Proofs. Indirect Proofs

Transcription

1 Math 2326 Fall 2008 Proofs. Idirect Proofs 9fte, it is very difficult to give a direct approach to! x (P(x) " C(x))D the coectio betwee P!x) ad C!x) might ot be suitable to this approach. For example, i class, we prove that the siuare of a eve umber is eve. It turs out that the coverse of this is also true. If is a iteger ad 2 is eve, the is eve. Cuatified,!! 2 is eve " is eve). If we were to taje a direct approach, we would start. Suppose tha 2 is eve. Lhe there is a iteger J for which 2 # 2J. At this poit, we would be stucj. It is hard to get from 2 # 2J to # 2N. Ay proof techiiue other tha a direct proof is ofte called a idirect proof. Lhe two most commo idirect methods are called Proof by Cotrapositio ad Proof by Cotradictio. p I p"i L L L L F F F L L F F L I the direct proof, the approach was to show the secod lie i the Lruth table does ot occur by showig that I has to be true whe p is true. I a proof by cotrapositio, we show that the secod lie does ot occur by showig that whe I is false, p is also false. Lhat is, i a direct proof, we assume p is true ad show that we have to be i the top lie of the truth table. For a proof by cotrapositio, we assume I is false ad show that we have to be i the bottom lie. Alteratively!ad bettero) a proof by cotrapositio is a direct proof of the cotrapositive. Lhat is,! x (P(x) " C(x)) #! x ( C(x) " P(x)). We start by assumig C is false, ad directly show that P is false. LetRs taje our example above. If 2 is eve, the is eve. Lhe cotrapositive of this is SWhe is odd, 2 is also odd.t We essetially proved this already i the last set of otes

2 page 2!as a corollary to provig that the product of odd umbers is odd) but letrs do it agai. Lhe proof would go. Let be odd. Lhe for some iteger J, # 2J % 1. Vow 2 #!2J % 1) 2 # 4J 2 % 4J % 1 # 2!2J 2 % 2J) % 1, so 2 is also odd. Xy cotrapositio, whe 2 is eve, must be eve. Here is a example from the!extra credit) from the last exam. Suppose that {u, v, w} is a liearly idepedet set of vectors. Prove that {u, w} is also a liearly idepedet set. Here is the solutio I gave. LhiJ of this as a p " I statemet, where p is the propositio that Zu, v, w[ is idepedet ad I is the propositio Zu, w[ idepedet. We said i class that p " I # I " p. I will show that istead. Here, we ca assume Zu, w[ is liearly depedet, so for some x, y, ot both \ero, xu ] yw = 0. With this, xu ] 0v ] yw = 0 is a depedece relatio for Zu, v, w[, so Zu, v, w[ is depedet as well. As you ca see, this is exactly a direct proof of the cotrapositive. Aother example. Prove the followig. If x % y > 100 the x > 50 or y > 50. I Iuatified form,! x! y!x % y > 100 "!x > 50 $ y > 50) ). Lhis ca probably be doe directly, but it is ot straightforward. It is almost embarrassigly easy by cotrapositio, provided the details are doe right. First, the cotrapositive of x % y > 100 "!x > 50 $ y > 50) is!x > 50 $ y > 50) "!x % y > 100). Lo use this we apply De MorgaRs law ad carefully egate the ieiualities.!x! 50 $ y! 50) "!x % y! 100). Lhe hard part is doe. Vow the proof. Assumig x! 50 ad y! 50, we have x % y! 50 % 50 # 100. Lhat is, x % y! 100. A complicatio. bosider the problem. Prove that the sum of a ratioal umber ad a irratioal umber is irratioal. Here, irratioal meas ot ratioal. I Iuatified form,! x! y!!x is ratioal $ y is irratioal) "!x % y is irratioal) Lhe cotrapositive is ot much better. If x % y is ratioal, the x is irratioal or y

3 page 3 is ratioal. Lhere is a tricj for this Jid of problem. Some hypotheses ca be SsucJedT ito the uiverse of discourse. I claim! x! y!!x is ratioal $ y is irratioal) "!x % y is irratioal) is logically eiuivalet to! r! y!y is irratioal " r % y is irratioal), where the uiverse of discourse for r is the set of ratioal umbers ad the uiverse for y is the set of all real umbers. If you believe this, the cotrapositive claim is! r! y!r % y is ratioal " y is ratioal). Here is how the proof would go. Sice r is ratioal, for some itegers m, with " 0, r # m. Sice r % y is ratioal, there are itegers p, I with I " 0 ad r % y # p I. Vow y #!r % y) % r # p I % m p % Im #, which is a Iuotiet of two itegers I with o\ero deomiator so y is ratioal. I geeral,! x! y!!p!x) $ C!y) $ R!x, y)) " S!x, y)) is logically eiuivalet to!!x with P!x) true)!!y with C!y) true)!r!x, y)) " S!x, y)). Aother example. If x ad y are positive ad xy > 100 the x > 10 or y > 10. Cuatified.! x! y!!x > 0 $ y > 0 $ xy > 100) "!x > 10 & y > 10) ) Modified Iuatified versio.! x > 0! y > 0!xy > 100 "!x > 10 & y > 10)), cotrapositive.! x > 0! y > 0!!x! 10 $ y! 10) " xy! 100).

4 page 4 I that last form, the result is easy to prove. It goes lije this. Let x ad y be positive real umbers with x! 10 ad y! 10. Lhe xy! 10&10 so xy! 100, as desired. Lechically, somethig lije this should be writte. Problem. Prove that if x ad y are positive ad xy > 100, the x > 10 or y > 10. Solutio. Suppose x ad y are positive, x! 10 ad y! 10. Lhe xy! 10&10, or xy! 100. Xy way of cotrapositio, the origial result is proved. Fially, for this problem, ote that the positivity of x ad y is eeded. bertaily if x # %3 ad y # %50, xy # 150 > 100, but x > 10 is false ad y > 10 is false. Lhe reaso. multiplyig by a egative umber chages the directio of a ieiuality. Lo Jeep this from happeig, we eeded x ad y to be positive. Lhere is aother idirect approach, called Proof by Cotradictio. Lhe techiiue applies more geerally tha to Nust! x (P(x) " C(x)). It ca also be used occasioally for the simpler form! x P(x). Lhe idea is this. Whe is a " F a true statemet? Sice L " F is false, we eed a to be false. Said differetly, if we Jow p " F is a true statemet, the we Jow p is false so p must be true. Lhe techiiue gets its ame from that fact that a propositio which is always false is called a cotradictio. Restated, to prove! x P(x), show that P!x) implies a cotradictio. For the more complicated! x (P(x) " C(x)), recall that (P(x) " C(x)) # P!x) $ C!x). How do we get the cotradictio? Usually it is of the form!somethig) $!its egatio). Sice a $ a # F, this does it. As a example, taje our previous example yet agai. If x ad y are positive ad xy > 100 the x > 10 or y > 10. Cuatified.! x! y!!x > 0 $ y > 0 $ xy > 100) "!x > 10 & y > 10) ).

5 page 5 Lhis time, we will V9L modify the uiverse of discourse. Istead of usig cotrapositio, we give a proof by cotradictio. Lhe egatio of!x > 0 $ y > 0 $ xy > 100) "!x > 10 & y > 10) is!x > 0 $ y > 0 $ xy > 100) $!x! 10 $ y! 10). We will show this is false!a cotradictio). We have doe all the worj already. If x > 0 ad y > 0 ad x! 10 ad y! 10, the xy! 10&10, or xy! 100. Lhis meas that xy > 100 AND xy! 100, ad this is our cotradictio. Ay proof by cotrapositio ca be doe by cotradictio, but it is ofte cosidered bad form to do so. Lhe exceptio beig cases where a proper proof by cotrapositio worjs best by chagig the uiverse of discourse!as i the last two examples.) Lhe reaso cotrapositio is ofte preferred is that a proof by cotradictio will be almost idetical to the cotrapositio proof, but loger. For example, we previously proved that if 2 is eve, the is eve. Lhe proof wet. If is odd, the for some iteger J, # 2J % 1. Vow 2 #!2J % 1) 2 # 4J 2 % 4J % 1 # 2!2J 2 % 2J) % 1, so 2 is also odd. Xy cotrapositio, whe 2 is eve, must be eve. Here is how the proof would go if we used a proof by cotradictio istead. Suppose 2 is eve but is odd. Sice is odd, # 2J % 1 for some iteger J. Vow 2 #!2J % 1) 2 # 4J 2 % 4J % 1 # 2!2J 2 % 2J) % 1, so 2 is also odd, cotradictig the assumptio that is eve. Here, we did exactly the same calculatio, ad showed that 2 must be odd. I the cotrapositio proof, we were essetially doe. I the cotradictio proof, we had to say this cotradicted the origial assumptio that 2 was eve. I geeral, whe provig p " I by cotrapositio, you prove I " p. With a proof by cotradictio, you prove p $ I " F. If the false statemet is of

6 page 6 the form p $ p, the cotrapositio proof is to be desired. Here is the classical proof by cotradictio. Lhis is oe case where there is o choice%%cotrapostio Nust does ot worj. 2 is irratioal. Lhis might ot looj lije a propositio of the form! x (P(x) " C(x)) or eve oe of the form! x P(x), but appearaces are deceivig. Remember the defiitio of ratioal. x is ratioal if there are itegers m, with " 0 ad x # m. Lo be irratioal meas there are o such itegers. We might put this i Iuatified form (! m! ( " 0) ) 2 " m +, ' ),'*. Lhe uiverse of discourse for the deomiator really should exclude 0 as above. Ufortuately, it is hard to prove 2 is irratioal from here. Rather tha usig the usual defiitio of ratioal, it helps greatly to use a modified versio. x is ratioal if. there are itegers m, with " 0, x # m, AND at least oe of m ad is odd. Lhe reaso this is true is that oce we have m ad with x # m, we ca divide out commo factors of 2. Lhat is, 66f meas 66. Lhis is obviously ratioal 100!Iuotiet of two itegers) ad we ca divide out a 2 to get 66f # 33. With these 50 prelimiaries, here goes the proof. Suppose 2 is ratioal!that is, assume p). Lhe there are itegers m ad with " 0 ad at least oe of m ad is odd with 2 # m. Multiply by. 2 # m. Vow siuare. 2 2 # m 2. We see that m 2 is twice a iteger so m 2 is eve. We have show previously that if the siuare of a iteger is eve, the the iteger is eve. Lhat is, sice m 2 is eve, we Jow that m is eve. Lhus, there is a iteger J so that m # 2J. Vow 2 2 # m 2 #!2J) 2 # 4J 2. Dividig by

7 page 7 2, 2 # 2J 2. Lhis meas that 2 is also eve, ad hece, that is eve. Xut the m ad are both eve cotradictig the assumptio that at least oe of them is odd. May people fid the above proof usatisfyig ad eve ucovicig. It certaily is ot the origial GreeJ proof. Lheir proof was closer to this. If 2 # m, the as we saw above, both m ad are eve. Divide out a 2 to get 2 #!mh2)!h2) # m 1. Xy exactly the same reasoig, m 1 ad 1 are also both 1 eve. We ca cacel out aother factor of 2 to get 2 # m 2 2. Agai, we are forced to coclude that m 2 ad 2 are eve. blearly we ca factor out 2Rs forever, but this is absurd!ie a cotradictio). A fial example. Vo positive siuares!of itegers) ca differ by 2. Lhat is, it is ever the case that m 2 % 2 # 2. Here is the proof. Assume the result is wrog, that there are positive itegers m ad with m 2 % 2 # 2. If we factor this,!m % )!m % ) # 2. If the product of two positive itegers is 2, the the larger is 2 ad the smaller is 1, so m % # 2 ad m % # 1. Addig, 2m # 3, so m # 3, cotradictig the 2 assumptio that m is a positive iteger. Summary of proof techiiues. Lo prove! x (P(x) " C(x)) Directly. Assume P!x) is true for a geeric x, try to prove C!x) is also true by botrapositio. Assume C!x) is false for x, prove that P!x) is also false by botradictio. Assume P!x) is true ad C!x) is false, obtai a cotradictio