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Chapter 5 Existence and Proof by contradiction

Chapter 5 Existence and Proof by contradiction. 5.1 Counterexamples 5.2 Proof by Contradiction 5.3* A Review of Three Proof Techniques 5.4 Existence Proofs 5.5 Disproving Existence Statements. Section 5.1 Counterexamples.

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Chapter 5 Existence and Proof by contradiction

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  1. Chapter 5 Existence and Proof by contradiction • 5.1 Counterexamples • 5.2 Proof by Contradiction • 5.3* A Review of Three Proof Techniques • 5.4 Existence Proofs • 5.5 Disproving Existence Statements

  2. Section 5.1 Counterexamples Notice that some quantified statements of the type xS, R(x)are false. We have seen that  (x  S, R(x))   x  S,  R(x), that is, xS, R(x)is false, then there exists some element xS for which R(x) is false. Such an element x is called a counterexample of the statement xS, R(x). Finding a counterexample verifies that xS, R(x) is false.

  3. Examples Example: Consider the statement: If x  R, then (x2-1)2>0. Show that this is false by exhibiting a counterexample. Solution: For x=1, (x2-1)2=(12-1)2=0. Thus x=1 is a counterexample. If a statement P is shown to be false in some manner, then P is said to be disproved. Example Disprove the statement: For every odd positive integer n, 3 | (n2-1). Solution: Since 3 (32-1), it follows that n=3 is a counterexample.

  4. Counterexample for x  S, P(x)Q(x) We see that a quantified statement of the type x  S, R(x) is false if  x  S,  R(x) is true. Similarly, the quantified statement x  S, P(x)Q(x) is false if  x  S,  (P(x)Q(x)) is true, Where  (P(x)Q(x))  P(x)  ( Q(x)). Thus, x  S, P(x)Q(x) is false if  x  S, (P(x)  ( Q(x))). That is, to show that x  S, P(x)Q(x) is false, we need to exhibit a counterexample, which is then an element x  S for which P(x) is true and Q(x) is false.

  5. Examples Example Disprove the statement: Let n  Z. If n2+3n is even, then n is odd. Solution: If n=2, then n2+3n= 22+3(2)=10 is even and 2 is even. Thus n=2 is a counterexample. Example Show that the statement: Let n  Z. If 4 | (n2-1), then 4 | (n-1) is false. Solution: Since 4 | (32-1) but 4 (3-1), it follows that n=3 is a counterexample.

  6. Section 5.2 Proof by Contradiction Suppose that we would like to show that a certain mathematical statement R is true. We may assume R is false statement and, from this assumption, we are able to arrive at or deduce a statement that contradicts some assumption we made in the proof or some known fact. Therefore, we have established the truth of the implication ( R) C However, because ( R) C is true and C is false, it follows that  R is false and so R is true, as desired. This technique is called proof by contradiction. Often a proof by contradiction begins with Suppose that R is false. or Assume, to the contrary, that R is false.

  7. Proof by Contradiction Cont. If R is the statement x  S, P(x)Q(x), then a proof by contradiction of the statement consists of verifying the implication (x  S, P(x)Q(x))  C for some contradiction C. Note that (x  S, P(x)Q(x))   x  S, (P(x)  ( Q(x))), it follows that a proof by contradiction of x  S, P(x)Q(x) might begin with Assume, to the contrary, that there exists some element x  S for which P(x) is true and Q(x) is false. The remainder of the proof then consists of showing that this assumption leads to a contradition.

  8. Example Result: There is no smallest positive real number. Proof: Assume, to the contrary, that there is a smallest positive real number, say r. Since 0<r/2<r, it follows that r/2 is a positive real number that is smaller than r. This, however, is a contradiction. #

  9. Example Result: If a is an even integer and b is an odd integer, then 4 (a2+2b2). Proof: Assume, to the contrary, that there exists an even integer a and an odd integer b such that 4|(a2+2b2). Thus a=2x, b=2y+1, and a2+2b2=4z for some integer x, y, and z. Hence (2x)2+2(2y+1)2=4z. Simplifying, we obtain 4x2+8y2+8y+2=4z or, equivalently, 2=4z-4x2-8y2-8y= 4(z-x2-2y2-2y). Since z-x2-2y2-2y is an integer, 4 | 2, which is impossible. #

  10. Example The following results concern irrational numbers. Recall that a real number is rational if it can be expressed as m/n for some m, n  Z, where n0. A real number is irrational if it is not rational. Result: The sum of a rational number and an irrational number is irrational. Proof: Assume, to the contrary, that there exist a rational number x and an irrational number y whose sum is a rational number z. Thus x+y=z, where x=a/b and z=c/d for some integers a, b, c, d  Z and b, d 0. This implies that y=c/d-a/b=(bc-ad)/bd. Since bc-ad and bd are integers and bd 0 it follows that y is rational, which is a contradiction. #

  11. Section 5.3 A Review of Three Proof Techniques In order to prove the truth of a statement x  S, P(x)Q(x), we have been introduced to three proof techniques: direct proof, proof by contrapositive, and proof by contradiction. Prove x  S, P(x)Q(x). • If we use a direct proof, then Assume that P is true, and show that Q is true. • If we use a proof by contrapositive, then Assume that Q is false, and show that P is false. • If we use a proof by contradiction, then Assume that P Q is false, and obtain a contradiction.

  12. Section 5.4 Existence Proofs An existence theorem concerning an open sentence R(x) over a domain S can be expressed as a quantified statement  x  S, R(x): There exists x  S such that R(x). Such a statement is true provided that R(x) is true for some x  S. A proof of existence theorem is called an existence proof. An existence proof may consist of displaying or constructing an example of such an object or perhaps, with the aid of known results, verifying that such objects must exist without every producing a single example of the desired example.

  13. Example Result. There exists an integer whose cube equals its square. Proof: Since 13=12=1, the integer 1 has the desired property. # Result. There exist real numbers a and b such that (a+b)2=a2+b2. Proof: Let a=1 and b=0. Then (a+b)2=(1+0)2=12=12+02=a2+b2. #

  14. Example Result: There exist irrational numbers a and b such that ab is rational. Proof. Consider the number . This is either rational or irrational. We consider two cases. Case 1: is rational. Then we can take a=b= , and we have the desired result. Case 2: is irrational. In this case, consider ab, where a= and b= . Observe that ab= Which is rational. #

  15. Existence of Solutions Recall Intermediate Value Theorem of Calculus If f is a function that is continuous on the closed interval [a, b] and k is a number between f(a) and f(b), then there exists a number c  (a, b) such that f(c)=k.

  16. Example Result. The equation x5+2x-5=0 has a real number solution between x=1 and x=2. Proof. Let f(x)=x5+2x-5. Since f is a polynomial function, it is continuous on R and so f is continuous on [1, 2]. Now f(1)=-2 and f(2)=31. Since 0 is between f(1) and f(2), it follows by the Intermediate Value Theorem of Calculus that there is a number c between 1 and 2 such that f(c)= c5+2c-5=0. Hence c is a solution. #

  17. Uniqueness An element belonging to some prescribed set A and possessing a certain property P is unique if it is the only element of A having property P. Typically, to prove that only one element of A having property P, we proceed in one of two ways: • We assume that a and b are elements of A possessing property P and show that a=b. (direct proof) (2) We assume that a and b are distinct elements of A possessing property P and show that a=b. (proof by contradiction)

  18. Example Result. The equation x5+2x-5=0 has a unique real number solution between x=1 and x=2. Proof. Assume, to the contrary, that the equation x5+2x-5=0 has two distinct real number solutions a and b between x=1 and x=2. We may assume that a<b. Since 1<a<b<2, it follows that a5+2a-5< b5+2b-5. On the other hand, a5+2a-5= b5+2b-5=0, which produces a contradiction. #

  19. Section 5.5 Disproving Existence Statements To disprove a quantified statement of the type  x  S, R(x) requires a different approach. Note ( x  S, R(x))   x  S,  R(x), It follows that the statement  x  S, R(x) is false if R(x) is false for every x  S. Result: Disprove the statement: There is a real number x such that x6+2x4+x2+2=0. Solution: Let x  R. Since x6 ,x4, and x2 are all even powers of the real number x, it follows that x60, x4 0, and x2 0. Therefore, x6+2x4+x2+22 and so x6+2x4+x2+20. Hence the equation x6+2x4+x2+2=0 has no real number solution. #

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