Mathematical Induction (selected questions)

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1 Mtheticl Iductio (selected questios). () Let P() e the propositio : For P(), L.H.S. R.H.S., P() is true. Assue P() is true for soe turl uer, tht is, () For P( ),, y () By the Priciple of Mtheticl Iductio, P() is true for ll turl uers.. () Let P() e the propositio : For P(), L.H.S. R.H.S., P() is true. Assue P() is true for soe N, tht is, () For P( ),, y () (c) Let P() e the propositio : y is divisile y y for y itegers, y d positive iteger. or y yg,y, where g,y is polyoil i,y d, y Z, N. For P(), y yyyg,y, P() is true. Assue P() is true for soe N, tht is, y yg,y, where g,y Z[, y] For P( ), y y y yg,y y yg,y, where g,yg,y y,y. (d) Let P() e the propositio : For P(), L.H.S. R.H.S., P() is true. Assue P() is true for soe N, tht is, ()

2 For P( ),, y (). (e) Let P() e the propositio : 5 For P(),, P() is true. Assue P() is true for soe N, tht is, 5 For P( ), 5, y ().. (f) Let P() e the propositio : For P(),, P() is true. Assue P() is true for soe N, tht is, For P( ),, y ().. (g) Let P() e the propositio : 5 where ϵ. For P(), 55, where ϵ. P() is true. Assue P() is true for soe N, tht is, 5, where ϵ. For P( ), , y () 56 5, where ϵ.. (h) Let P() e the propositio : The o. of digols i cove -sided polygo is f() ( ) ( ). For P(), There is o digol i trigle. f() 0. P() is true.

3 Assue P() is true for soe N \{,}, tht is, The o. of digols i cove -sided polygo is f() ( ) ( ) () For P( ), Let P, P,., P, P e the vertices i clocwise directio of the ()-sided polygo. By (), The o. of digols i cove -sided polygo P P.P is f() ( ) There re dditiol ( ) digols coected to the the poit P, ely, P P, P P,, P P. The o. of digols i cove ()-sided polygo P P.P P is { } f() ( ) ( ) () { ( ) () } { } ( ) [( ) ] f( ).. (i) Let P e the propositio:,,,,,,,,, R, For P(), Also, P() is true. Assue P() is true for soe N\{}, For P( ), Also, where,,,, y () where,, By the Priciple of Mtheticl Iductio, P() is true N\{}.,. (j) Let P() e the propositio : - 5 for 0. For P(0), 0-8 ( ) P(0) is true. Assue P() is true for soe N, where 0, tht is, - 5 () For P( ), ()- ( ) ( 5 ), y ()

4 5 ( ) (0) 0.(0) 0.(0) 0.(0) 0. (0) 5, sice ( ) 5 By the Priciple of Mtheticl Iductio, P() is true N, where 0.. () Let P() e the propositio : > for 5. For P(5), >55 P(5) is true. Assue P() is true for soe N, where 5, tht is > For P( ), > 5 5> By the Priciple of Mtheticl Iductio, P() is true N, where 5.. (l) Let P() e the propositio :!! For P(),!! P() is true. Assue P() is true for soe N, tht is!! For P( ),!!!!!, y () p!! By the Priciple of Mtheticl Iductio, P() is true N.. () Let P() e the propositio : C!!! is lwys iteger, where 0 r For P(0), C, which is iteger. Assue P() is true for soe {0}, tht is, C!!! is lwys iteger. () For P( ), C C re itegers. For r, C C C is the su of two itegers y (), d is iteger. By the Priciple of Mtheticl Iductio, P() is true N {0}.. () Let P() e the propositio :, where

5 For P(), 09 9 P() is true. Assue P() is true for soe N, tht is,, where For P(), 7 6 6, y () 6, where 6.. (o) Let P() e the propositio :, where. For P(), 9 7 P() is true. Assue P() is true for soe N, tht is,, where For P(), 6, y (), where. (p) Let P() e the propositio : < (We chge the propositio d prove this first.) For P(), < P() is true. Assue P() is true for soe N, tht is, For P(), <, y () < < < <. (q) Let P() e the propositio : The uer of pirs of o-egtive itegers (, y) stisfyig y is f() ( ) [ ( ) ] For P(), (, y) (, 0), f() P() is true. For P(), (, y) (, 0) d (0, ) re solutios. Also, f() P() is true. Assue P() is true for soe N, tht is, The uer of pirs of o-egtive itegers (, y) stisfyig y is 5

6 () f() ( ) [ ( ) ] For P( ), Cosider the equtio z (), y () The uer of pirs of o-egtive itegers (, z) stisfyig () is f() ( ) [ ( ) ] Put z y, () ecoes (y ) or y () Oviously (, y) (, 0) stisfies (), ut (, z) (, y ) (, ) does ot stisfy (). The totl uer of pirs of o-egtive itegers (, y) stisfyig () is f() ( ) [ ( ) ] [( ) ] [ ( ) ] f ( ) P( ) is true., where Z.. () Let P() e the propositio : ( ) ( ) For P(0), ( ) ( ) 0, where 0 For P(), ( ) ( ) ( ) [ ] 0, where 5. Assue P( ) d P() re true for soe N ( ), tht is, ( ) ( ), ( ) ( ).(*) For P( ), ( ) ( ) [( ) ( ) ][( ) ( ) ] ( ) ( ) ( ) ( ) 8 [ ] ( ) ( ) ( ) ( ) 8 ( ), where - Z. By the Priciple of Mtheticl Iductio, P() is true N {0}., y (*). () Let α 5, β 5, the αβ6, αβ. Let P() e the propositio : α β, where Z. For P(), αβ6, where N. For P(), α β αβ αβ6 8, where 7 Z. Assue P() d P( ) re true, tht is, α β, α β, where, Z. For P( ), α β α β αβαβα β 6 6

7 7, where. P() is true.. Let P() e the propositio : There re two collectios of rles of the se qutity,. Accordig to the rules of tig rles set y the questio, the first plyer wis. For P(), The first plyer te rle fro oe collectio d the secod plyer tes the lst rle d wis. P() is true. Assue P(), P(),, P() re true. So if there re two collectios of rles of the se qutity less th or equl to, the secod plyer wis. For P( ), There re two collectios of rles of the se qutity,. If the first plyer tes wy y uer of rles, sy p, p fro oe collectio. The secod plyer tes wy lso p rles fro the other collectio. So ow we hve two collectios of rles ech of. Thus y the iductive hypothesis, the secod plyer wis.. () Let P() e the propositio : ( ), where, Z, N. For P(),, where,. For P(), ( ), where, ( ). P(), P() re true. Assue P() is true for soe N, tht is, ( ) () For P( ), ( ) ( ) ( ) ( )( ) ( ) ( ) [ ] ) ( where ( ) ( ),.() () Fro (), ( ) ( ),.() Fro (), ( ) ( ) ) ( ) ( [ ] ( ) [ ]... ) ( ) ( ) ( [ ] [ ] ) ( ) ( ) ( ) ( ) (.() Fro (),

8 (i) If is eve, < 0, < 0. ( ) ( ) ( ).(5) ( ) ( ) N N, y (5). (ii) If is odd, > 0, > 0. ( ) ( ) ( ).(6) ( ) ( ) N N, N N, y (6) 6. First, show y Mth. Iductio the propositio P() : ( ) N.() The, : : : : : ( ) ( ) [ ( )] The lst ter of ( ). Addig up ll equlities, r ( ) ( ) r 5... ( ), y (). (o. of ter ), (o. of ters ), 5 7 (o. of ters ), The lst ter of [ ] ( )( ).() 6 By (), S By (), The lst ter of - ( )()(). 6 6 ( )( ).() 6 Hece S S - 5 ( )( ) ()()() [ ] Hece S r r r ( ).() r r r 5 Copre () d (), we get r ( )( ) ( ) ( ) ( ) r 6 7. (Bcwrd Mtheticl Iductio)... (i) () I() : If i [, ], i,,,, the f ()... f ( ) f For I( ), sice it is give tht f () f ( ) f. I( ) is true. 8

9 Assue I(... ) is true. i.e. f ()... f ( ) f.() For I( ),... f... f f () Assue I() is true ( ), i.e. Put f ( )... f (... f ( )... f (... f f ( )... f ( ) f ( )... f ( )... ) f ) f... f, the f f f I( ) is true f ( )... f ( ) ( )f I( ) is lso true. (c) N, ( N d r N) such tht r., y (), y I() By (), I( ) is true I( ) is true I( ) is true I( r) I() is true. (ii) si si si cos si, π π. f() si is cove o [0, π]. Lst prt follows fro (i).... f... f 9. Let P() e the propositio : For P(), L.H.S., R.H.S., P() is true. Assue P() is true for soe N, tht is, () For P( ), 9

10 ,. Let By the Priciple of Mtheticl Iductio, P() is true for ll turl uers. q q q F (z) ( z) ( z)( qz)... ( z)( qz)...( q z), q q q Let P() e the propositio : F (z) F (qz) ( qz)( q z) ( q z). For P(), F zf qz z qzqz P() is true. Assue P() is true for soe N, tht is, F zf qz qz q z q z () For P( ), But, F zf z zqz q z d F qzf qz qzq z q z F zf qz F zf qz zqz q z qzq z q z qz q z q z zqz q z qzq z q z, y () qz q z q z qz q z q z q z qz q z q zq qz q z qz q z q z q z., Assue tht Addig the ter to oth sides, we get: 0

11 7. Let P() e the propositio : For P(), L.H.S () R.H.S. P() is true. Assue P() is true for soe N, tht is, or () For P( ), () [( )] ( ), y (*) [( )] ( ) ( )( )( ) ( ) ( )( ) ( )( )( ) ( )( ) ( ) ( ) 0 P() is true. ( )( )( ) 8. Let P() e the propositio : ( ) (), ( ) () For P(), P() is true. ( ), ( ) / ( (*) ) Assue P() is true for soe N, i.e., ( ) (), ( ) () For P( ), ( ) ( ), y () d () ( ) ( ).(5) ( ) ( ),y (5) d () ( ) ( )

12 9. Let P() e the propositio : p - > p > > q > q - > 0 where p 0 > > q 0, p ( p q ) d pq For P(), Note tht p >>0,q >0.. p p q < p p p, q > q p p q >p q, >0, q < Also, p >>0,q >0 p >p >>q >q >0 d P() is true. Assue P() is true for soe N, i.e., p >p >>q >q >0 For P( ), p p q < p p p, q > q p p q >p q, >0, q < Also, p >>0,q >0 P( ) is true. 0. () u! Let S(p) e the stteet : u u For S(), u u, u! S() is true.! Assue S() is true for soe N, i.e., u u For S( ), u u u u u, y ()!!!! u, S( ) is true. By the Priciple of Mtheticl Iductio, S(p) is true p N. () Oviously, u >0!

13 Let P() e the propositio : u < For P(), L.H.S.u, R.H.S , P() is true.! Assue P() is true for soe N, i.e., u < For P( ), u <!! < Sice 0<u <, hece <u < Now, fro (), put, u u u < u <! < <!! < < > >. Let P() e the propositio : l < d l - < l. For P(), l < d l < l P() is true. Assue P() is true for soe N, i.e., l < () d l < l () For P( ), l < < l d l l > l l -. () Prove of i ( )( ) is oitted. 6 i

14 ( ) ( ) i i ( i) ( ) ( ) ( ) ( )( ) ( )( ).() 6 6 () Let P() e the propositio :....() For P(), L.H.S. R.H.S. P() is true. Assue P() is true for soe N, i.e.,....(*) For P( ), i i..., y (*) i ( ) ( ) ( ) ( ) ( ) By (), ( 6. ( ) ( ).... )( ) > [(.)(.( ))...(.) ] / (! ) i [ ] /, y A.M.>G.M.! < ( )( ) 6 By (),... >... /! / <!. Let P() e the propositio : where is rithetic sequece. For P(), L.H.S. R.H.S., P() is true. Assue P() is true for soe N\{}, tht is, () For P( ),, y (), where d is the coo differece.

15 By the Priciple of Mtheticl Iductio, P() is true N\{}. 5

[ 20 ] 1. Inequality exists only between two real numbers (not complex numbers). 2. If a be any real number then one and only one of there hold.

[ 20 ] 1. Inequality exists only between two real numbers (not complex numbers). 2. If a be any real number then one and only one of there hold. [ 0 ]. Iequlity eists oly betwee two rel umbers (ot comple umbers).. If be y rel umber the oe d oly oe of there hold.. If, b 0 the b 0, b 0.. (i) b if b 0 (ii) (iii) (iv) b if b b if either b or b b if