# Proof By Contradiction Examples And Solutions

It is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well. Here I introduce you to, two other methods of proof. Therefore n is even. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. Proofs and refutations: standard techniques for constructing proofs; counter-examples. A number of things could be wrong. These videos go through the basics of each of the topics with so. For a more detailed explanation, please first read the Theory Guides above. This contradicts the assumption. METHODS OF PROOF 74 2. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. Use proof by contradiction to show that for all integers n, 3 n + 2 is not divisible by 3. , p ⇒ q is proved. The three forms are (1) (Direct) If n2 is even, n is even. This method assumes that the statement is false and then shows that this leads to something we know to be false (a contradiction). Proof root is irrational by contradiction (this is mentioned. 7 pg 91 # 27. Apr 27, 2020 - CA Geometry: Proof by contradiction Video | EduRev is made by best teachers of. Lecture sheet; version with solutions. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. This is true. That's what proofs are about in mathematics and in computer science. If logic is inconsistent then proof by contradiction is still very much a valid rule of reasoning, but so is its negation, and the rule which says that from $1 + 1 = 2$ we can conclude that you are the next pope. Then n = 2k for some k. contradiction proofs tend to be less convincing and harder to write than direct proofs or proofs by contrapositive. Finding a contradiction means that your assumption is false and therefore the statement is true. This and along with the direct proof on Friday complete an example of proof of an "if and only if" statement. The proof on the board. It is usually not as neat as a two-column proof but is far easier to organize. Then n2 = 2m + 1, so by definition n2 is even. Then (x + y)(x y) = 1, so x y and x + y are divisors of 1. In these cases, when you assume the contrary, you negate the original. For example, to prove that ot all triangles are obtuse", we give the following counter example: the equilateral triangle having all angles equal to sixty. 2 Equivalent Statements. Do the same for an iterative algorithm. Let M = N + 1. (Direct) If n2 = 0, then n = 0 and n is even. Here are a couple examples of proofs by contradiction: Example. It is not clear how to prove it directly since we can not con-. 4 EXAM 2 SOLUTIONS Problem 22. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m’ > m • m and w are not blocking Case #1 and #2 exhaust space. In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. The command \newtheorem{theorem}{Theorem} has two parameters, the first one is the name of the environment that is defined, the second one is the word that will be printed, in boldface font, at the beginning of the environment. If n+1 objects are put into n boxes, then at least one box contains two or more objects. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Proof by Contradiction is another important proof technique. The contrapositive of the above statement is: If x is not even, then x 2 is not even. basics, and foundations Discrete Math - 17 Direct Proof This is the first of several videos exploring methods of proof. " Sir Arthur Conan Doyle. Copious Examples of Proofs 19 Rewrite it in each of the three forms and prove each. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. Proof by Contradiction. An example is "Prove that the product of two nonzero real numbers is nonzero. A direct proof will attempt to lay out the shortest number of steps between p and q. Chapter 3 Foundations of Geometry 1: Points, Lines, Segments, Angles 3. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. An alternative proof is obtained by excluding all possible ways in which the propositions may fail to be equivalent. John 19:14)—Was Jesus crucified in the third hour or the sixth hour? Problem: Mark’s Gospel account says that it was the third hour (9 a. But if a/b = √ 2, then a 2 = 2b 2. Example: A Diophantine Equation. 3 Review the proof techniques on page 116−−118 Here is a result that is proved by three diﬀerent proof techniques. ----- ----- EXAMPLE 2 Give a proof by contradiction of the theorem "If 3n + 2 is odd, then n is odd. Thus admitting one solution gives rise to an infinite descent, so there can be no solutions. This problem is taken from the Putnam competition and is a good example for demonstrating logical thinking and mathematical proof. Proof by contradiction is often used when you wish to prove the impossibility of something. ” Solution: We give a proof by contradiction. This is also known as proof by assuming the opposite. 8 (a) Prove that if n is even, then (3n)2 is even. Example 1 Prove there is no largest prime, i. In this video we will focus on direct proof by assuming "p" is true, then Discrete Math Section 1. Thus the quality of your solution is at least as great as that of any other solution. Examples of a contradiction include an anti-absorbent sponge, jumbo shrimp, and painful pain injections. This document draws some content from each of the following. Solution: Suppose √2 is rational. Proof: Suppose not. Fill in the truth table for ((A implies B) and (B implies C) implies (A implies C)). This is the technique of proof by maximal counterexample, in this case applied to perfect matchings in very dense graphs. Thus, 3n + 2 is even. 4- Bacic Proof Methods I- Direct Proof, Proof by Cases, and Proof by Working Backward In this section we will introduce specific types or methods of proof of mathematical statements. If logic is inconsistent then proof by contradiction is still very much a valid rule of reasoning, but so is its negation, and the rule which says that from $1 + 1 = 2$ we can conclude that you are the next pope. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Prove that Proof: By contradiction, we obtain Suppose , then (given). In logic, it is a fundamental law- the law of non contradiction- that a statement and its denial cannot both be true at the same time. Just as Gillman’s proof has variations, which are based on grouping larger collections of terms, so there are variations on Cusumano’s. Example: A Diophantine Equation. A proof by contradiction also known as an indirect proof, establishes the truth of a given proposition by the supposition that it is false and the subsequent drawing of a conclusion that is contradictory to something that is proven to be true. Then you manipulate and simplify, and try to rearrange things to get the right. The statement P1 says that x1 = 1 < 4, which is true. Relation between Proof by Contradiction and Proof by Contraposition 2) proof by contradiction, you suppose there is an x in D such that P (x) and ~Q (x). Section 4-7 : The Mean Value Theorem. We argue by contradiction. A classic proof: $\sqrt{2}$ is irrational. The solution (c7=8 etc. 2 Quadratic Inequalities. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. Which proof technique? Direct proof –express x2 as 2k for some k, i. What does this language mean? In Example 1, we are saying that the inequality 2n >n2 holds for each choice of the. Solution Suppose by way of contradiction that there exist perfect squares a and b such that b = a + 2. This is a contradiction because there are a total of N objects. Examples In mathematics Irrationality of the square root of 2. Proof by Contradiction. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. Thus the quality of your solution is at least as great as that of any other solution. Chapter 6: Formal Proofs and Boolean Logic The Fitch program, like the system F, uses "introduction" and "elimination" rules. The more work you show the easier it will be to assign partial credit. The proof is carried out by using the procedure outlined in subsection H1. Example: A Diophantine Equation. So let's just assume that a rational times an irrational gives us a rational number. But this one it true because for x<0 x+ 1 x <0 and 0 <2. Example of a Proof by Contradiction Theorem 4. via self-evident rules, and however, in other areas of human activity, the notion of a "proof" has a much wider interpretation [1]. Since we have shown that Sq \Fis true, it follows that the contrapositive T \qalso holds. 2 Incorrect variable use: don’t use the same variable name twice for two different variables. 9 ∀x [(Cube(x) ∧ Large(x)) ∨ (Tet(x) ∧ Small(x))] ∀x [Tet(x) → BackOf(x, c)]. Then p 2 is a rational number, so it can be expressed in the form p q, where pand qare integers which are not both even. Solving it explicitly came later. ” Then use other things you know to try to reach a. Mathematical Proofs: A Transition to Advanced Mathematics, 4th Edition introduces students to proof techniques, analyzing proofs, and writing proofs of their own that are not only mathematically correct but clearly written. If it were rational, it would be expressible as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. Sometimes the negation of a statement is easier to disprove (leads to a contradiction) than the original statement is to prove. Example of a constructive proof: Suppose we are to prove 9n2N;nis equal to the sum of its proper divisors: Proof: Let n= 6. Since a statement is true or false, all statements therefore belong in the set of true statements. The proof is by contradiction. Thus, It is not raining. Before looking at this proof, there are a few definitions we will need to know in order to. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. Closed Form Identities 6 5. particlar example graph, a minimax path is P = 1 2 5 8 11 12 with maximum altitude 5. The material is organized around five types of thinking: logical, relational, recursive, quantitative, and analytical. Proof by Contradiction aka reductio ad absurdum, i. Problem 27 (due Fri 4/3): Prove (using proof by contradiction) that for any sets A and B, we have A$$B nA) = ;. Proof: (direct proof) Assume that n is an even integer. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. Some examples of how to define a recursive function. solution's quality. We would rather proof the contrapositive: x<0 implies x+ 1 x <2. In this case, there are in nitely. The only way out of this situation is that the assumption was wrong. Solution:. identical to that which must follow in your solution) follows. The first known proof is believed to have been given by the Greek philosopher and mathematician Thales. Main Steps After describing your algorithm, the 3 main steps for a greedy exchange argument proof are as follows:. Another way to write is using its equivalence, which is Example: Given A and B are sets satisfying. H1: Introduction to Proof by Contradiction. Example 11 Show that 3√2 is irrational. Proof by Contradiction: (AKA reductio ad absurdum). In all the elementary examples, there are only two options (eg rational/irrational, infinite/finite), so you assume the opposite, show it cannot be true and then conclude the result. It explores properties of odd, even and consecutive numbers- both numerically and algebraically- and also covers properties of the sum, difference and product of these numbers, again these are explored numerically and algebraically. lm 2=so can assume 2 2 4m l= 22 2ln = so n is even. Thus, the statement is proved using an indirect proof. When proving an IF AND ONLY IF proof directly, you must make sure that the equivalence you are proving holds in all steps of the proof. (b) Prove that the square root of 3 is irrational. Example: A Diophantine Equation. It includes disproof by counterexample, proof by deduction, proof by exhaustion and proof by contradiction, with examples for each. Then n = 2k for some k. " Solution: Let p be "3n + 2 is odd" and q be "n is odd. Unlike the earlier examples, I will not describe the thought process that lead to the proof; in each case, I followed the basic outline on page 7. Cube(b) ∧ a = b 2. Example #2. It explains the standard “moves” in mathematical proofs: direct computation, expanding definitions, proof by contradiction, proof by induction, as well as choosing notation and strategies. So before moving on to the next chapter, let’s try our hand at some informal proofs. ] Suppose not. This resource is designed for UK teachers. " Begin the proof with "Assume that a ≠ 0 and b ≠ 0. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. In summary:. You can put this solution on YOUR website! "The product of a non-zero rational number and an irrational number is irrational. Relation between Proof by Contradiction and Proof by Contraposition As an example, here is a proof by contradiction of Proposition 4. 8 (a) Prove that if n is even, then (3n)2 is even. Then P being false implies something that. What does this language mean? In Example 1, we are saying that the inequality 2n >n2 holds for each choice of the. Examples In mathematics Irrationality of the square root of 2. If it leads to a contradiction, then the statement must be true. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Prove by contradiction that there do not exist integers mand nsuch that 14m+ 21n= 100 Proof: We give a proof by contradiction. Worked solutions for some past years Regents Exam. An Indirect Proof. Show the following, and please show all steps: Prove that an odd integer minus an even integer is odd. One well-known use of this method is in the proof that \sqrt{2} is irrational. 2 2 2 4n l= Proof by Contradiction Theorem: is irrational. You can put this solution on YOUR website!. 2 More Methods of Proof A proof by contradiction establishes that p is true by assuming that p is false and arriving at a contradiction, which is any proposition of the form r ^:r. Daniel Solow’s How to Read and Do Proofs begins with the simpler methods of mathematical proof-writing and gradually works toward the more advanced techniques typically presented in an introduction to advanced mathematics. Again, we do not offer this example as the best proof of this fact about even and odd numbers, but rather it is a simple illustration of a proof by contradiction. This works because if \(C$$ is a contradiction and $$\neg P \to. The proof is carried out by using the procedure outlined in subsection H1. This means that the Indirect Proof has been accomplished: by showing that the assumption led to a self-contradiction, one has shown that the assumption was false, and hence that its negation (the conclusion) is true. Type 2: To prove p → q Assume p and ¬q are true. Contradictive Proof Example Prove the following: No odd integer can be expressed as the sum of three even integers. Either the triangles are congruent or they are not. 1 Conjunction rules Conjunction Elimination (∧ Elim). Thus, 3n + 2 is even. " For example, the set E above is the set of all values the expression 2 nthat satisfy the rule 2 Z. Proof time. The idea of proving by contradiction is: we ﬂrst. 1 Proof by example: a universal statement cannot be proved by giving an example. Example 4 is the classic proof of this kind. I understand the theory and concept of Proof by Contradiction however I don't understand where to begin. A PowerPoint covering the Proof section of the new A-level (both years). 2 Incorrect variable use: don’t use the same variable name twice for two different variables. learn geometry proofs and how to use CPCTC, Two-Column Proofs, FlowChart Proofs and Proof by Contradiction, videos, worksheets, games and activities that are suitable for Grade 9 & 10, examples and step by step solutions, complete two column proofs from word problems, Using flowcharts in proofs for Geometry, How to write an Indirect Proof or Proof by Contradiction. 3 - Proof by Contrapositive proof by contrapositive A proof by contrapositive proves a conditional theorem of the form p → c by showing that the contrapositive ¬c → ¬p is true. Proof for other series. A historical example. Then we discussed an alternative to the direct proof, proof by contradiction. Concepts that you will need to know for the Regents Math - Algebra, Geometry, Measurement, Probability, Statistics, Trigonometry. Print Proof by Contradiction: Definition & Examples Worksheet 1. The next group of rules deals with the Boolean connectives contradiction. Once a mathematical statement has been proved with a rigorous argument, it counts as true throughout the universe and for all time. nThese have the following structure: ¥Start with the given fact(s). A number of things could be wrong. This is really a special case of proof by contrapositive (where your \if" is all of mathematics, and your \then" is the statement you are trying to prove). Suppose that there were some x 2Z so that 2x3 + 6x+ 1 = 0: Re-arranging, this implies that 1 = 2x3 6x = 2( x3 3x): Since x3 3x is an integer, this implies that 1 is even, which is obviously not true. The Proof Page. Proof by contradiction (also known as indirect proof or the method of reductio ad absurdum) is a technique which can be used to prove any kind of statement. Section 4-7 : The Mean Value Theorem. One typical application is to show that a given equation has no solutions. The (Pedagogically) First Induction Proof 4 3. I p divides both x =p1 p2 pk and q, and divides x q, I =)pjx q =)p x q. representative-case proof. It explores properties of odd, even and consecutive numbers- both numerically and algebraically- and also covers properties of the sum, difference and product of these numbers, again these are explored numerically and algebraically. 2 2 2 4n l= Proof by Contradiction Theorem: is irrational. We will use the following well known facts : : 1. then ﬁnd a logical contradiction stemming from this assumption. CLARK Contents 1. An easy-to-use guide that shows how to read, understand, and do proofs. Call this integer n. Shows how and when to use each technique such as the contrapositive, induction and proof by contradiction. That is, suppose there is an integer n. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m’ > m • m and w are not blocking Case #1 and #2 exhaust space. Hence, from the proof by cases, (r ∨ s ∨ t) ⇒ q is proved, i. Many of the statements we prove have the form P )Q which, when negated, has the form P )˘Q. 2 Exercise 21) Let n= abbe the product of positive integers a and b. If it were rational, it would be expressible as a fraction a/b in lowest terms, where a and b are integers, at least one of which is odd. It is a logical law that IF A THEN B is always equivalent to IF NOT B THEN NOT A (this is called the contrapositive, and is the basis to proof by contrapositive), so A ONLY IF B is equivalent to IF A THEN B as well. This is a contradiction as x and y should be positive. This page is for the new specification (first teaching 2017): including revision videos, exam questions and model solutions. We obtain the desired conclusion in both cases, so the original statement is true. We have to prove 3√2 is irrational Let us assume the opposite, i. To prove a statement P is true, we begin by assuming P false and show that this leads to a contradiction; something that always false. The idea of the proof is really quite simple. Here we will look at some examples of proofs and non-proofs. PROOFS BY INDUCTION PER ALEXANDERSSON Introduction This is a collection of various proofs using induction. Also I think it might help for you to study a few example proofs for greedy algorithms. Prove that Proof: By contradiction, we obtain Suppose , then (given). 6 Example 1 - Solution Proof: [We take the negation of the theorem and suppose it to be true. For example E ˘ ' 2 n: 2 Z " ˘ ' n : n isaneveninteger " ˘ ' n : n ˘ 2k,k 2 Z ". We have n3 n= (n 1)n(n+ 1). We will use proof by contradiction. Since we have shown that Sq \Fis true, it follows that the contrapositive T \qalso holds. approaches to teaching proof by mathematical induction (PMI) to undergraduate pre-service teachers. 1 Ifyouconsidertheexamplesofproofsinthelastsection,youwillnoticethatsometermsandrulesofinferenceare speciﬁctothesubjectmatterathand. com/site/tlmaths314/ Like my Facebook Page: https://www. If you try to do this, you will find that if you make your hexagon very large, then you can get somewhat close to. A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. So let's look hard at the above example. The Mathematician's Toolbox. ()): Assume [a] = [b]. re·duc·ti·o·nes ad absurdum Disproof of a proposition by showing that it leads to absurd or untenable conclusions. We can prove A is not true by finding a counter example. Then b = b1 = b(ac) = (ab)c = [0] c = 0 : But this contradicts our original hypothesis that b is a nonzero solution of ax = [0]. However, the principle of explosion ( ex falso quodlibet ) has been accepted in some varieties of constructive mathematics, including intuitionism. it must be that Bis true, and we have a proof by contradiction. Outline Theorem 2. Proof by Contradiction (Example 1) •Show that if 3n + 2 is an odd integer, then n is odd. Thus, the statement is proved using an indirect proof. Then (x y) = (x + y) = 1. But this is clearly impossible, since n2 is even. Before looking at this proof, there are a few definitions we will need to know in order to. He supposed there were a finite number and showed that that led to an absurdity — just as we’ve done in our examples. Proof by Exhaustion. The "proof" by josgarithmetic" is wrong starting from his second line. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. ] (12) So, there exists p,q such that: v = p 2 w = q 2 (13) And, we have our. (Otherwise, it would be zero everywhere. Proof by contradiction in logic and mathematics is a proof that determines the truth of a statement by assuming the proposition is false, then working to show its falsity until the result of that assumption is a contradiction. Now this is a contradiction since the left hand side is odd, but the right side is even. Because x is positive, we can multiply. Then x = (y z)(y +z). Indirect Proofs. The proof is by contradiction. n2 odd ⇒ n odd For (1), if n is odd, it is of the form 2k + 1. Prove that if aand bare real numbers with aa. The idea behind proof by contradiction is that a statement must be. Navigate all of my videos at https://sites. In this example it all seems a bit long winded to prove something so obvious, but in more complicated examples it is useful to state exactly what we are assuming and where our contradiction is found. Thus the quality of your solution is at least as great as that of any other solution. Part III: More on Proof. lm 2=so can assume 2 2 4m l= 22 2ln = so n is even. Proof is the primary vehicle for knowledge generation in mathematics. State that the proof is by contradiction. Below are several more examples of this proof strategy. For example E ˘ ' 2 n: 2 Z " ˘ ' n : n isaneveninteger " ˘ ' n : n ˘ 2k,k 2 Z ". , the further out you must go for the approximation to be valid within ǫ. A direct proof, or even a proof of the contrapositive, may seem more satisfying. Start of proof: Let \(n$$ be an integer. Assume $$n$$ is a multiple of 3. Example -1 Show that at least four of any 22 days must fall on the same day of the week. Proof by Contradiction Date: 04/29/2003 at 07:07:29 From: Ajay Subject: Proof by Contradiction Is there any specific mathematical theory that states that Proof by Contradiction is a valid proof? E. There can be many ways to express the same set. One Theorem of Graph. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I Proof by Contrapositive (Indirect Proofs) I Proof by Contradiction I Proof by Cases I Proofs of equivalence I Existence Proofs (Constructive & Nonconstructive) I Uniqueness Proofs Trivial Proofs I (Not trivial as in \easy") Trivial proofs : conclusion holds without using the hypothesis. This is true. Therefore, when the proof contradicts itself, it proves that the opposite must be true. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m' > m • m and w are not blocking Case #1 and #2 exhaust space. Every integer greater than one has a prime divisor. These videos go through the basics of each of the topics with so. Example2 1. ) X is B Example: Let’s think about an example. \Help! I don’t know how to write a proof!" Well, did anyone ever tell you what a proof is, and how to go about writing one? Maybe not. A Simple Proof by Contradiction Theorem: If n2 is even, then n is even. Then there exists integers a and b. course of its proof. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Watch more videos and sign up for a FREE. Solution We formulate this statement as an. 1 Writing mathematics - Exercise Solutions 3 4. Use the method of proof by contradiction to prove the following statements. In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. Aristotle’s discussion of the principle of non-contradiction also raises thorny issues in many areas of modern philosophy, for example, questions about what we are committed to by our beliefs, the relationship between language, thought and the world, and the status of transcendental arguments. This example illustrates an alternative to using truth tables to establish the equiv-alence of two propositions. Problem Set 4 Solutions Section 3. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. Then this even number N is a multiple of 2. Such examples are called counter examples. Maths Genie - A Level Maths revision page. With such a wide target area, that's often a much easier task. sqrt(2) is irrational is normally proved using contradiction. Villanova CSC 1300 - Dr Papalaskari Proofs, examples, and counterexamples ∃x P(x) For existential statements: • A single example suffices to prove the theorem (constructive proof). Lecture Slides By Adil Aslam 32. 4 Some Words of Advice. Theorem (Euclid): For positive numbers a and b (with a > 2 b) the quadratic equation x^2 + b^2 = a x has a solution. Justify all of your decisions as clearly as possible. For all natural numbers n, n3 nis a multiple of 3. Here are a couple examples of proofs by contradiction: Example. 3 - Proof by Contrapositive proof by contrapositive A proof by contrapositive proves a conditional theorem of the form p → c by showing that the contrapositive ¬c → ¬p is true. Prove that if x is even, then x2 + 3 is odd. A classic proof: $\sqrt{2}$ is irrational. Proposition. Prove that A[B = A\B if and only if A = B. proof in terms of induction. [We must deduce the contradiction. Theorem: There is no greatest integer. If 3n+2 is odd then. Start studying [3] Proofs. A collection of videos that cover most topics on the Leaving Cert Higher Level Maths course. Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Direct Proof: Example Theorem: 1 + 2 +h3 +rÉ + n =e n(n+1. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. The solution (c7=8 etc. (2) (Contrapositive) If n is odd, n2 is odd. The more work you show the easier it will be to assign partial credit. It is a contradiction of rational numbers but is a type of real numbers. This is a contradiction as x and y should be positive. (Otherwise, it would be zero everywhere. have no common factors (see Chapter 4). The X-Wing prooves that gk8!= 7. 1) using proof by contradiction, one follows an indirect route: derive r Ù Ør, then conclude that (7. The correct proof is this: Let assume that the product of two odd numbers, m and n, is an even number N: N = m*n. 104 Proof by Contradiction 6. Proof: Assume 0 < c < d. I love you and I don't love you. (2) 6) Using proof by contradiction show that there are no positive integer solutions to the Diophantine equation − =. Of course we can't just have HALTS simulate P on input D, since if P doesn't halt, we'll never know exactly when to quit the simulation and answer no. 4 (a) Prove that A ⊆ B iﬀ A∩B = A. Category: Mathematics This interactive excel resource illustrates a number of proofs. Mathematical proof is the gold standard of knowledge. By De Morgans law, we have a jb and a j(b+ 1). Multi-level views of proofs that hide or reveal details as required ─ indispensible for writing longer proofs! Top Symbolic Logic and The Basic Methods of Proof. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. [1 mark] Assume positive integer solutions. Suppose by contradiction that there is a greatest even integer. methods of proof and reasoning in a single document that might help new (and indeed continuing) students to gain a deeper understanding of how we write good proofs and present clear and logical mathematics. X∞ n=1 1 1+ √ n. I said we will do it through a proof by contradiction. A complete chapter is dedicated to the different methods of proof such as forward direct proofs, proof by contrapositive, proof by contradiction, mathematical induction, and existence proofs. Example Theorem: For every integer x, if x2is even, then x is even. Formal Proofs A proof is equivalent to establishing a logical implication chain Given premises (hypotheses) h1 , h2 , … , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing h1 ∧h2 ∧… ∧hn ⇒c MSU/CSE 260 Fall 2009 6 Formal Proof. There are infinitely many prime numbers. nThese have the following structure: ¥Start with the given fact(s). ] Suppose not. Statement Reason. The simplicity. Example: Parity Here is a simple example that illustrates the method. Simply put, we assume that the math statement is false and then show that this will lead to a contradiction. Example: Prove that if 푛푛 is an integer and 푛푛 3 + 5 is odd, then 푛푛 is even using a. Assume that [~Prove] is true. For proof by contradiction, suppose there are positive integers, greater than one, with no prime divisors. Among 13 people there are two who have their birthdays in the same month. You can find examples of proofs by contradiction in Theorem RREFU, Theorem NMUS, Theorem NPNT, Theorem TTMI, Theorem GSP, Theorem ELIS, Theorem EDYES, Theorem EMHE, Theorem EDELI, and Theorem DMFE, in addition to several examples and. Proof by Exhaustion. Formally, this \indirect proof" method is justiﬂed by the logical equivalence: p · ((:p)! (r ^:r)): Prove for all integers n, if n2 is divisible by 5 then so is n. " For example, the set E above is the set of all values the expression 2 nthat satisfy the rule 2 Z. Villanova CSC 1300 - Dr Papalaskari Proofs, examples, and counterexamples ∃x P(x) For existential statements: • A single example suffices to prove the theorem (constructive proof). Proof by Contradiction. Still, there seems to be no way to avoid proof by contradiction. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. (noun) An example of proof is someone returning to eat at the same restaurant many times showing they enjoy the food. Suppose L were regular. Therefore y = 0, contradicting that it is positive. Both of these methods are called constructive proofs of existence. This means a b is in lowest terms. Proof by contradiction: Assume P(x) is true but Q(x) is false. PRACTICE EXAM 1 SOLUTIONS Problem 1. Example: ! Prove that an integer n is even, if n2 is even. Quiz Proof #4 February 19, 2018 Theorem 1. , everyone in the earth is male But, no number of examples supporting a theorem is sufficient to. , 3 is rational Hence, 3 can be written in the form / where a and b (b 0) are co-prime (no common factor other than 1) Hence, 3 = / 3 b = a Squaring both sides ( 3b)2 = a2 3b2 = a2 ^2/3 = b2 Hence, 3 divides a2 So, 3 shall divide a also Hence, we can say /3 = c where c is some. Solution: By contradiction. I love you and I don't love you. Prove that if x is even, then x2 + 3 is odd. " Indirect Proof (Proof by Contradiction) of the statement: Assume the opposite of what you want to prove, and show it leads to a contradiction of a known fact. Here is a proof found off a very nice math history website. Then we discussed an alternative to the direct proof, proof by contradiction. Prove the statement using a proof by contradiction. Why can't we use one counterexample as the contradiction to the contradicting statement? Example: Let a statement be A where a-->b. Another way to write up the above proof is: Since seven numbers are selected, the Pigeonhole Principle guarantees that two of them are selected from one of the six sets {1,11},{2,10},{3,9}, {4,8}, {5,7},{6}. Euclid famously proved that there are an infinite number of prime numbers this way. n m =2 mn =2 22 2 mn = so m is even. This means that each step in the proof must. Nice introduction to the concept of recursion in terms of programming. I said we will do it through a proof by contradiction. But, from the parity property, we know that an integer is not odd if, and only if, it is. Start of proof: Assume, for the sake of contradiction, that there are integers $$x$$ and $$y$$ such that $$x$$ is a prime greater than 5 and $$x = 6y + 3\text{. Let s = 0 1 in L. Assume the triangles are congruent and reason to a contradiction. Tindle, who. Solution We formulate this statement as an. To prove that L is not a regular language, we will use a proof by contradiction. For starters, let's negate our original statement: The sum of two even numbers is not always even. The idea behind proof by contradiction is that a statement must be. Then, by Lemma 4, there is a such that, up to a subsequence, in. Given: ΔABC is scalene. Solutions to In-Class Problems Week 1, Fri. A proof by contradiction starts with assuming :q (and p). (4) The method of mathematical induction is necessary to prove some theorems that we studied so far in this course. If we wanted to prove the following statement using proof by contradiction, what assumption would we start our proof with?. Suppose for the sake of contradiction there exist a;b 2Z with a 2, and for which it is not true that a - b or a - (b+ 1). Bhaskara's First Proof Bhaskara's proof is also a dissection proof. Proof by contradiction. Bhaskara was born in India. From this assumption, p 2 can be writ-ten in terms of a b, where a and b have no common factor. Then let p be the pumping length given by the pumping lemma. Example 4: Prove the following statement by contradiction: For all integers n, if n 2 is odd, then n is odd. Proof: This is easy to prove by induction. Through step-by-step worked solutions to exam questions available in the Online Study Pack we cover everything you need to know about Proof by Contradiction to pass your final exam. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. Proof by contradiction "When you have eliminated the impossible, what ever remains, however improbable must be the truth. The classic example is the following proof that the square root of 2 is irrational: 1. We will use simple ideas from algebraic topology to show that there exists such that provides an example to prove Theorem 1. • This amounts to proving ¬Y ⇒ ¬X 1 Example Theorem n is odd iﬀ (in and only if) n2 is odd, for n ∈ Z. Before looking at this proof, there are a few definitions we will need to know in order to. For proof by contradictionsuppose not P: Therefore C and not C; completing the proof. 1provides an optimal solution for the fractional knapsack problem. It turns out we won’t need to resort to classical logic for this theorem, but just to make things easier in our first pass we’ll go ahead and use it. Example of proof by contradiction and more on proof by induction. 2 Proof (by contradiction): Want to prove both m and n are even. Deduce that if the hypotheses are true, the conclusion must be true too. Inequalities. Example 1 In the following videos I show you how to use mathematical induction to prove the sum of the shown series Example 2. Solution We formulate this statement as an. He supposed there were a finite number and showed that that led to an absurdity — just as we've done in our examples. Proofs by Contradiction 2. In addition, the author has supplied many clear and detailed algorithms that outline these proofs. , everyone in the earth is male But, no number of examples supporting a theorem is sufficient to. Specifically, the way they teach both Proof by Contradiction and Proof by Mathematical Induction, two techniques that are vital to any upper level analysis/algebra/geometry class, is phenomenal. 2 Nonconstructive: we do not nd a witness a directly. In the following, I cover only a single example, which combines induction with the common proof technique of proof by contradiction. Lecture Slides By Adil Aslam 32. [We take the negation of the given statement and suppose it to be true. The possible truth values of a statement are often given in a table, called a truth table. However, there is an approach that is vaguely similar to disproving by counter-example, called proof by contradiction. We have and thus. Although a direct proof can be given, we choose to prove this statement by contraposition. For example, the statement "the equation 4x^2-y^2 = 1 has no integer solutions for x and y" has a simple contradiction proof. This document draws some content from each of the following. A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. Example from the text: square root of 2 is irrational ; Careful: When using proof by contradiction, mistakes can lead to apparent contradictions. That is, there is a natural number x and natural numbers y and z such that x = y2 z2. I am looking for some examples of when proof by contradiction is used in a problem with more than one case. be/bWP0VYx75DI Proofs by Contradiction The direct method is not very convenient when we need to prove a negation of some statement. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. The Gödel number of formula \(\forall x_0 ( eg \in x_0 x_0)$$ using (*) is. , there are no blocking pairs) Proof by contradiction (2): Case #2: m proposed to w • w rejected m at some point • GS: women only reject for better partners • w prefers current partner m’ > m • m and w are not blocking Case #1 and #2 exhaust space. Relation between Proof by Contradiction and Proof by Contraposition. First, we'll look at it in the propositional case, then in the first-order case. (3) (Contradiction) If n2 is even and n is odd, then n2 is odd. Indirect Proof is foolproof. Prove that if u is an odd integer, then the equation x2 + x u = 0 has no solution that is an integer. Example 4: Prove the following statement by contradiction: For all integers n, if n 2 is odd, then n is odd. The proof of this corollary illustrates an important technique called 'proof by contradiction'. Proof by contradiction "When you have eliminated the impossible, what ever remains, however improbable must be the truth. What is interesting, is that Atiyah was not directly looking at the Riemann Hypothesis, but was studying something else. TheCartesianProduct8 1. Problem 28 (due Fri 4/10): Prove that x = 3 is the unique solution to x 2 + 9 = 6x. The possible truth values of a statement are often given in a table, called a truth table. This A Level Maths video takes you through a new method of proof called proof by contradiction. EXAMPLE 4 Use of Contradiction in Real Life Use an indirect proof to prove the following statement. The proof of this result provides a proof of the sine rule that is independent of the proof given in the module, Further Trigonometry. The number 2 is a prime number. The third part provides more examples of common proofs, such as proving non-conditional statements, proofs involving sets, and disproving statements, and also introduces mathematical induction. For starters, let's negate our original statement: The sum of two even numbers is not always even. A proof by contradiction might be useful if the statement of a theorem is a negation--- for example, the theorem says that a certain thing doesn't exist, that an object doesn't have a certain property, or that something can't happen. Thus this element x belongs to A∪B but does not belong to B. In computer science, proof has found an additional use: verifying that a particular system (or component, or algorithm) has certain desirable properties. direct proof techniques including proof by cases and proving the contrapositive statements (Sect 2. Hence, we can represent it as R\Q, where the backward slash symbol denotes ‘set minus’ or it can also be denoted as R – Q, which means set of real numbers minus set of rational numbers. 9 ∀x [(Cube(x) ∧ Large(x)) ∨ (Tet(x) ∧ Small(x))] ∀x [Tet(x) → BackOf(x, c)]. Practice questions Use the following figure to answer the questions regarding this indirect proof. Therefore a 2 must be even. We will use a proof by contradiction. If you can do that, that example is called a. A classic proof by contradiction from mathematics is the proof that the square root of 2 is irrational. All the same principles apply, however. Often proof by contradiction has the form. n m =2 mn =2 22 2 mn = so m is even. For example E ˘ ' 2 n: 2 Z " ˘ ' n : n isaneveninteger " ˘ ' n : n ˘ 2k,k 2 Z ". Another common way of writingitis E ˘ ' n2Z:n iseven ". He supposed there were a finite number and showed that that led to an absurdity — just as we’ve done in our examples. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. TheCartesianProduct8 1. Inequalities 10 7. Then n= 2k. for many problems there may be many di erent optimal solutions. Below are several more examples of this proof strategy. This is true. If we can prove that $$\neg P$$ leads to a contradiction, then the only conclusion is that $$\neg P$$ is false, so $$P$$ is true. Proof: By contradiction; assume n2 is even but n is odd. Nice introduction to the concept of recursion in terms of programming. Alternatively, you can do a proof by contradiction: As-sume that Y is false, and show that X is false. A PowerPoint covering the Proof section of the new A-level (both years). Example of a Proof by Contradiction Theorem 4. Then the total number of objects is at most $1+1+\cdots+1=n$, a contradiction. This is a "proof by contradiction", a reductio ad absurdum. There are infinitely many prime numbers. If you make an assumption, and that assumption produces a statement that does not make sense, then you must conclude that your assumption is wrong. We will prove this by contradiction. Inequalities 10 7. This is proof by contradiction. For example, — n is always divisible by 3" n(n + 1)„ "The sum of the first n integers is The first of these makes a different statement for each natural number n. These videos go through the basics of each of the topics with so. This method assumes that the statement is false and then shows that this leads to something we know to be false (a contradiction). proof by cases/enumeration proof by chain of i s proof by contradiction proof by contrapositive For any algorithm, we must prove that it always returns the desired output for all legal instances of the problem. But this is clearly impossible, since n2 is even. Then N ≥ n, for every integer n. If linearly dependent, then. This method sets out to prove a proposition P by assuming it is false and deriving a contradiction. Files included (2) Proof questions. Content Accuracy rating: 5 The content is accurate, error-free, and unbiased. Part III: More on Proof. Prove that if lim n→∞ a n b n = 0 then P a n is convergent. For sorting, this means even if the input is already sorted or it contains repeated elements. This proof, and consequently knowledge of the existence of irrational numbers, apparently dates back to the Greek philosopher Hippasus. ] Suppose not. 3 Proof by contradiction We end with a description of proof by contradiction. (You should give direct proof!) If A 6⊂B, then there an element x ∈ A but x/∈ B. Run M on hPi. Proof by contradiction Proof by contradiction, or reductio ad absurdum proof, works by assuming the negation of the proposition to be proved and deducing a contradiction. 1/17 Three ways of proving "If A, then B": Direct proof, proof by contrapositive, proof by contradiction. (2) 6) Using proof by contradiction show that there are no positive integer solutions to the Diophantine equation − =. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Ex: p∧~p Claim:Suppose c is a contradiction. Choose s to be 0p1p. Reading, Discovering and Writing Proofs Version 0. [1 mark] Assume positive integer solutions. Simple proof by contradiction. Let’s take a look at two examples. This should be straightforward. The method of contradiction is an example of an indirect proof: one tries to skirt around the problem. Proof by contradiction makes some people uneasy—it seems a little like magic, perhaps because throughout the proof we appear to be `proving' false statements. This completes the proof. Solution: Suppose √2 is rational. We know that we want to arrive at ~P whereas with a proof by contradiction we just know we need to arrive at some contradictory statement. NEGATION 3 We have seen that p and q are statements, where p has truth value T and q has truth value F. Formal Proofs A proof is equivalent to establishing a logical implication chain Given premises (hypotheses) h1 , h2 , … , hn and conclusion c, to give a formal proof that the hypotheses imply the conclusion, entails establishing h1 ∧h2 ∧… ∧hn ⇒c MSU/CSE 260 Fall 2009 6 Formal Proof. Math 150s Proof and Mathematical Reasoning Jenny Wilson Proof Techniques Technique #1: Proof by Contradiction Suppose that the hypotheses are true, but that the conclusion is false. Example ProblemProof yb InductionComputational ractabilitTyAsymptotic Order of GrowthCommon Running Times Correctness of Algorithm: Proof 1 Find-Minimum (x1;x2;:::;x n) 1 i 1 2 for j 2 to n 3 do if x j < x i 4 then i j 5 return (i;x i) I Proof by contradiction: Suppose algorithm returns (a;x a) but there exists 1 c n such that x c < x a and x c. Proof by contradiction is often used when you wish to prove the impossibility of something.
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# How does one fit the curve $y = ae^{bx} + c$?

How does one fit the curve $y = ae^{bx} + c$?

The method of taking logarithms of both sides does not simplify to allow linear regression.

I can take the three equations derived from setting the gradient to zero and solve for $a$ and $c$ in terms of $b$, but then I'm left with a non-linear equation in $b$ which I would have to solve numerically.

Is there a better way? It seems like this is a trivial modification to the case where $c$ is zero...

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I would say that's actually the best way to do your equation, since you have exploited the fact that your parameters $a$ and $c$ are linear parameters in your model. (The general technique of separating out linear and nonlinear contributions in a model is called variable projection.) Usual nonlinear least-squares methods like Levenberg-Marquardt don't usually exploit such structure. Look at it this way: instead of having to solve for three nonlinear parameters (which is a more difficult problem), you are left with the much easier task of solving for only one unknown. – J. M. May 30 '12 at 14:04

Given the overall structure of your question, I'll assume that you have given data and by "fit the curve", you mean to find values of $a$, $b$, and $c$ so that the function $ae^{bt}+c$ is a good fit to that data.

In general, given data $\{x_i,y_i\}_{i=1}^n$ and a function univariate function $f_{a,b,c}(x)$ that depends on parameters $a$, $b$, and $c$, we fit the data by finding values of $a$, $b$, and $c$ that minimize $$\sum_{i=1}^n (f_{a,b,c}(x_i) - y_i)^2.$$ In the case where the expression $f_{a,b,c}(x)$ is linear in the parameters $a$, $b$, and $c$, this is a linear optimization problem and nice matrix methods can be applied. Otherwise, it's a non-linear optimization problem. Sometimes, this non-linear problem can be translated to a linear problem but sometimes strictly non-linear techniques must be used.

As an example, you might try the following input in WolframAlpha:

FindFit[{{-3,-1},{-2,0},{-1,0},{0,1},{1,2},{2,4},{3,8}},
a*exp(b*t)+c, {a,b,c}, t]

You should find that $f(t)=1.74*e^{0.54t}-0.96$ is a reasonable fit to this data. The result is given as a numerical approximation (decimal numbers, rather than exact), because numerical techniques are used. A plot of the function and the data looks like so:

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A straightforward method (no need for initial guessed values, no iterative process) is shown with numerical example in pages 16-18 in the paper "Régressions et équations intégrales" published on Scribd :

http://www.scribd.com/JJacquelin/documents

With the data set given by Mark McClure, the result is shown on the figure below. The fitting of the curve to the data is quite the same, although the values of the parameters are slightly different. For practical use, the difference is negigible. This small discripency is a consequence of the too low number of experimental points.

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# Given two blank rulers, measure any length

Say, you are given ruler $A$ of length $72.84 \text{ cm}$ and ruler $B$ of length $86.63\text{ cm}$. Neither of them have any marking/gradation of any sort on them. They are blank, except for their total length written on them.

Using only A and B, can you measure a length of $31.23\text{ cm}$?

Also, is it possible to generalize this concept further?

That is to say:

Given any two blank rulers, measure any given length.

• An observation which may or may not be helpful: $\gcd (7284, 8663) = 1$ Jul 17, 2018 at 17:24
• I think a better title would be "measure a given length". No pair of rulers can measure every length. Jul 17, 2018 at 20:26
• @BallpointBen do you mean irrational and complex measures?
– Nick
Jul 18, 2018 at 4:13
• Jul 18, 2018 at 18:00

Yes. It is Possible

Here's a useful fact: If $a$ and $b$ are integers with $\gcd(a,b) = d$, then there exist integers $x$ and $y$ such that $ax + by =d$. In fact, one can compute exactly what $x$ and $y$ are by the extended Euclidean algorithm.

In this case, we have $$3539 \times 8663 - 4209 \times 7284 = 1$$

Or, $$3539 \times 86.63 - 4209 \times 72.84 = 0.01$$

So, in theory, you could measure $0.01$ cm by marking off $3539 \times 86.63$ cm along a line, and then marking off $4209 \times 72.84$ cm along the same line; the difference in the markings will be $0.01$ cm.

Of course, now that you can measure $0.01$ cm, you can measure any multiple thereof.

For your generalization, given two blank rulers, you can measure any length that is a multiple of the gcd of their lengths. (Make sure you choose units where the lengths of the rulers are integers. Here, we chose 0.01 cm)

Edit: As noted by Silverfish in the comments, the interesting fact I mention above is Bézout's identity

• This question gives a great picture for the extended euclid's algorithm that i've never seen before. Jul 17, 2018 at 18:03
• A question out of curiosity: how did you find out that $3539\times 8663 - 4208\times 7284 = -1$? Did you search it with a computer program, did you have a hitch, did you already know...? Jul 17, 2018 at 20:41
• @AndreaDiBiagio I'd assume it was calculated through the linked extended Euclidean algorithm. Which might well have been, itself, run through a computer program, but it wasn't a "search".
– Nic
Jul 17, 2018 at 21:24
• @NicHartley makes total sense. I didn’t read through the answer properly. Jul 17, 2018 at 21:43
• For "here's a useful fact" it might be worth stating its name - it's Bézout's identity Jul 17, 2018 at 23:28

This all boils down to (using 0.01 cm as the unit) as

Does $\gcd(7284,8663)$ divide 3123?

The answer is yes: the gcd turns out to be 1. The generalisation is obvious: two blank rulers of lengths $a$ and $b$ units ($a,b$ are natural numbers) can measure any length that is a multiple of $\gcd(a,b)$ units.

• Actually, if a/b is rational and therefore gcd(a,b) exists then you can measure any multiple of gcd(a,b) exactly. If a/b is irrational then you can approximate any number with arbitrary precision, but not exactly. Jul 18, 2018 at 22:26

The other answers are in the affirmative and consider measuring multiples of the GCD of the two rulers, or $0.01\text{ cm}$ in the original problem.

For really small problems, it would be better to have an irrational ratio between the two lengths, e.g. a ruler of length $1$ and a ruler of length $\pi$. One can get multiples of $\pi$ arbitrarily close to the integers and thus construct basic units of measurement arbitrarily small.

Doing so, you can measure ANY non-negative real length to ANY desired positive degree of accuracy. Contrast such a ruler with the $0.01\text{ cm}$ ruler mentioned above which can only measure any non-negative real within $0.01\text{ cm}$ of accuracy ($0.005$ really, but there's an uncertainty in the measurement in which side it is actually closer to, so you wind up with up to $0.01$ from the further side).

• brilliant. I should have asked a ruler length $\pi$ and another of length $e$. I didn't think about irrationals.
– Nick
Jul 18, 2018 at 7:41
• Nice observation. A fancy way to say this is that the subgroup of $(\mathbb R,+)$ generated by $\{a, b\}$ is discrete if $\frac{a}{b}\in \mathbb Q$ and dense otherwise. Jul 18, 2018 at 12:22
• Note it’s not proven that $\pi/e$ is irrational. Jul 18, 2018 at 22:11
• @RomanOdaisky True, but either $\frac{e\pi+1}{e}$ or $\frac\pi e$ must be irrational. Jul 18, 2018 at 22:54
• @HansMusgrave Irrelevant — in the (highly unlikely) event $\pi/e$ is rational, $\pi$ and $e$ would have a GCD and better accuracy than the GCD would be impossible. Jul 19, 2018 at 14:31

We can write it as the equation $7284y - 8663x = 3123$

or $y = \frac{8663x + 3123}{7284}$

From this we can see that $8663x\equiv 7284-3123 \mod 7284$

$8663x \equiv 4161 \mod 7284$

$8663 \equiv 1379 \mod 7284$

So, $x \equiv \frac{4161}{1379} \mod 7284$

$x \equiv 4161\cdot \frac{1}{1379} \mod 7284$

$x = 4161\cdot 3539 = 14725779$........ where $3539$ is the multiplicative inverse of $1379$

So, $x = 14725779$ and $y = 17513650$

So, if we measure out $17513650$ lengths of the $72.84$ stick and then measure back with $14725779$ lengths of the $86.63$ stick, we achieve a distance of $31.23$ from our starting point.

• At just over 12.5 thousand km, I’m not sure the answer to the question is “yes”. Jul 17, 2018 at 21:20
• $5727 \times 72.84 - 4815 \times 86.63 = 31.23$ as well; you can always subtract $k \times 8663$ instances of the short stick and $k \times 7284$ instances of the long stick to get the same total length, and for this $k = 2021$ works. Granted this is still $4\text{km}$ or so of measuring to get something about $1/12000$ that length... Jul 17, 2018 at 21:37
• I took your solution and got $x \% 8663$ and $y \% 7284$. Note that obviously $8663 \times 72.84 - 7284 \times 86.63 = 0$, so adding or removing that many copies of each stick will not change the final length at all. Actually, if we go one step further, removing more sticks than are there and measuring in the other direction $2469 \times 86.63 - 2936 \times 72.84$ also works and is somewhat smaller! Jul 18, 2018 at 5:50
• @Ian I suppose you could sequence the positive and negative measurements cleverly to avoid going around the world, you go back and forth through the origin instead. In fact you should be able to do it given no more than a couple hundred meters of space. Jul 18, 2018 at 16:50
• I meant a couple hundred centimeters. I could not see the units while commenting. Could not correct the comment due to the time limit, which major sucks. Jul 18, 2018 at 16:59


# Math Help - Need Help With a Tangent Line Problem

1. ## Need Help With a Tangent Line Problem

1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)

2) Determine where this line intersects the y-axis

2. Originally Posted by erimat89

1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)

2) Determine where this line intersects the y-axis
First find $y'$:

$y'=2Ax$.

Thus, the slope of the line at $x=c$ is $2Ac$.

Thus, the tangent has the equation $y-Ac^2=2Ac(x-c)\implies y=2Acx-Ac^2$

Now find where it crosses the x axis:

$0=2Acx-Ac^2\implies x=\dots$.

It crosses the y axis when x=0.

So $y=\dots$

Can you take it from here?

--Chris

Maybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?

4. Originally Posted by erimat89

1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)

2) Determine where this line intersects the y-axis
Originally Posted by erimat89
Maybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?
It asks you to show that the line intersects the x-axis at the point (c/2,0).

Since I got the equation of the line to be $y=2Acx-Ac^2$, at y=0, the line crosses the x-axis.

Thus $0=2Acx-Ac^2\implies 2Acx=Ac^2\implies x=\frac{Ac^2}{2Ac}\implies x=\frac{c}{2}$. So we see that it crosses the x-axis at $\left(\tfrac{1}{2}c,0\right)$

$\mathbb{Q.E.D.}$

Now, it intersects the y axis when x=0. Thus, we see that $y=2Ac(0)-Ac^2\implies y=-Ac^2$.

Thus, the y intercept is $\left(0,-Ac^2\right)$.

Does this make sense?

--Chris

5. ## Thanks

Makes alot of sense you clarified things for me. I've never been good with any sort of math problem that involves words for some reason I get thrown off.


# Characterization of convex set with empty interior on Hilbert spaces

Is the following statement true?

"Let $H$ be a Hilbert space and $C\subset H$ a convex set. If $C$ has empty interior then there exist $a$ and a proper subspace $V\subset H$ such that $C\subset(a+V)$."

I guess this is false due to the following counterexample. Let $$C=\{(x_n)_{n\in\mathbb N}:|x_n|\leq 1/2^n,\ \forall n\in\mathbb N\}\subset \ell_2(\mathbb N).$$ Clearly, $C$ is convex. Moreover, it is easy to see that $C^\perp=\{0\}$, and consequently, there are no proper subspace $V$ and $a$ such that $C\subset(a+V)$.

Finally, given some $(y_n)_{n\in\mathbb N}\in C$ and $r>0$, fix $n_0$ such that $1/2^{n_0}<r$. Then $$|y_{n_0+2}-1/2^{n_0}|\geq1/2^{n_0}-1/2^{n_0+2}>1/2^{n_0+2}.$$ Define $$z_n=\left\{\begin{array}{r} y_n,\ if\ n\neq n_0+2 \\ y_{n_0+2}-1/2^{n_0},\ if\ n= n_0+2\end{array}\right..$$ Then $(z_n)_{n\in\mathbb{N}}\in B((y_n)_{n\in\mathbb N},r)$ and $(z_n)_{n\in\mathbb{N}}\notin C$. This way we proved that no ball is contained in $C$, so $C$ has empty interior.

Is everything correct? Am I missing something here?

• There are easier counterexamples. What about the line $x=1$ in $\mathbb{R}^2$? – ChocolateAndCheese Mar 28 '17 at 23:55
• @ChocolateAndCheese That's true! I'm so sorry, there was a mistake in my question. Actually I meant "in the translation of a proper subspace of $H$". – André Porto Mar 29 '17 at 0:16
• Your proof that int (C) is empty is fine. – DanielWainfleet Mar 29 '17 at 6:58

Your counterexample is correct, assuming that "subspace" (as often) means a closed subspace of $H$. Otherwise, it would not work because the set $C$ is contained in $\ell^1(\mathbb{N})$ which is a dense subspace of $\ell^2(\mathbb{N})$. So I proceed by assuming the subspaces are closed.

The justification is slightly lacking in the following: the property $C^\perp = \{0\}$ only implies that $C$ is not contained in any proper linear subspace; it does not preclude $C$ from being contained in a proper affine subspace. For example, the set $A=e_1+e_1^\perp$, which is an affine hyperplane, satisfies $A^\perp = \{0\}$ since its linear span is all of $\ell^2$.

However, the above is easy to repair: since $C$ contains $0$, any affine subspace containing it would be a linear subspace.

### A simpler example

Let $C$ be the set of all sequences $x$ such that $x_n=0$ except for finitely many $n$. Then $C$ is convex and has empty interior, since adding an arbitrarily small multiple of the vector $(1/2^n)_{n\in\mathbb{N}}$ to an element of $C$ takes one out of $C$. It's also dense in $\ell^2$, so can't be contained in a proper closed subset of any kind.

(Your example has the additional property of being closed, which however wasn't required.)

• the statement "the property $C^\perp={0}$ only implies that $C$ is not contained in any proper linear subspace" is not true. The counterexample is the set $C$ defined in the counterexample gave by you. Clearly, $C^\perp=\{0\}$ and it is a proper linear subspace of $H$. – André Porto Apr 17 '17 at 19:38
• @AndréPorto See the first paragraph: "I proceed by assuming the subspaces are closed." – user357151 Apr 17 '17 at 19:43
• Yeah, ok then. My bad. – André Porto Apr 17 '17 at 20:37

As @Gerry pointed out, $C^\perp=\{0\}$ does not imply that $C$ is not contained in a proper affine subspace of $H$, so the example I gave in my question may not hold.

Fortunately, I found another example that really works this time. It is a much simpler one and is a counterexample to the statement even when $C$ is assumed to be closed.

Let $C$ be the set of vectors in $\ell_2(\mathbb N)$ with all its coordinates greater than or equal to $0$.

Clearly, $C$ is a closed convex. Let us see that $C$ has empty interior. Given $a=(a_n)_{n\in\mathbb N}\in C$ and $r>0$, since $a_n\to 0$, fix $m$ such that $a_m<r/2$ and define $b=(b_n)_{n\in\mathbb N}$ to be the sequence obtained by changing $a_m$ by $a_m-r/2$. Clearly, $b\in B(a,r)$ and, since $b_m=a_m-r/2<0$, $b\notin C$. Then $B(a,r)$ is not contained in $C$.

Finally, let us see that $C$ is not contained in a proper affine subspace. Pick $a\in \ell_2(\mathbb N)$ and a subspace $V$ such that $C\subset(a+V)$. Since $0\in C$, we get that $a\in V$, so actually $C\subset V$. We will conclude below that $V=\ell_2(\mathbb N)$.

Given $a=(a_n)_{n\in\mathbb N}\in \ell_2(\mathbb N)$, define $a^+=(a^+_n)_{n\in\mathbb N}$ by $$a^+_n=\left\{\begin{array}{r}a_n,\ \mbox{if}\ a_n\geq0 \\ 0,\ \mbox{if}\ a_n<0\end{array}\right..$$ and $a^-=(a^-_n)_{n\in\mathbb N}$ by $$a^-_n=\left\{\begin{array}{r}-a_n,\ \mbox{if}\ a_n<0 \\ 0,\ \mbox{if}\ a_n\geq0\end{array}\right..$$ Then, $a^+,a^-\in C\subset V$ and consequently $a=a^+-a^-\in V$.


# First Order ODEs¶

Last time, after a general introduction, we met the general first order ODE:

$$y' = f(t,y).$$

We also examined the very simple, yet interesting such ODE $y'=r y$ and solved it qualitatively (using a slope field) and symbolically (using separation of variables).

Today, we'll generalize those ideas further and explore more challenging examples.

Much of today's material is explored in sections 1.1.3 and 1.3.1 of our text.

## The logistic equation¶

Just as the exponential equation $y'=ry$ was a motivating example for us last time, the logistic equation will be a motivating example today:

$$y' = r y (1-y/K).$$

The logistic equation can be interpretted as a growth model, just as the exponential equation can. As we'll see, though, the last $(1-y/K)$ term has the effect of slowing the growth down though as the quantity $y$ approaches the carrying capacity $K$.

### Equilibrium solutions¶

Like the exponential equation, the logistic equation is an autonomous equation - not depending explicitly on $t$. When we write an autonomous equation in the form $$y'=f(y),$$ the roots of $f$ form constant or equilibrium solutions of the differential equation.

That is if $y_0$ is a real number with $f(y_0)=0$, then the constant function $y(t) \equiv y_0$ is a solution of the ODE since both sides of $y'=f(y)$ are zero.

Furthermore, if we plot the equilibrium solutions, then they contrain any other solutions.

### Logistic equilibrium solutions¶

The logistic equation $y' = r y (1-y/K)$ has two equilibrium solutions, namely

$$y(t) \equiv 0 \: \text{ and } \: y(t) \equiv K.$$

Let's plot those:

### Logistic solutions¶

Other solutions to the logistic equation are constrained by the equilibrium solutions. Furthermore, bewtween $0$ and $K$, the right side of the logistic equation

$$y' = r y (1-y/K)$$

is positive (assuming $r$ is positive). Thus, we expect a solution to increase from the bottom equilibrium to the top. Solutions must look something like so:

### Logistic slope field¶

We should get a similar sort of picture from a slope field:

## Separation of variables¶

Let's apply separation of variables to solve the logistic equation $$\frac{dy}{dt} = r y (1-y/K)$$ symbolically.

This, is problem #17 from section 1.3.1 of our text.

I guess we'd start with $$\frac{1}{y(1-y/K)}dy = r\,dt.$$

### Separation of variables (cont)¶

The book suggest that we use the partial fractions decomposition

$$\frac{1}{y(1-K/y)} = \frac{1}{y}+\frac{1}{K-y},$$

$$\int\left(\frac{1}{y}+\frac{1}{K-y}\right)dy = \int r\,dt$$

or

$$\log\left(y\right) - \log\left(K-y\right) = r\,t+c_1.$$

### Separation of variables (cont 2)¶

Combining the logarithms in that last equation yields

$$\log\left(\frac{y}{K-y}\right) = r\,t+c_1,$$

which allows us to apply the exponential to both sides to get

$$\frac{y}{K-y} = e^{r\,t+c_1} = c e^{r\,t}.$$

Note that we've pulled the old $e^{a+c_1} = e^ae^{c_1} = e^ac$ trick.

### Separation of variables (cont 3)¶

Now, we can solve that last equation for $y$ to get

$$y = \frac{K\,c\,e^{r\,t}}{1+c\,e^{r\,t}} = \frac{K}{1+\frac{1}{c}e^{-r\,t}}.$$

That is our solution:

$$y(t) = \frac{K}{1+\frac{1}{c}e^{-r\,t}}.$$

### Interpretation¶

Given our solution,

$$y(t) = \frac{K}{1+\frac{1}{c}e^{-r\,t}},$$

It's not hard to see that

$$\lim_{t\to -\infty}y(t) = 0 \: \text{ and } \: \lim_{t\to\infty}y(t)=K$$

and that $y$ increases from $0$ to $K$ as $t$ ranges from $-\infty$ to $\infty$.

That agrees with our qualitative analysis!

### An IVP¶

In order to plot a specific solution, we'll need values for the parameters $r$ and $K$, as well as an initial condition. Let's assume that

$$K=100, \: r=3, \text{ and } y(0)=4.$$

Thus, our solution becomes

$$y(t) = \frac{100}{1+\frac{1}{c}e^{-3\,t}}.$$

### An IVP (cont)¶

To find $c$ in

$$y(t) = \frac{100}{1+\frac{1}{c}e^{-3\,t}},$$

we use the initial condition $y(0)=4$ to get

$$4 = \frac{100}{1+\frac{1}{c}}.$$

We solve this to get $c=1/24$ so that

$$y(t) = \frac{100}{1+24e^{-3\,t}}.$$

Here's how to plot that solution in Python:

import matplotlib.pyplot as plt
import numpy as np
def y(t): return 100/(1+24*np.exp(-3*t))
ts = np.linspace(-1,5)
ys =  y(ts)
plt.plot(ts,ys);

## Slope fields for autonomous equations¶

The logistic equation is somewhat indicative of the way that the general autonomous equation $y'=f(y)$ works.

We first find the equilibrium solutions; then, assuming that $f$ is continous, the slopes don't change sign between the equilibria.

Then, it's easy to sketch the solutions.

### Example¶

For example, here's the slope field for $y'=y^2-1$, together with a solution between the two equilibria of $\pm 1$, which must be decreasing:

### Finally...¶

How about I do one by hand?


# The Union of Two Subspaces is Not a Subspace in a Vector Space

## Problem 274

Let $U$ and $V$ be subspaces of the vector space $\R^n$.
If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$.

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## Proof.

Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but $\mathbf{u} \not \in V$.
Similarly, since $V$ is not contained in $U$, there exists a vector $\mathbf{v} \in V$ but $\mathbf{v} \not \in U$.

Seeking a contradiction, let us assume that the union is $U \cup V$ is a subspace of $\R^n$.
The vectors $\mathbf{u}, \mathbf{v}$ lie in the vector space $U \cup V$.
Thus their sum $\mathbf{u}+\mathbf{v}$ is also in $U\cup V$.
This implies that we have either
$\mathbf{u}+\mathbf{v} \in U \text{ or } \mathbf{u}+\mathbf{v}\in V.$

If $\mathbf{u}+\mathbf{v} \in U$, then there exists $\mathbf{u}’\in U$ such that
$\mathbf{u}+\mathbf{v}=\mathbf{u}’.$ Since the vectors $\mathbf{u}$ and $\mathbf{u}’$ are both in the subspace $U$, their difference $\mathbf{u}’-\mathbf{u}$ is also in $U$. Hence we have
$\mathbf{v}=\mathbf{u}’-\mathbf{u} \in U.$

However, this contradicts the choice of the vector $\mathbf{v} \not \in U$.

Thus, we must have $\mathbf{u}+\mathbf{v}\in V$.
In this case, there exists $\mathbf{v}’ \in V$ such that
$\mathbf{u}+\mathbf{v}=\mathbf{v}’.$

Since both $\mathbf{v}, \mathbf{v}’$ are vectors of $V$, it follows that
$\mathbf{u}=\mathbf{v}’-\mathbf{v}\in V,$ which contradicts the choice of $\mathbf{u} \not\in V$.

Therefore, we have reached a contradiction. Thus, the union $U \cup V$ cannot be a subspace of $\R^n$.

## Related Question.

In fact, the converse of this problem is true.

Problem. Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.

For a proof, see the post “Union of Subspaces is a Subspace if and only if One is Included in Another“.

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##### Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017.

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# If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.

If $$\sec \theta + \tan \theta =x$$, then find the value of $$\sin \theta$$.

$$\sec \theta + \tan \theta = x$$ $$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$ $$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$ $$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$ $$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta$$ $$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$ $$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$

Here is a different approach: Since $1 + \tan^2\theta = \sec^2\theta$, we have $$\sec^2\theta - \tan^2\theta = 1$$ Factoring yields $$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$$ Since we are given that $\sec\theta + \tan\theta = x$, we obtain $$x(\sec\theta - \tan\theta) = 1$$ Therefore, $$\sec\theta - \tan\theta = \frac{1}{x}$$ This yields the system of equations \begin{align*} \sec\theta + \tan\theta & = x \tag{1}\\ \sec\theta - \tan\theta & = \frac{1}{x} \tag{2} \end{align*} Adding equations 1 and 2 and solving for $\sec\theta$ yields \begin{align*} 2\sec\theta & = x + \frac{1}{x}\\ 2\sec\theta & = \frac{x^2 + 1}{x}\\ \sec\theta & = \frac{x^2 + 1}{2x} \end{align*} Therefore, $$\cos\theta = \frac{1}{\sec\theta} = \frac{2x}{x^2 + 1}$$ Subtracting equation 2 from equation 1 and solving for $\tan\theta$ yields \begin{align*} 2\tan\theta & = x - \frac{1}{x}\\ 2\tan\theta & = \frac{x^2 - 1}{x}\\ \tan\theta & = \frac{x^2 - 1}{2x} \end{align*} Thus, $$\sin\theta = \tan\theta\cos\theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}$$

• I didn't get the answer. The answer is $\dfrac {1-x^2}{1+x^2}$. – Aryabhatta May 14 '17 at 13:15
• What did you get for $\sec\theta$ and $\tan\theta$? – N. F. Taussig May 14 '17 at 13:19
• I got $\sec \theta=\dfrac {x^2+1}{2x}$ and $\tan \theta =\dfrac {2x-x^2-1}{2x}$. – Aryabhatta May 14 '17 at 13:21
• I agree with your answer for $\sec\theta$. When I solved for $\tan\theta$, I obtained $$\frac{x^2 - 1}{2x}$$ which led to the answer $$\frac{x^2 - 1}{1 + x^2}$$ – N. F. Taussig May 14 '17 at 13:26
• I have added the details of my calculations. I checked my answer for the angles $\pi/6$, $\pi/4$, and $\pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign. – N. F. Taussig May 14 '17 at 13:50

The equation becomes $$1+\sin\theta=x\cos\theta$$ Set $X=\cos\theta$ and $Y=\sin\theta$, so the equation becomes $$\begin{cases} X^2+Y^2=1 \\[4px] 1+Y=xX \end{cases}$$ Note that $x\ne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting $$(1+Y)^2+x^2Y^2=x^2$$ that simplifies to $$(1+x^2)Y^2+2Y+1-x^2=0$$ that yields $$Y=-1 \qquad\text{or}\qquad Y=\frac{x^2-1}{x^2+1}$$ Is $Y=-1$ a solution for the problem?

By the way, you also get $\cos\theta$, since $$X=\frac{1}{x}(1+Y)=\frac{1}{x}\frac{x^2+1+x^2-1}{x^2+1}=\frac{2x}{x^2+1}$$

From where you are:

You obtained a quadratic function in $\sin(\theta)$. Perform the substitution $u =\sin(\theta)$.

We obtain the quadratic (in $u$):

$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$

$$\Rightarrow u_{1,2} = \frac{- 2 \pm \sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$

$$= \frac{- 2 \pm \sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$

$$= \frac{- 2 \pm \sqrt{4x^4}}{2(x^2+1)}$$

$$= \frac{- 2 \pm 2x^2}{2(x^2+1)}$$

$$= \frac{- 1 \pm x^2}{x^2+1}$$

Therefore,

$$\sin(\theta)_{1,2} = \frac{- 1 \pm x^2}{x^2+1}$$

One of those solutions will not work out.

This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.

• What is discriminant method? – Aryabhatta May 14 '17 at 13:05
• en.wikipedia.org/wiki/Discriminant#Degree_2 – user370967 May 14 '17 at 13:06
• I didn't understand... – Aryabhatta May 14 '17 at 13:14
• If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ – user370967 May 14 '17 at 13:21
• Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$ – user370967 May 14 '17 at 13:22

WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$

$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$