Discrete Mathematics

4. Find the domain and range of these functions. Note that in each case, to find the domain, determine the set of elements assigned values by the function.

Let’s first review the relevant definition to solve this exercise. Definition: “If f is a function from A to B, we say that A is the domain of f and B is the codomain of f. If f (a) = b, we say that b is the image of a and a is a preimage

4. Find the domain and range of these functions. Note that in each case, to find the domain, determine the set of elements assigned values by the function. Read More »

9. Find these values

In this exercise, we are asked to calculate the values of the functions floor and ceiling. Definitions: “The floor function is the function that takes as input a real number x, and gives as output the greatest integer less than or equal to x, denoted ⌊x⌋ or floor(x). Similarly, the ceiling function maps x to

9. Find these values Read More »

Prove that ∑(-1/2)^j = [2^(n+1) + (-1)^n]/3×2^n whenever n is a nonnegative integer

In this case, we will use Mathematical Induction. PRINCIPLE OF MATHEMATICAL INDUCTION: “To prove that P(n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: BASIS STEP: We verify that P (1) is true. INDUCTIVE STEP: We show that the conditional statement P (k) → P

Prove that ∑(-1/2)^j = [2^(n+1) + (-1)^n]/3×2^n whenever n is a nonnegative integer Read More »

a) Find a formula for 1/(1×2) + 1/(2×3) + 1/n(n+1) by examining the values of this expression for small values of n. b)Prove the formula you conjectured in part (a)

Let’s solve this exercise! a) Find a formula by examining the values of this expression for small values of n n=0 it is undefined. n=1: 1/(1×2) = 1/2 n=2: 1/(1×2) + 1/(2×3) = 1/2 + 1/6 = 4/6=2/3 n=3: 1/2 + 1/6 +1/12 = 3/4 n=4: 3/4 +1/20 = 4/5 1/(1×2) + 1/(2×3) + 1/n(n+1)

a) Find a formula for 1/(1×2) + 1/(2×3) + 1/n(n+1) by examining the values of this expression for small values of n. b)Prove the formula you conjectured in part (a) Read More »