrary, we have proven P A
INFS 501 - Solution to 5.2.4 Problem 5.2.4 asks us to fill in the missing parts of an "is an element of" proof of an obvious fact about sets - It's obvious when you draw a picture. She is asking for a rigorous proof, using the definitions of union and sub
INFS 501 Problem 7.4.17
Suppose f : X Y and g : Y Z are functions and g f is onto.
Question 1: Must g be onto? Answer 1: Yes. Proof: Choose any z Z. Since g f is onto, we can find x X such that g f x = z. This makes f x Y the pre-image we need under g bec
INFS 501 - Solution to Problem 8.1.14 Claim: Define d k = 3k 2k , k such that k 0. Then,
d k = 5d k 1 6d k 2 k such that k 2.
Proof: d k 1 and d k 2 are defined because k 2.
Applying the definition of d * to d k1 and d k 2 , 5d k 16d k2 = 5 3k 1 2k 1 6 3k
INFS 501 - Hint on 4.1.16 Note that in the sequence: a1 , a2 , a3 , . , ak , . = 3, 6, 12, 24, 48, 96 , . , 32k 1 , . , terms are doubled as you move from left to right. In such a situation, start with the initial approximation: ak 2k . correctly at k = 1
INFS 501 - Problem 10.4.27
Problem: Express gcd(1568,4158) as a linear combination of 1,568 & 4,158 Step 1 x 1,568 x 1,022 x 546 x 476 x 70 x 56 x 14 Step 2 so, 1022 = 4158 - 2 x 1568 so, 546 = 1568 - 1 x 1022 so, 476 = 1022 - 1 x 546 so, 70 = 546 - 1 x 4
INFS 501 - 10.4.20 Solution Step 1: Translate "Welcome" to its numeric equivalent: W E L C O M E -> -> -> -> -> -> -> 23 5 12 3 15 13 5.
Step 2: Encode those numeric equivalents using RSA(55,3): W E L C O M E -> -> -> -> -> -> -> 23 5 12 3 15 13 5 -> -> -
INFS 501 10.2.19 Here x A y if and only if x A y x = y. This problem seems
hard only because of the formalities using the definitions. Actually, reflexivity, symmetry, and transitivity follow simply from the fact that equality between numbers is reflexive
INFS 501 10.2.13 Consider the circle relation x , y , x C y x2 y 2 = 1. Remember we may use the alternative notation, x , y C , instead of
Is C Reflexive? Answer: No, C is not reflexive. For example, 02 02 = 0 1, so 0, 0 C. For another example, 152
INFS501 Homework 8.2.2.d Method 1: Match the problem to the formula in Chapter 4 by reversing the summation order, then factoring out 1n : 1 n 1 n 12 . 2n 3 2 n2 2n 1 2n = 1 n[ 1 2 4 . 2 n ] , a geometric sequence with r = -2,
2 n 1
INFS 501 - Problem 5.2.13 The following is a proof that for all sets A, B, and C, if A B , then A C B C. Proof: Suppose A, B, and C are any sets and A B. Let x A C. By the definition of the union A C , either: (i) x A , or ii) x C.
In the first case, x A
INFS 501 - Problem 5.2.9 Suppose A, B, and C are sets. Show: A B C B = A C B. (*) Solution 1: The textbook solution is a logical but lengthy and, for the novice, a confusing 2-step process: 1st, show the set on the left hand side (LHS) of (*) is a subset