Repeated multiplication is represented using exponential notation, for example:

Transcription

1 Appedix A: The Lws of Expoets Expoets re short-hd ottio used to represet my fctors multiplied together All of the rules for mipultig expoets my be deduced from the lws of multiplictio d divisio tht you re lredy fmilir with Expoetil ottio Repeted multiplictio is represeted usig expoetil ottio, for exmple: = 4 There re four fctors i the product, ech of which is I the mthemticl expressio 4, 4 is clled the expoet d is usully clled the bse For rel umber d positive iteger, expoetils re defied by: =, totl of fctors, ech of which is A specil cse of expoetil ottio to ote: 1 = This mkes sese, sice 1 should hve oly oe fctor of Rules for combiig expoets Suppose tht d b re rel umbers d tht m d re positive itegers 1 m = + m m = m ( ) m = m 4 ( b) = b 5 b = b Justifictios of these lws re bsed o the lws of multiplictio d divisio tht you re fmilir with Rules 1, d 4 re justified below, d the ides of the justifictios for Rules d 4 described (The justifictios of Rules d 4 re exercises for this ppedix)

2 For the first expoetil rule, m = ( ) ( ) totl of fctors totl of m fctors = totl of + m fctors = + m For the third expoetil rule, ( ) m = ( ) ( ) ( ) totl of m sets of brckets ech brcket hs copies of = the totl umber of copies of is m = m For the fourth expoetil rule, ( b) = ( b) ( b) ( b) ( b) copies of ( b) so there re copies of d copies of b = ( ) (b b b b) copies of copies of b = b The ide for justifyig secod expoetil rule is this: there re copies of i the umertor of the frctio, d m copies of i the deomitor of the frctio Some of these copies of will ccel out Lstly, the ide for the justifictio of the fifth expoetil rule is this: there re copies of the frctio b Thus, the umertor will cosist of copies of multiplied together, so the umertor is The deomitor will cosist of copies of b multiplied together, so the deomitor is b (To review cocepts such s umertor, deomitor d multiplictio of frctios, see Appedix E) My of these rules lso work whe m d re ot positive itegers - for exmple, if m d re frctios Some specil cosidertios pply whe m d re ot positive itegers i order to mke sure tht the expressio mkes good mthemticl sese For exmple, while it is possible to write dow collectio of symbols like:

3 1/ ( ) = The collectio of symbols ( ) 1/ should represet the umber tht is the squre root of egtive two However, sice squrig y rel umber gives result tht is greter th or equl to zero, there is o rel umber tht, whe squred, gives While it is possible to write dow the collectio of symbols: ( ) 1/, there is o rel umber tht equls this collectio of symbols Exmple A1 Use the lws of expoets to simplify the followig lgebric expressios whe possible ) 4b (b) 5 b) (x ) x 4 Solutio: ) 4b (b) 5 = 4b 5 b 5 = 4 4 b b 5 = 97 b 6 b) (x ) x 4 = (x ) 4 x 4 ( ) = 4 81 x6 1 8 x = 4 x6 8 = 4 x Prt (b) illustrtes iterestig poit - the pplictio of the expoet rules c geerte lgebric expressios such s x, which we hve o wy of iterpretig t this poit Iterpretig d simplifyig frctiol d egtive expoets The wy tht we defied expoetil ottio s short-hd for repeted multiplictio mkes complete sese for expoets tht re positive itegers However, exctly wht expressio like x should me is ot so cler Let be rel umber, d let m d be positive itegers 0 = 1 (provided tht 0) = 1 (provided tht 0) 1/ = (if is eve iteger, the this oly mkes sese whe 0) m/ m = m = ( ) (if is eve iteger, the this oly mkes sese whe 0)

4 Exmple A Usig the lws of expoets, simplify the followig expressios s much s possible ) 1 L L 1 ( ) b) wy ( ) c) + r Solutio: 1 y 8 ) ( ) ( ) 1 1 L L = L L 1 = ( ) L 6 L 6 9 = L = 7 L b) ( wy) = w y w y w y w y 8 ( ) ( ) = = ( ) = 8 8 y y y c) ( + r ) = ( + r ) + r r r ( ) = The importt poit i Prt (c) is tht there i o expoetil rule tht is vlid whe the terms iside the brcket re dded (rther th multiplied or divided) together I prticulr, ( + b) + b The importt thig to remember here tht tht ( + b) d + b re ot the sme (To review the expsio of complicted lgebric expressios, see Appedix B) Exmple A ) 8w 18 b) 64k 9 9T c) Solutio: ) w = w = w

5 b) 64k 9T 9 64k = = 9 9T 64 9 k T 8k = / T 9 9 c) The simplifictio of or zero, the 1/ 1 = ( ) = = is ot completely stright-forwrd If is positive umber s you might expect If is egtive umber, the the situtio is more complicted If is odd umber, the = To illustrte this cosider the cse where = d =, ( ) = 8 = If is eve umber, the = To illustrte this cosider the cse where = d =, ( ) = 4 = Exercises for Appedix A For Problems 1-10, evlute the qutity (if possible) without usig clcultor ( 1) 4 6 ( 1) 5 7 ( ) 1 ( 1) ( 8)

6 For Problems 11-0, simplify the expressio s much s possible 11 e e t e t e 1 x ( y) ( x y) / ( ) + 1 b b 14 (T w 4 ) 1/ 15 e e r 16 / w y w+ y 17 ( C e wt ) 18 (5xy) 1 (xy ) 19 6uv 0 8 P 4 4P For Problems 1-5, decide whether ech of the sttemets re true or flse 1 ( + b) = + b 10q = 1 10q b = + b 4 x x m = x + m 5 z + z = z 4 I problems 6 d 7 you will justify the lst two expoetil rules 6 Suppose tht is rel umber, d tht d m re positive itegers Expli why: m = m 7 Suppose tht d b re rel umbers, d tht is positive iteger Expli why:

7 b = b Aswers to Exercises for Appedix A / e e t xy 1 ( + b) 14 T w 15 e r (4w 4 y )/(w + y) 17 C e wt 18 (1/5)xy 19 6 u (v) 1/ 0 P 1 Flse Flse Flse 4 Flse 5 Flse 6 The bsic ide here is ccelltio of commo fctors from the umertor d deomitor There re fctors of i the umertor d m fctors of i the deomitor I the cse where > m, ll of the fctors of i the deomitor will ccel with fctors of i the umertor, levig - m fctors of remiig i the umertor I the cse where < m, the fctors of i the umertor will ccel with fctors i the deomitor, levig m - fctors of i the

8 deomitor Thus, the simplified expressios will be: (1/) m - -1(m - ) = = - m 7 The bsic ide here is tht you will hve fctors, ech of which is /b Thus, the overll umertor will cosist of copies of ll multiplied together, ie the overll umertor is Similrly, the overll deomitor is b