Moder Algebra 1 Sectio 1 Assigmet 1 JOHN PERRY Eercise 1 (pg 11 Warm-up c) Suppose we have a ifiite row of domioes, set up o ed What sort of iductio argumet would covice us that ocig dow the first domio will oc them all dow? We have to show that if you oc dow ay oe domio, the it ocs dow the oe behid it Eercise 2 (pg 11 Eercise 2) Prove usig mathematical iductio that for all positive itegers, (2 1)( 1) (1) 1 2 2 2 3 2 2 For the iductive base, assume 1 The left had side of (1) simplifes to ad the right had side simplifies to 1 2 1 1(2 1 1)(1 1) 1 3 2 Sice the two sides are equal, (1) is true for 1 Now assume > 1, ad assume that 1 2 2 2 2 (2 1)( 1)/ for all : 1 < O the left had side of (1), we get 1 1 2 2 2 2 1 2 2 2 () 2 2 By the iductive hypothesis, this simplifies as 1 2 2 2 () 2 ()(2() 1)(() 1) 2 2 ()(2) 2 23 3 2 1
Moder Algebra 1 Sectio 1 Assigmet 1 Joh Perry 2 O the right had side, we get (2 1)( 1) 2 2 ( 1) 23 3 2 We see that the left ad right had sides of (1) are equal Hece 1 2 2 2 2 (2 1)( 1)/ Eercise 3 (pg 11 Eercise 3) You probably recall from your previous mathematical wor the triagle iequality: for ay real umbers ad y, (2) y y Accept this as give (or see a calculus tet to recall how it is proved) Geeralize the triagle iequality, by provig that (3) 1 1 2, for ay positive iteger For the iductive base, assume 1 Let 1 be arbitrary, but fied It is obvious that 1 1 We have show that the assertio is true for 1 Assume > 1 ad 1 1 2 for all : 1 < Let 1, 2,, be arbitrary, but fied The left had side of (3) is (4) 1 1 1 This is a sum of two itegers By the basic triagle iequality (2), 1 1 1 1 By the iductive hypothesis, 1 1 1 2 1 Substitutig this fact ito (4), we obtai (3) Eercise 4 (pg 12 Eercise 4) Give a positive iteger, recall that 1 2 3 (this is read as factorial) Provide a iductive defiitio for (It is customary to actually start this defiitio at, settig! 1) Defie! 1 ad ()! Eercise 5 (pg 12 Eercise 5) Prove that 2 < for all 4 For the iductive base, assume 4
Moder Algebra 1 Sectio 1 Assigmet 1 Joh Perry 3 The 2 4 1 < 24 4! Assume > 4, ad that 2 <! for all : 4 < We have ()! > 2 1 > 2 1 2 2 Eercise (pg 13 Eercise 14) I this problem you will prove some results about the biomial coefficiets, usig iductio Recall that ( )!!, where is a positive iteger, ad (a) Prove that (5) 1 for 2 ad < Hit: You do ot eed iductio to prove this Bear i mid that! 1 (b) Verify that 1 ad 1 Use these facts, together with part a, to prove by iductio o that is a iteger, for all with (c) Use part a ad iductio to prove the Biomial Theorem: For o-egative ad variables, y, ( y) y (a) We start with the right had side of (5): ()! 1!( )! ()! ( 1)!( )! Get a commo deomiator; the 1, ()! ( ) ()!!( )!!( )! ()! ( ) ()!!( )! ()! (( ) )!( )! ()!!( )!!( )!
Moder Algebra 1 Sectio 1 Assigmet 1 Joh Perry 4 (b) First, by defiitio () (7)!( )! 1 1 ( )! 1! 1 Let be arbitrary, but fied We prove that is a iteger by iductio o For the iductive base, assume If, equatio () shows that is a iteger By defiitio,, so there is othig more to show for i Assume <, ad assume that is a iteger for all i, : i < We ow from part (a) that 1 Sice <, the iductive hypothesis implies that ad are both itegers By closure, the sum of two itegers is also a iteger Thus is a iteger 1 (c) We proceed by iductio o For the iductive base, assume The ( y) 1 ad y y 1 The two sides are equal Assume >, ad assume ( y) i i i y i for all : <
Moder Algebra 1 Sectio 1 Assigmet 1 Joh Perry 5 By the iductive hypothesis, ( y) ( y)( y) 1 ( y) 1 y y 1 y 1 y 2 Reide the secod summatio, so that ( y) 1 1 1 1 y 1 y 1 y y y (What happeed here? I added 1 to the s that ide the sum, which requires me to subtract 1 from the s i the formula This is called reideig, ad is a useful tool You ca verify that it is correct by writig out a few terms of the sum before ad after reideig I do ot epect ayoe to have writte the proof this way I fact I showed studets a differet way of writig the same thig myself, but you eed to see it sooer or later, so there it is!) Combiig lie terms, we have ( y) 1 1 y 1 y y y y