Proof by induction ME 8 n Let f ( n) 9, where n. f () 9 8, which is divisible by 8. f ( n) is divisible by 8 when n =. Assume that for n =, f ( ) 9 is divisible by 8 for. f ( ) 9 9.9 9(9 ) f ( ) f ( ) [9(9 ) ] [9 ] 9(9 ) 9 8(9 ) f ( ) f ( ) 8(9 ) As both f () and 8(9 ) are divisible by 8 then the sum of these two terms must also be divisible by 8. Therefore f (n) is divisible by 8 when n = +. If f (n) is divisible by 8 when n =, then it has been shown that f (n) is also divisible by 8 when n = +. As f (n) is divisible by 8 when n =, f (n) is also divisible by 8 for all n and n by mathematical induction. 0 0 0 0 0 0 a B BB 0 30 3 0 0 0 9 0 9 3 0 0 0 0 0 0 B B B 0 90 3 0 0 0 7 0 7 b As 0 0 3 B and 0, 0 3 3 B we suggest that 3 n B 0 n. 0 3 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
c 0 0 n ; LHS 0 3 0 3 0 0 RHS 0 3 0 3 As LHS=RHS, the matrix equation is true for n =. Assume that the matrix equation is true for n =. 0 0 i.e. 0 3 0 3 With n = + the matrix equation becomes 0 0 0 0 3 0 3 0 3 0 0 0 3 0 3 0 0 0 0 0 0 3(3 ) 0 0 3 Therefore the matrix is true when n = +. If the matrix equation is true for n =, then it is shown to be true for n = +. As the matrix equation is true for n =, it is now also true for all n and n by mathematical induction. 3 Basis: n = : LHS = 3 + 4 = 7; RHS = (3 + ) = 7 3r 4 3 r r r 3 4 3 4 3 4 r r 3 3 3 4 3 7 4 So if the statement holds for n =, it holds for n = +. Conclusion: The statement holds for all n Z +. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
4 a 9 6 9 6 n ;LHS 4 7 4 7 8() 6() 9 6 4() 8() 4 7 As LHS = RHS, the matrix equation is true for n =. Assume that the matrix equation is true for n =. 9 6 8 6 i.e.. 4 7 4 8 With n = + the matrix equation becomes 9 6 9 6 9 6 4 7 4 7 4 7 8 6 9 6 4 8 4 7 7 9 64 8 6 36 4 3 64 7 56 8 9 6 6 4 4 8 7 8( ) 6( ) 4( ) 8( ) Therefore the matrix equation is true when n = +. If the matrix equation is true for n =, then it is shown to be true for n = +. As the matrix equation is true for n =, it is now also true for all n and n by mathematical induction. n b det( A ) (8n )( 8 n) 64n B 8n 64n 8n 64n 8n 6n 4n 8n n ( A ) 8n 6n SoB 4n 8n Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 3
( n) 5 a f ( n ) 5 n 5 n 5.5 n 4(5 ) n n f ( n ) f ( n) [5(5 ) ] [5 ] Therefore, µ = 4. n n 5(5 ) (5 ) n 4(5 ) b n f ( n) 5, where n. () f () 5 5 6, which is divisible by 6. f(n) is divisible by 6 when n =. Assume that for n =, f ( ) 5 is divisible by 6 for. Using (a) f ( ) f ( ) 4(5 ) f ( ) f ( ) 4(5 ) As both f() and 4(5 ) are divisible by 6 then the sum of these two terms must also be divisible by 6. Therefore f(n) is divisible by 6 when n = +. If f(n) is divisible by 6 when n =, then it has been shown that f(n) is also divisible by 6 when n = +. As f(n) is divisible by 6 when n = n, f(n) is also divisible by 6 for all n and n by mathematical induction. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4
n n 6 Let f ( n) 7 4, where n. f () 7 4 7 4, which is divisible by 6. f(n) is divisible by 6 when n =. Assume that for n =, f ( ) 7 4 is divisible by 6 for. f ( ) 7 4 7.7 4.4 7(7 ) 4(4 ) f ( ) f ( ) [7(7 ) 4(4 ) ] [7 4 ] 7(7 ) 4(4 ) 7 4 6(7 ) 3(4 ) 6(7 ) 3(4 ).4 6(7 ) (4 ) 6[7 (4) ] f ( ) f ( ) 6[7 (4) ] As both f() and divisible by 6. 6[7 (4) ] are divisible by 6 then the sum of these two terms must also be Therefore f(n) is divisible by 6 when n = +. If f(n) is divisible by 6 when n =, then it has been shown that f(n) is also divisible by 6 when n = +. As f(n) is divisible by 6 when n =, f(n) is also divisible by 6 for all n and n by mathematical induction. 7 Basis: n = : LHS = 5 = 5; RHS = ( + 3) = 5 6 r r 4 3 6 r r r 4 r r 4 5 r r = ( + )( + 3) + ( + )( + 5) 6 = 6 (3 + + 49 + 30) = ( + )( + )(( + ) + 3) 6 So if the statement holds for n =, it holds for n = +. Conclusion: The statement holds for all n. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5
8 a Basis: n = : LHS = + 4 = 5; RHS = 3 3 5 = 5 r 4 3 r r r r r 3 3 8 30 37 5 3 4 3 4 So if the statement holds for n =, it holds for n = +. Conclusion: The statement holds for all n Z +. b Using a and the formula for n r nn 4n nn n 6 6 n(n + ) + (4n + ) = n(n + )(n + ) 6n 3 + n 3 + n + n = (n 3 + 3n ) r n 8n 6n n 4n 8n n n 3n n n n n 8n n 8 n 8 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
9 a Basis: n = : LHS = RHS = M c c 0 c 0 c c M M 0 c c c c c c c 0 0 c 0 0 c c c c 0 0 So if the statement holds for n =, it holds for n = +. Conclusion: The statement holds for all n Z +. b Consider n = : det M = 50 c = 50 So c = 5, since c is +ve. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
Challenge a Basis: n = : LHS = RHS = cos sin sin cos cos sin M sin cos cos sin cos sin cos sin M M sin cos sin cos sin cos cos cos sin sin cos sin sin cos sin cos cos sin sin sin cos cos cos sin sin cos So if the statement holds for n =, it holds for n = +. Conclusion: The statement holds for all n Z +. b The matrix M represents a rotation through angle θ, and so M n represents a rotation through angle nθ. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8