Week 4 - Hasse diagrams and logical statements

Tuesday, 12 March 2019

10:51 PM

Another important relation is called a partial order. We say that a relation is a partial order when it has the three characteristic qualities: • every value should be less than or equal to itself • ifa < bandb < athenb = a • ifa < bandb < cthena < c. Typically, we take a set together with partial ordering such as "less than or equal to" S, "subset" C or "divides" l. Together, the partial ordering and set are called a partially ordered set or poset. Instead of an arrow diagram, we use a Hasse Diagram to visualise a poset. This communicates the relation between elements of the set with as few edges as possible. Consider the set Together with the partial ordering C.

 

Make a Hasse Diagram for this poset. Click on two verticies to make or destroy edge between them. Show steps {2} (Your score will not be affected.) {3,4} {3} You have created all of the correct edges. You were awarded 4 marks. You scored 4 marks for this part. Score: 4/4

 

In 1742, Christian Goldbach wrote in a letter to Leonhard Euler that Every even integer greater than 2 can be written as the sum of two primes. This is now called Goldbach's conjecture, and is the oldest and best known unsolved problem in number theory. While the proof has remained elusive, the conjecture has been computer verified for numbers up to 4 x 1018 Prove the following weakened version of the Goldbach conjecture: Every even integer between 12 and 19 is the sum of two primes. The even integers between 12 and 19 are the numbers 12, 14, 16 and 18. Prove that it is possible to write these numbers as a sum of two primes, by actually writing them as sum of two primes: • 12 • 14 16 • 18 5+7 11+5 11+7

 

Gap 1 Good, 12 Gap 2 Good, 14 Gap 3 Good, 16 Gap 4 Good, 18 5 + 7. You were awarded 1 mark. 7 + 7. You were awarded 1 mark. 11 + 5. You were awarded 1 mark. 11 + 7. You were awarded 1 mark. You scored 4 marks for this part. Score: 4/4 Answered

 

In this question we introduce the notion of existence (denoted A) through the familiar concept of odd and even numbers. If n e Z then we say n is even when there exists some integer k such that n = 2k, n is odd when there exists some integer k such that n = 2k -k 1. Prove the following statements. a) Prove 18 is even. By definition 18 is even when there exists some integer k such that 18 because we can show exactly what it is: k 2k. It is easy to demonstrate that there is some integer k with this property, Submit part Gap 1 Your answer is correct. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

b) Prove 7 is odd. By definition 7 is odd when there exists some integer k such that 7 because we can show exactly what it is: k 2k + 1. It is easy to demonstrate that there is some integer k with this property, Submit part Gap 1 Your answer is correct. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

c) There exists some numbers which are even, and some numbers which are odd. But all numbers are even or odd. Formally we write this as e Z, n is even or n is odd. This is a statement about every possible integer. So we can not consider each integer individually and instead we consider an unspecified n e Z. There are an infinite number of individual existance proofs to do here, but we can reduce the workload to two possibilities by considering n (mod 2). If n If n O (mod 2) then Ak e Z such that n = 2k, hence n is even. 1 (mod 2) then Ak e Z such that n 1 + 2k, hence n is odd. Since these are the only possible values forn (mod 2), every integern is either even or odd. Submit part Gap 1 Your answer is correct. You were awarded 1 mark. Gap 2 Your answer is correct. You were awarded 1 mark. You scored 2 marks for this part. Score: 2/2 Answered

 

A function f(x) is said to be unbounded if VIV, ax, > M. Consider the function f(a:) = V"7• a) Before we tackle the general statement. Let's consider some examples. For M = 3 there exists a number such that > M. Prove this by finding such an 3 2 1 32 + I Submit part > 3. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

b) For M = 19, there exists a number such that > M. Prove this by finding such an x. Submit part > 19. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

c) Now prove that f(x) is unbounded. Which is to say that for any M we must show that there exists a number a: such that > M. Select all expressions for that make > M? 1 Submit part You chose a correct answer. You were awarded 1 mark. You chose a correct answer. You were awarded 1 mark. You scored 2 marks for this part. Score: 2/2 Answered

 

d) A function is bounded if it is not unbounded. What logical expression corresponds to being bounded? Submit part You chose a correct answer. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

A consequence of the Mean Value Theorem applied to the parabola f (x) = is that for every interval [a, b] there exists some c e [a, b] such that f(b) — f(a) f'(c) = a) Before we tackle the general statement. Let's consider some examples. For the interval [0, 3], find the value of c whose existence was foretold by the Mean Value Theorem. (Your score will not be affected.) Show steps Submit part You revealed the steps. Your answer is correct. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

b) For the interval [4, 5], find the value of c whose existence was foretold by the Mean Value Theorem. 4.5 Submit part Your answer is correct. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

Machine generated alternative text: c) Now prove that for every interval [a, b] there exists some c e [a, b] such that f(b) f'(c) — b ¯ f(a) a Suppose [a, b] is some unknown interval. There exists some c e [a, b] such that f' 2 , because the value of cis Submit part Your answer is numerically correct. You were awarded 2 marks. You scored 2 marks for this part. Score: 2/2 Answered

 

Machine generated alternative text: The statement 'all integers are prime' can be expressed formally as Vn e Z, n is prime. Of course, this is false. We can prove this by contradiction or by proving that the negation is true. The negation of 'all integers are prime' is 'some integers are not prime', which can be expressed formally as An e Z, nisnotprime. a) Enter a positive integer (less than 100) which is not prime, and threby prove with counterexample that not all integers are prime. Submit part The number 4 is not prime because it is divisible by 2. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

Machine generated alternative text: b) Which expression is the negation of Vm e Q, an e Q, m + n (Your score will not be affected.) Show steps O Vm e Q, an e Q, m + n m x n. @ 3m e Q, Vn e Q, m + n m x n. 0 3m e Q, Vn e Q, m + n m x n. O Vrn e Q, An e n Submit part You chose a correct answer. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

Machine generated alternative text: c) The statement is false, and we can prove this by contradiction. Suppose the statement is true. Then it is true for m But ifwe subtract n this means that 1 O 1, which means there exists an n e Q such that e, which is the contradiction we were expecting. Submit part Gap 1 Your answer is correct. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

Machine generated alternative text: d) Prove that the expression Vm e Q is true. Consider some fixed m e Q where m 1. Then choose An e Q, m n 1 This choice of n has the descired property that m + n m/ (m—l) m m x n, because m + n(l mn = 0 When you are doing this yourself it is helpful to start with m + n = m x n and then deduce n. But when presenting your proof it is important that you preserves the order of the expression you are proving. Submit part Gap 1 Your answer is numerically correct. You were awarded 1 mark. You scored 1 mark for this part. Score: 1/1 Answered

 

 

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