Blog

How many solutions are there to the equation  x1 + x2 + x3 + x4 + x5 + x6 = 37, where each xi is a non-negative integer?  

Consider the problem of purchasing cans of juice. Suppose that juice is sold in 2-packs or 5-packs. Let Q(n) be the statement that it is possible to buy n cans of juice by combining 2-packs and 5-packs. This problem discusses using strong induction to prove that for any n ≥ 4, Q(n) is true. 1) The inductive step will prove that for k ≥ 5, Q(k+1) is true. What is the inductive hypothesis? (a) Q(j) is true for any j = 4, 5, …, k. (b) Q(j) is true for any j = 4, 5, …, k+1. (c) Q(j) is true for any j = 5, …, k+1. [Q1]   2) The inductive step will prove that for k ≥ 5, Q(k+1) is true. What part of the inductive hypothesis is used in the proof? (a) Q(k-2) (b) Q(k-1) (c) Q(k) [Q2]   3)In the base case, for which values of n should Q(n) be proven directly? (a) n = 3, n = 4, and n = 5. (b) n = 4 and n = 5. (c) n = 4, n = 5, and n = 6. [Q3]

Determine the truth value of each expression below. The domain is the set of all real numbers. ∀x∃y (xy > 0)  [Q1] ∃x∀y (xy = 0)  [Q2] ∀x∀y∃z (z = (x – y)/3)  [Q3] ∀x∃y∀z (z = (x – y)/3)  [Q4] ∀x∃y y2 = x   [Q5] ∀x∃y (x < 0 ∨ y2 = x)   [Q6] ∃x ∃y (x2 = y2 ∧ x ≠ y)  [Q7] ∀x ∀y (x2 ≠ y2 ∨ |x| = |y|)   [Q8]

Indicate whether each of the following arguments is valid or invalid.   If is an irrational number, then    is an irrational number. is an irrational number.∴   is an irrational number.   [Q1] p ↔ qp ∨ q∴ p   [Q2] ¬(p → q)q → p∴ ¬q    [Q3] q → pp∴ ¬(p → q)    [Q4] The patient has high blood pressure or diabetes or both.The patient has diabetes or high cholesterol or both.∴ The patient has high blood pressure or high cholesterol.  [Q5] ∀x (P(x) → Q(x))∃x ¬P(x)∴ ∃x ¬Q(x)   [Q6]

How many vertices does a full 4-ary tree with 50 internal vertices have?

The following two statements are logically equivalent.  If x is a rational number and y is an irrational number then x-y is an irrational number. If x is a rational number and x-y is an rational number then y is a rational number.

Below are the steps for a proof by contradiction of the following theorem: Theorem: There is no smallest positive real number. Put the steps of the proof in the correct order so that each step follows from previous steps in the proof. (a) This contradicts the assumption that r is the smallest positive real number. (b) Assume there is a smallest positive real number called r. (c) Consider r/2, which is a positive real number since r is positive. (d) Moreover, r/2 < r,  since both are positive. Proof: [Step1]; [Step2]; [Step3]; [Step4].    

Decide whether each logical expression is a tautology, contradiction or neither. (p ∨ q) ∨ (q → p)     [Q1] (p → q) ↔ (p ∧ ¬q)  [Q2] (p → q) ∨ p   [Q3] (¬p ∨ q) ↔ (¬p ∧ q)   [Q4]

Below is an outline of the steps of the proof that  is irrational. Put the steps in the outline in the correct order. (a) Since d2 is even, d is even. (b) Squaring both sides of the equation =n/d leads to n2 = 2d2. (c) Since n is even, and n2 = 2d2, d2 is also even. (d) Then can be written as the ratio n/d of two integers n and d ≠ 0, such that n and d have no common factors. (e) Since n and d are both even,  n and d are both divisible by 2. (f) This a contradiction as n and d have a common factor. Therefore  must be irrational. (g) Since n2 is even, n is also even. (h) Suppose that  is rational. Proof: Step 1 [Step1]; Step 2 [Step2]; Step 3 [Step3]; Step 4 [Step4]; Step 5 [Step5]; Step 6 [Step6]; Step 7 [Step7]; Step 8 [Step8]  

A group of students is selected from a class. Every student in the class is either in the 3rd grade or the 4th grade, and no student is in both the 3rd and the 4th grades. How many students must be selected in order to guarantee that at least five 3rd graders or at least five 4th graders are selected?