# Fall 2002 CSCI 333 Homework 1 Solution

## Problem 1

For each of the following relations, prove if the relation is reflexive, symmetric, antisymmetric, and transitive. All of these relations are taken over the set of positive integers.

1. For integers a and b, ab if and only if a+b equals 10.

It's not reflexive because ~(3 ≡ 3).
If's not antisymmetric because 3 ≡ 7 and 7 ≡ 3.
It's not transitive because 3 ≡ 7 and 7 ≡ 3 but ~(3 ≡ 3).
It is symmetric because if a+b equals 10 then b+a equals 10.

2. For integers a and b, ab if and only if a+b is greater than 10.

It's not reflexive because ~(3 ≡ 3).
If's not antisymmetric because 3 ≡ 8 and 8 ≡ 3.
It's not transitive because 3 ≡ 8 and 8 ≡ 4 but ~(3 ≡ 4).
It is symmetric because if a+b is greater than 10 then b+a is also greater than 10.

3. For integers a and b, ab if and only if a < b+3.

It's not symmetic because 3 ≡ 7 but ~(7 ≡ 3).
It's not antisymmetric 3 ≡ 4 and 4 ≡ 3.
It's not transitive because 9 ≡ 7 and 7 ≡ 5 but ~9 ≡ 5.
It is reflexive because a < a+3.

4. For integers a and b, ab if and only if a+b is divisible by 3.

It's not reflexive because ~(4 ≡ 4).
If's not antisymmetric because 4 ≡ 8 and 8 ≡ 4.
It's not transitive because 4 ≡ 8 and 8 ≡ 10 but ~(4 ≡ 10).
If is symmetric because if a+b is divisible by 3 the equivalent b+a.

5. For integers a and b, ab if and only if a equals 10.

It's not reflexive because ~(4 ≡ 4).
If's not symmetric because 10 ≡ 4 and ~(4 ≡ 10).
It is antisymmetric because if ab and ba, then both a and b are equal to 10.
It is transitive because if ab and bc, then a must equal 10 and consequently ac.

6. For integers a and b, ab if and only if a*b < a+b.

It's not reflexive because ~(4 ≡ 4).
It's not antisymmetric because 1 ≡ 5 and 5 ≡ 1.
It's not transitive because 10 ≡ 1 and 1 ≡ 3 but ~(10 ≡ 3).
It is symmetric.

## Problem 2

Use induction to prove that the function defined by the following recurrence relation:

• T(0) = 1
• T(n) = T(n-1) + 2, for n ≥ 1

has the closed form solution

• T(n) = 2 n + 1

In the base case, we see that T(0) = 1 = 2*0+1.

In the induction case, we assume T(n) = 2 n + 1. By the recurrence relation T(n+1) = T(n) + 2. By induction this means that T(n+1) = (2 n + 1) + 2 which can be rearranged to obtain T(n+1) = 2 (n + 1) + 1.