208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.2 Terms and Symbols 1 How to Read and Do Proofs [1] Symbols mostly used in mathematical proofs are as follows: 1.1 Introduction Symbol ⇒ ⇔ ∈ ⊆ ∅ ∼ ∀ ∃ ∋ A proof is a convincing argument expressed in the language of mathematics. Given two statements A and B, each of which may be either true or false, a fundamental problem of interest in mathematics is to show that the following statement, called an implication is true: If A is true, then B is true. If A, then B. A implies B. A ⇒ B. : ∧ ∨ Statement A is called hypothesis and B is called conclusion. The truth table of “A implies B” is as follows: A B A⇒ B True True False False True False True False True False True True Meaning implies if and only if is an element of subset empty set not for all (for each, for any, for every) there is (there are, there exists) such that such that and or Q.E.D. (which was to be determined) Some useful terms are given as follows: Axiom -- A statement whose truth is accepted without a proof. Corollary -- A proposition whose truth follows almost immediately from a theorem. Definition -- An agreement, by all parties concerned, as to the meaning of a particular term. Lemma -- A proposition that is used in the proof of a subsequent theorem. Proposition -- A true statement of interest. Theorem -- An important proposition. From the truth table, if we want to prove that A implies B is true. The only case we need to do is to assume that A is true then our job is to conclude that B is true. A statement A is true if and only if B is true A⇔ B is identical to A ⇒ B AND B ⇒ A, therefore its proof involves proving the implication in both directions. 1 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.3 Forward-Backward Method All other proof methods rely on this method. As a general rule, the forward-backward method is probably the first method to try on a problem unless you have reason to use a different approach based on the form of B. The following table summarizes this proof method When to use it As a first attempt, or when B does not have a recognizable form. What to assume What to conclude How to do it A B Solution The backward process starts from asking “how can I conclude that B is true?”. The answer is a triangle is isosceles when its two sides are equal, that is, x = y or x − y = 0. So we have Work forward using the facts given in A and apply the backward process to B by asking “How or when can I conclude that the statement B is true?” B1: x − y = 0. The forward process uses the facts given in A as follows: A1: xy / 2 = z / 4. ♥ Example 1: [1] Prove the following proposition. Proposition 1: If the right triangle XYZ with the sides of lengths x and y 2 and hypotenuse of length z has an area of z / 4 , then the triangle XYZ is isosceles. 2 A2: x + y = z . 2 2 2 ( A3: xy / 2 = x + y 2 2 ) / 4. A4: x − 2 xy + y = 0. 2 2 A5: ( x − y ) = 0. 2 A6: x − y = 0. The condense proof may be given as follows: Proof of Proposition 1: The hypothesis together with the Pythagorean theorem yield x + y = 2 xy and hence 2 2 ( x − y) 2 = 0. Thus the triangle is isosceles, as required. 2 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.4 Construction Method The following table summarizes this proof method When to use it When B has the quantifiers “there is, there are, there exists.” B is in the form: There is an “object” with a “certain property” such that “something happens.” What to assume What to conclude How to do it A That there is the desired object. Construct the object, then show that it has the certain property and that the something happens. Solution ♥ Example 2: [1] Prove the following proposition. Proposition 2: If m < n are consecutive integers and m is even, then 4 B can be rewritten as “There is an integer p such that m + n − 1 = 4 p. Therefore we can use construction method where 2 divides m + n − 1. 2 2 2 Object: p Certain property: an integer 2 2 Something happens: m + n − 1 = 4 p By using the facts in A, we need to find an object p that has the certain property and that something happens. We use the following steps: S1: Let n = m + 1. S2: Then m + n − 1 = m + ( m + 1) − 1 = 2m ( m + 1) . 2 2 2 2 S3: Because m is even, there is an interger k such that m = 2k . S4: Letting p = k ( m + 1) , it follows that m 2 + n 2 − 1 = 2m ( m + 1) = 4k ( m + 1) = 4 p, and so 4 divides m + n − 1. This completes the proof. 2 3 2 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.5 Choose Method The following table summarizes this proof method When to use it When B has the quantifiers “for all, for each, for every, for any.” B is in the form: What to assume What to conclude How to do it For every “object” with a “certain property,” “something happens.” A, and choose a general object with the certain property. That the something happens. Work forward from A and the fact that the general object has the certain property to conclude that something happens. Solution To show that S = T is to show that S is a subset of T and T is a subset of S . In this proof, we will only prove that S is a subset of T . The other direction can be done similarly. S is a subset of T can be rewritten as Be sure that the general object can be replaced with any object to satisfy the “for all” quantifier. “For all elements x ∈ S , x ∈ T . ” ♥ Example 3: [1] Prove the following proposition. Proposition 3: If S and T are the two sets defined by Therefore we can use choose method where Object: x Certain property: x ∈ S Something happens: x ∈ T S = {real numbers x : x 2 − 3 x + 2 ≤ 0} T = {real numbers x :1 ≤ x ≤ 2} , By using the facts in A, we need to find a general object x that has the certain property and that something happens and the general object can be extended to all objects in the set. We use the following steps: then S = T . S1: x − 3 x + 2 ≤ 0. S2: ( x − 2 )( x − 1) ≤ 0. 2 S3: x ≤ 2 and x ≥ 1. S4: Therefore for all elements x ∈ S , x ∈ T . 4 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.6 Specialization Method The following table summarizes this proof method When A has the quantifiers “for all, for each, for every, for any.” When to use it A is in the form: For every “object” with a “certain property,” “something happens.” What to assume What to conclude How to do it A B Identify, in the for-all statement, the object, the certain property, and the something that happens. Look for one particular object to apply specialization to. (Usually it is the same object as when the choose method is used in the backward process.) Verify that this object has the certain property and that something happens for this particular object. Solution From the backward process, we have ♥ Example 4: [1] Use the following definition ∗ ∗ ∗ ∗ B1: v ≤ w and w ≤ v . Definition 1: A real number u is a least upper bound for a set S of real numbers if and only if (1) u is an upper bound for S and (2) for every upper bound v for S , u ≤ v. Using the forward process, we first rewrite A from the Definition 1 ∗ A1: “For every upper bound u for T , v ≤ u. ” ∗ ∗ A3: “For every upper bound u for T , w ≤ u. ” to prove the following proposition. ∗ Proposition 4: If v ∗ ∗ A2: Using specialization method, we have v ≤ w . ∗ A4: Using specialization method, we have w ≤ v . ∗ and w are least upper bounds for a set T , then ∗ ∗ ∗ ∗ A5: Therefore v ≤ w and w ≤ v . v∗ = w∗ . The proof is then completed. 5 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.7 Nested Quantifiers When a statement contains more than one quantifier, we say that the statement has nested quantifiers. When processing such statements, always work from left to right. Solution ♥ Example 5: [1] Use the following definition B is identical to “For every real number y , there is a real number x such that mx + b = y. ” Since B contains nested quantifiers, we process Definition 2: A function f of one variable is onto if and only if for every from left to right to have real number y , there is a real number x such that f ( x ) = y. Object: y Certain property: real number Something happens: there is a real number x such that mx + b = y. to prove the following proposition. Proposition 5: If m and b are real numbers with m ≠ 0, then the function Object: x Certain property: real number Something happens: mx + b = y. f ( x ) = mx + b is onto. Using the forward-backward method, we have B1: For every real number y , there is a real number x such that mx + b = y. A1: Choose a real number y (Choose method.) B2: There is a real number x such that mx + b = y. A2: From m ≠ 0, construct the real number x = ( y − b ) / m (Construction method.) B3: mx + b = y. A3: mx + b = m ( y − b ) / m + b = ( y − b ) + b = y. Since A3 = B3, the forward-backward method is finished and the proof is completed. A condensed proof may look like Proof of Proposition 5: To show that f is onto, let y be a real number. Because, by hypothesis, m ≠ 0, let x = ( y − b ) / m. It is easy to see that f ( x ) = mx + b = y, and so the proof is complete. 6 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 ♥ Example 6: [1] Prove the following proposition. Proposition 6: If a, b, and c are real numbers with a < 0, then there is a real number y such that for every real number x , ax + bx + c ≤ y. 2 Solution B contains nested quantifiers, we process from left to right to have Object: y Certain property: real number 2 Something happens: for every real number x , ax + bx + c ≤ y. Object x Certain property: real number 2 Something happens: ax + bx + c ≤ y. Using the forward-backward method, we have 4ac − b 2 (Construction method.) 4a 2 B1: For every real number x, ax + bx + c ≤ y. A2: Choose a real number x (Choose method.) 2 B2: ax + bx + c ≤ y. A1: Construct y = A3: It follows that 2 b 4ac − b 2 ax + bx + c = a x + + . 2a 4a A4: Because a < 0, we have 4ac − b 2 ax 2 + bx + c ≤ = y. 4a 2 Since A4 = B2, the forward-backward method is finished and the proof is completed. 7 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 ♥ Example 7: [1] Prove the following proposition. 2 Proposition 7: If r is a real number such that r = 2, then r is irrational. 1.8 Contradiction Method Consider the following truth table A B True True False False True False True False NOT B False True False True NOT A False False True True A⇒ B True False True True NOT B ⇒ NOT A True False True True We can see that the truth of A AND NOT B are opposite to that of A ⇒ B . This leads to a proof method called “contradiction.” The proof begins by assuming that A and NOT B are true and then tries to conclude some contradiction. Proofs done by the contradiction method are shorter and easier than those done by the forward-backward method because you do not have to create the desired object (as in the construction method.) However, a disadvantage of this method is that you do not know exactly what the contradiction is going to be. Another disadvantage is that this method produces no meaningful result since the outcome is some contradiction not a constructed object. The following table summarizes this proof method When to use it Solution The proof contains the following steps. 2 S1: r = 2. S2: r is a rational number. S3: From the definition of a rational number, there are integers p and q with q ≠ 0 such that r = p / q. p and q must have no common divisor. S4: r = p / q . 2 2 2 S5: 2 = p / q . 2 When NOT B gives useful information. 2 S6: 2q = p . 2 2 When the statement B is one of the two possible alternatives. S7: p is even. S8: p is even. S9: p = 2k , for some integer k . What to assume When the statement B contains the key word “no” or “not.” A and NOT B What to conclude How to do it Some contradiction Work forward from A and NOT B to 2 S10: 2q = ( 2k ) = 4k . 2 2 2 S11: q = 2k . 2 2 2 S12: q is even. S13: q is even. S14: Since both p and q are even. They have 2 as their common divisor and we have reached a contradiction with S3. reach a contradiction. 8 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 ♥ Example 8: [1] Prove the following proposition. Proposition 8: Assume that a and b are integers with a ≠ 0. If a does 1.9 Contrapositive Method Consider the following truth table A B True True False False True False True False NOT B False True False True NOT A False False True True not divide b, then ax + bx + b − a has no positive integer root. 2 A⇒ B True False True True NOT B ⇒ NOT A True False True True We can see that NOT B ⇒ NOT A has the same truth as A ⇒ B . This leads to a proof method called “contrapositive.” The contrapositive method assumes that NOT B is true then tries Solution The proof uses the forward-backward method and contains the following steps. to conclude that NOT A is true. The disadvantage of the contrapositive method compared to the contradiction method is that you work forward from only one statement ( NOT B ) instead of two. However, the advantage is that you know what A1: ( NOT B ) x > 0 is an integer with ax + bx + b − a = 0. 2 you are looking for ( NOT A .) Thus, you can apply the forward-backward B1: ( NOT A ) a divides b. method. The following table summarizes this proof method When to use it A2: x= When NOT A and NOT B give useful information. What to conclude How to do it Work forward from NOT B and backward −b ± ( b − 2a ) , 2a b . a B2: There is an integer c such that b = ca. A3: Since x > 0, x = 1 − When the statement A and B contain the key word “no” or “not.” NOT B NOT A 2a = b x = −1 and x = 1 − . a When the statement A and B are one of the two possible alternatives. What to assume −b ± b 2 − 4 a ( b − a ) A4: b = (1 − x ) a, c = 1 − x. from NOT A. Since A4 = B2, the forward-backward method is finished and the proof is completed. 9 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 c) For every real number x between -1 and 1, there is a real number y 1.10 NOT of Statements between -1 and 1 such that x + y ≤ 1. 2 Both the contrapositive and contradiction methods require that you be able to write the NOT of a statement. The following list summarizes the rules for taking the NOT of statements that have a special form. [ ] 2. NOT [ A AND B ] becomes ( NOT A ) OR ( NOT B ) . 3. NOT [ A OR B ] becomes ( NOT A ) AND ( NOT B ) . 4. NOT [ there is an object with a certain property such that something happens ] becomes “For all objects with the certain property, the 2 1. NOT NOT A becomes A. something does not happen.” 5. NOT for all objects with a certain property, something happens [ Solution There is a real number x between -1 and 1 such that, for all real numbers y between -1 and 1, x + y > 1. 2 2 d) There is a real number x between -1 and 1 such that, for all real numbers y between -1 and 1, x + y ≤ 1. 2 2 ] becomes “There is an object with the certain property such that the something does not happen.” Solution For all real numbers x between -1 and 1, there is a real number y between -1 and 1 such that x + y > 1. 2 2 ♥ Example 9: [1] Find the NOT of the following statements. a) For every real number x ≥ 2, x + x − 6 ≥ 0. 2 Solution 2 There is a real number x ≥ 2 such that x + x − 6 < 0. b) There is a real number x ≥ 2 such that x + x − 6 ≥ 0. 2 Solution 2 For all real numbers x ≥ 2, x + x − 6 < 0. 10 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.11 Uniqueness Methods These methods are used when you want to show that there is only one object with a certain property such that something happens. There are two methods: direct and indirect. The following table summarizes the direct uniqueness method. When to use it What to assume What to conclude How to do it When B has the word “unique” in it. There are two such objects, and A. That the two objects are equal. Work forward using A and the properties of the objects. Also work backward to show the objects are equal. ♥ Example 10: [1] Prove the following proposition. Proposition 9: If a, b, c, d , e, and f are real numbers such that ad − bc ≠ 0, then there are unique real numbers x and y such that ax + by = e and cx + dy = f . Solution We need to prove both “there are” and “unique”. We can use the construction method to prove the “there are” part and therefore is omitted here. In this problem, we will only prove uniqueness. Using the direct uniqueness method, we assume that ( x1 , y1 ) and ( x2 , y2 ) are two objects. Using the forward-backward method, we have A1: ax1 + by1 = e and cx1 + dy1 = f . A2: ax2 + by2 = e and cx2 + dy2 = f . B1: ( x1 , y1 ) = ( x2 , y2 ) . B2: x1 − x2 = 0 and y1 − y2 = 0. A3: a ( x1 − x2 ) + b ( y1 − y2 ) = 0, and c ( x1 − x2 ) + d ( y1 − y2 ) = 0. A4: ( ad − bc )( x1 − x2 ) = 0. A5: Because ad − bc ≠ 0, x1 − x2 = 0. A6: Similar derivation leads to y1 − y2 = 0. From B2, A5, and A6, the proof is completed. 11 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 The following table summarizes the indirect uniqueness method. When to use it What to assume What to conclude How to do it When B has the word “unique” in it. There are two different objects, and A. Some contradiction. Work forward from A using the properties of the two objects and the fact that they are different. ♥ Example 11: [1] Prove the following proposition. Proposition 10: If r is a positive real number, then there is a unique real 3 number x such that x = r. Solution We need to prove both “there is” and “unique”. We can use the construction method to prove the “there is” part and therefore is omitted here. In this problem, we will only prove uniqueness. A1: x and y are two different real numbers such that x = r and y = r. 3 3 A2: x = y , x − y = 0. 3 3 3 ( 3 A3: ( x − y ) x + xy + y 2 2 ) = 0. A4: Because x ≠ y , ( x − y ) ≠ 0, therefore x + xy + y = 0. 2 2 − y ± −3 y 2 A5: x = . 2 2 A6: Because x is real, the only possibility is −3 y = 0 , y = 0. A7: r = 0 , which leads to contradiction. 12 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 ♥ Example 12: [1] Prove the following proposition. 1.12 Induction Method Induction is a method for proving that each of the statements in an infinite list is true. Use induction when the statement you try to prove has the form, “For every integer n ≥ n0 , P ( n ) , ” where P ( n ) is some statement that Proposition 11: For every integer n ≥ 1, n ∑k = k =1 n ( n + 1) . 2 depends on n. The key words to look for are “integer” and “ n ≥ n0 .” The method can be summarized as follows. When to use it When a statement P ( n ) is true for each integer n beginning with n0 . What to assume P ( n ) is true for n. What to conclude That P ( n + 1) is true; also show that P ( n0 ) is true. How to do it First show that P ( n0 ) is true by replacing n in P ( n ) by n0 . Write P ( n + 1) and relate P ( n + 1) to Solution P ( n ) then use the fact that P ( n ) is true Assume that P ( n ) : to show that P ( n + 1) is true. n ∑k = k =1 Then, P (1) : Note that induction does not help you to discover the correct form of the statement P ( n ) . Rather, induction only verifies that a given 1 ∑k = k =1 statement P ( n ) is true for all integers n greater than or equal to some Since, P ( n + 1) : initial one. One possible variation on induction is that P ( n ) , P ( n + 1) can be n +1 1(1 + 1) = 1 is true. 2 ∑k = k =1 n ( n + 1) is true. 2 ( n + 1) ( n + 1) + 1 ( n + 1)( n + 2 ) = . 2 2 Using the fact that P ( n ) is true, we can write replaced with P ( n − 1) , P ( n ) or with P ( j ) , P ( n ) , where j < n. P ( n + 1) : 13 n ( n + 1) ( n + 1)( n + 2 ) . n k = + ( n + 1) = ∑ ∑ k + ( n + 1) = 2 2 k =1 k =1 n +1 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 1.13 Elimination Method 1.14 Cases Method The method can be summarized as follows. When to use it The method can be summarized as follows. When B has the form “ C OR D ”, such as, “ A implies C OR D. ” What to assume The keyword to look for is “either/or.” A and NOT C (or A and NOT D ) What to conclude How to do it D (or C ) Work forward from A and NOT C , and When to use it What to assume What to conclude How to do it backward from D. (or work forward from A and NOT D, and backward from C.) When A has the form “ C OR D ”, such as, “ C OR D implies B .” The keyword to look for is “either/or.” Case 1: C Case 2: D Both cases must imply B . First prove that C implies B; then prove that D implies B. ♥ Example 14: [1] Prove the following proposition. Proposition 13: If a is a negative real number, then y = −b / ( 2a ) is a ♥ Example 13: [1] Prove the following proposition. 2 Proposition 12: If x − 5 x + 6 ≥ 0, then x ≤ 2 or x ≥ 3. maximum of the function ax + bx + c. 2 Solution Using elimination method, we have A1: x − 5 x + 6 ≥ 0. A2: ( NOT C ) x > 2. 2 B1: x ≥ 3. A3: ( x − 2 )( x − 3 ) ≥ 0. Solution Because there is no keyword in the proposition, we first use the forward-backward method. A4: ( x − 3 ) ≥ 0. B1: For every real number x, ay + by + c ≥ ax + bx + c. 2 Since A4 = B1, the proof is completed. 14 2 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 Since “For every” appears, we use the choose method. 1.15 Max/Min Method There are two max/min methods: max/min 1 and max/min 2. The max/min 1 method is used when we want to prove 1) All of set S is to the right of x. min {s : s ∈ S } ≥ x. A1: Choose a real number x. B2: Such that ay + by + c ≥ ax + bx + c. 2 B3: 2 ( y − x ) a ( y + x ) + b ≥ 0. 2) All of set S is to the left of x. max {s : s ∈ S } ≤ x. A2: If y − x = 0, the proposition is true. So, we then concentrate when y − x ≠ 0. A3: Therefore, either y − x > 0 or y − x < 0. S x 0 Since “either/or” appears, we now use cases method. A4: Case 1: assume that y − x > 0. S B4: a ( y + x ) + b ≥ 0. x 0 A5: From the fact that y = −b / 2a and a < 0, working from A4, we have 2ax + b > 0. A6: And so a ( y + x ) + b = ax + b / 2 = ( 2ax + b ) / 2 > 0. The max/min 2 method is used when we want to prove 1) Some of set S is to the right of x. min {s : s ∈ S } ≤ x. Since A6 = B4, the proof of case 1 is completed. 2) Some of set S is to the left of x. max {s : s ∈ S } ≥ x. A4: Case 2: assume that y − x < 0. B4: a ( y + x ) + b ≤ 0. S A5: From the fact that y = −b / 2a and a < 0, working from A4, we have 0 x 2ax + b < 0. A6: And so a ( y + x ) + b = ax + b / 2 = ( 2ax + b ) / 2 < 0. Since A6 = B4, the proof of case 2 is completed. Since both cases are proved, the proof is now completed. We can see that the proof of case 2 is almost identical to that of the first case. We can therefore omit the proof of case 2 by saying that “Assume, without loss of generality, that case 1 occurs…,” which means that the proof of case 1 can in general be applied with case 2 or any other cases. 15 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 ♥ Example 15: [1] Prove the following proposition. Proposition 14: If R is the set of all real numbers, then The max/min 1 method can be summarized as follows. When to use it min { x ( x − 2 ) : x ∈ R} ≥ −1. When B has the form “ min {s : s ∈ S } ≥ x ” or “ max {s : s ∈ S } ≤ x. ” What to assume Choose an s ∈ S , and A What to conclude How to do it s ≤ x or s ≥ x Convert B to “For all” statement. Use choose method. Work forward from A and the fact that s ∈ S . Also work backward. The max/min 2 method can be summarized as follows. When to use it When B has the form “ min {s : s ∈ S } ≤ x ” or “ max {s : s ∈ S } ≥ x. ” A What to assume What to conclude That there is an s ∈ S for which s ≥ x or How to do it Convert B to “There is” statement. Solution We use max/min 1 method. Converting B to “for all” statement, we have s≤x B1: For all real numbers x, x ( x − 2 ) ≥ −1. Use construction method to produce the desired s ∈ S . A1: Choose a real number x. B2: x ( x − 2 ) ≥ −1. B3: x − 2 x + 1 ≥ 0. 2 B4: ( x − 1) ≥ 0, which is always true for all real numbers. 2 Therefore, A1 = B4 and the proof is completed. 16 Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 Summary of proof methods Proof Method Forward - Backward When to Use It As a first attempt, or when B does not have a recognizable form. Contrapositive When NOT A and NOT B give useful What to Assume What to Conclude A B NOT B NOT A NOT A. information. When the statement A and two possible alternatives. Construction How to Do It Work forward using the facts given in A and apply the backward process to B by asking “How or when can I conclude that the statement B is true?” Work forward from NOT B and backward from B are one of the When the statement A and B contain the key word “no” or “not.” When B has the quantifiers “there is, there are, there exists.” A That there is the desired object. Construct the object, then show that it has the certain property and that the something happens. A, and choose a general object with the certain property. That the something happens. Work forward from A and the fact that the general object has the certain property to conclude that something happens. B is in the form: Choose Specialization There is an “object” with a “certain property” such that “something happens.” When B has the quantifiers “for all, for each, for every, for any.” B is in the form: For every “object” with a “certain property,” “something happens.” When A has the quantifiers “for all, for each, for every, for any.” A B Look for one particular object to apply specialization to. (Usually it is the same object as when the choose method is used in the backward process.) A is in the form: For every “object” with a “certain property,” “something happens.” Direct Uniqueness When B has the word “unique” in it. Be sure that the general object can be replaced with any object to satisfy the “for all” quantifier. Identify, in the for-all statement, the object, the certain property, and the something that happens. There are two such objects, and A. 17 That the two objects are equal. Verify that this object has the certain property and that something happens for this particular object. Work forward using A and the properties of the objects. Also work backward to show the objects are equal. Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 Proof Method Elimination Cases When to Use It When B has the form “ C OR “ A implies C OR D. ” D ”, such as, The keyword to look for is “either/or.” When A has the form “ C OR D ”, such as, “ C OR D implies B .” What to Assume A and NOT C A and NOT D ) Case 1: Case 2: When or “ max Max/Min 2 Contradiction has the form “ min B {s : s ∈ S } ≥ x ” {s : s ∈ S} ≤ x. ” B has the form “ min { s : s ∈ S } ≤ x ” or {s : s ∈ S} ≥ x. ” When NOT B s ∈ S, gives useful information. A and and (or work forward from NOT C , and backward from A and NOT D, and Both cases must imply B . C implies B; then prove that D implies B. s≤x Convert or First prove that s≥x A B to “For all” statement. Use choose method. A When “ max and A backward from C.) C D Choose an How to Do It Work forward from D. (or The keyword to look for is “either/or.” Max/Min 1 What to Conclude D (or C ) That there is an s ∈ S for which s ≥ x or s ≤ x NOT B Some contradiction Work forward from A and the fact that backward. Convert B to “There is” statement. s ∈ S. Also work Use construction method to produce the desired Work forward from A and NOT B s ∈ S. to reach a contradiction. When the statement possible alternatives. B is one of the two Indirect Uniqueness When the statement B contains the key word “no” or “not.” When B has the word “unique” in it. Induction When a statement integer n P (n) beginning with is true for each n0 . There are two different objects, and A. P (n) n. is true for Some contradiction. That P ( n + 1) true; also show that true. P ( n0 ) is Work forward from A using the properties of the two objects and the fact that they are different. First show that is true by replacing n in P (n) by n0 . is Write P ( n + 1) the fact that 18 P ( n0 ) and relate P (n) P ( n + 1) to is true to show that P (n) then use P ( n + 1) is true. Copyright 2007 by Withit Chatlatanagulchai 208581 Nonlinear Systems in Mechanical Engineering Lesson 4-1 Diagram of proof methods Forward-Backward Contrapositive Contradiction Induction Indirect Uniqueness Construction Choose Specialization Direct Uniqueness Elimination Cases Max/Min 1 Max/Min 2 References [1] How to Read and Do Proofs, by Daniel Solow, Wiley, 2002. 19 Copyright 2007 by Withit Chatlatanagulchai

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