HW 1 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson

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1 HW Solutios Math 5, Witer 2009, Prof. Yitzhak Katzelso.: Prove = ( + )(2 + ) for all atural umbers. The proof is by iductio. Call the th propositio P. The basis for iductio P is the statemet that 2 = ( + )(2 + ), which is true. For the iductio step, we assume that P is true. We d like to show that P + is true, amely that: ( + ) 2 = ( + )(( + ) + )(2( + ) + ). By the iductio step, the left-had side is equal to ( + )(2 + ) + ( + ) 2 ; we just replaced the first terms by usig P. The multiplyig eerythig out, we see that the left-had side is: ( ) = But the right-had side is equal to: ( + )( + 2)(2 + 3) = , ad so the left ad right had sides are equal. Thus P + is true. By mathematical iductio, this completes the proof..3: Prove = ( ) 2 for all atural umbers. The proof is agai by iductio; call the th propositio P. The basis for iductio P is the statemet that 3 = 2, which is obviously true. For the iductio step, we assume that P is true. We d like to show that P + is true, amely that: ( + ) 3 = ( ( + )) 2. The right-had side may be expaded usig the usual formula (a + b) 2 = a 2 + 2ab + b 2, with a = ad b = +. We see that the right-had side is equal to: ( ) 2 + 2( )( + ) + ( + ) 2. We use P (applied to the first term) to write this as: ( ) + 2( )( + ) + ( + ) 2.

2 2 The we use the result of Example to write this as: ( ) + 2( 2 ( + ))( + ) + ( + )2. Expadig out the last two terms ad combiig them, we see that they are equal to = ( + ) 3, so our right-had side is equal to: ( ) + ( + ) 3, ad so we ve show P + is true. By mathematical iductio, this completes the proof..4: Evaluatig (2 ): for = we get, for = 2 we get 4, for = 3 we get 9, ad for = 4 we get. This leads us to guess that (2 ) = 2. Now we prove this by iductio. Call the th propositio P. The the basis for iductio, P, was already proved (it s = 2 ). For the iductio step, assume P is true ad the we wat to show that P + is true; i.e. that (2 ) + (2( + ) ) = ( + ) 2. But the left-had side is, by applyig P, equal to 2 +(2(+) ), which equals = (+) 2. So P + is true. By mathematical iductio, this fiishes the proof..: Prove that 4 is divisible by 7 whe is a atural umber. The proof is by iductio; P is the statemet that 4 is divisible by 7. P is true because 4 = 7 is divisible by 7. For the iductio step, suppose that P is true; the we wat to show that is divisible by 7. Write (the first step here is by addig ad subtractig 4, ad the secod step is by the distributive law): = = ( 4 )+7 4. Now otice that by P, the first term is times somethig that is divisible by 7, ad is thus itself divisible by 7. The secod term is 7 times somethig, so it is also divisible by 7. So is a sum of two terms that are both divisible by 7, ad so it is itself divisble by 7. Thus P + is true; by mathematical iductio, this completes the proof..8: a) Prove that 2 > + for all itegers 2.

3 Oe may prove this by the modified versio of iductio, where the base case is P 2 istead of P. But the followig is much easier: If 2, the 2 2 = ( + ) ( + 2) > ( + ), so 2 > +. b) Prove that! > 2 for all itegers 4. The proof here is by iductio. The basis for iductio is P 4, which is the statemet that 4! > 4 2. This is true, because 4! = 24 > = 4 2. So ow for the iductive step, we assume that P is true for some 4 ad we wat to prove that P + is true; i.e. that ( + )! < ( + ) 2. By the iductive hypothesis P, we kow that (+)! = (+)! > ( + ) 2. By part a), 2 > ( + ) (remember that 4), so ( + ) 2 > ( + ) ( + ) = ( + ) 2. Puttig these statemets together, we see that ( + )! > ( + ) 2, as we wated to prove. So P + is true; by mathematical iductio this completes the proof..: For each N, let P deote the assertio that is a eve iteger. a) We wat to show that P + is true wheever P is true. So assume P is true; we wat to show that ( + ) 2 + 5( + ) + is eve. But that expressio is equal to = = ( ) + (2 + ). Now by P, the first term is eve. Ad the secod term is 2(+3), so it is eve. So their sum is eve. This meas P + is true. b) For which is P actually true? Well, cosider two cases. If is eve, the 2 is eve, 5 is eve, ad is odd, so their sum is odd. However, if is odd, the 2 is odd, 5 is odd, ad is odd, so their sum is still odd. Thus is ever eve, ad so P is ever true! The moral of the story here, of course, is that the basis for iductio matters ad oe ca get ito a lot of trouble by igorig it. 2. a) Verify the biomial theorem for =, 2, ad 3. ( ) of( all, simple calculatios prove the followig list of idetities: 0First =, ) ( =, ( ) =, ad ) =. From these applied to values of betwee ad 3, we see that the = case of the biomial theorem is just (a + b) = a + b, which is obvious. The = 2 case is (a + b) 2 = a 2 + 2ab + b 2, which is true, ad the = 3 case is (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3, which is also true. b) Show that ( ) ( k + ) ( k = + ) k for k =,2,...,. Write out the left-had side: it s! k!( k)! +! ( k + )! + k! = (k )!( k + )! k!( k + )! 3

4 4 ( + )! = k!( k + )! = ( + )! k!( + k)!. But this is just ( ) + k.. c) Prove the biomial theorem usig mathematical iductio ad part b). The basis case P has already bee cosidered i part a). So assume P ad try to prove P +. To prove P +, look at (a + b) + = (a + b) (a + b). By P, this is equal to: ( (a + b)( 0 ) a + ( ) ( ) a b ab + ( ) b ). Multiply this out by had ad collect like terms. There are two of each type of term except for the first ad last. We get: ( ) ( ) ( ) ( ) ( ) a + ( + )a b + ( + )a 2 b ( ) ( ) ( ) +( + )ab + b. By part b) applied to the coefficiets of the iterior terms, ad oticig that the first ad last coefficiets are, this equals: ( ) ( ) ( ) a + a b + a 2 b ab + b. 2 Sice ( ) ( + 0 = +) =, the outside coefficiets work ad this is exactly P +. By mathematical iductio, we have thus prove the biomial theorem. 2.3: Show that (2 + 2) /2 does ot represet a ratioal umber. We prove this statemet by cotradictio. Assume for cotradictio that (2 + 2) /2 is a ratioal umber; call it q. The q 2 = is also a ratioal umber. Ad so q 2 2 = 2 is also a ratioal umber. But we kow that 2 is irratioal, so this is a cotradictio. Hece our assumptio was wrog, ad so (2+ 2) /2 is ot a ratioal umber. 2.5 Show that (3 + 2) 2/3 does ot represet a ratioal umber. We prove this statemet agai by cotradictio. Assume for cotradictio that (3 + 2) 2/3 is a ratioal umber; call it q. The q 3 is also

5 ratioal; but q 3 = (3 + 2) 2 = = + 2. Sice q 3 is ratioal, q 3 = 2 is also ratioal. A ratioal umber divided by is still ratioal, so this meas that 2 itself is ratioal. But we kow 2 is irratioal. This is a cotradictio, ad so our assumptio was wrog; thus (3 + 2) 2/3 is irratioal. 5

1. MATHEMATICAL INDUCTION

1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1