Let’s refresh the relevant definition we need to know to solve this exercise. “A function f from A to B is called onto, or a surjection, if and only if for every element b∈B there is an element a∈A with f (a)=b. A function f is called surjective if it is onto.” Discrete Mathematics and […]
11. Which functions in Exercise 10 are onto? Read More »
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 »
Let’s revisit relevant definitions that will help us solve this exercise. “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
7. Find the domain and range of these functions Read More »
To solve this type of exercise, first, we need to know what is a function. “Let A and B be nonempty sets. A function f from A to B is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of
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 »
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)
To make this proof, we will use the principle of 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
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 (k + 1) is true for all positive
We will use the mathematical induction to answer b). So, let’s start with the definition of the principle of 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
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 (k + 1) is true for all positive
