James Mathews
Part I Problems, second set
12 13 14 16 17 18 20 21 25
12.
Prove by induction on n that, for all positive integers n, 3 divides 4n  5 .
13.
Prove by induction on n that n !  2 n for all integers n such that n  4
14.
Prove Bernoulli’s inequality (1  x)n  1  nx for all non-negative integers n and
real numbers x  1 .
n
1
n

16.
Prove by induction on n that 
for all positive integers n.
n 1
i 1 i (i  1)
17.
For a positive integer n the number an is defined inductively by
a1  1
6a  5
ak 1  k
, for k a positive integer.
ak  2
Prove by induction on n, that for all positive integers n, (i) an  0 and (ii) an  5 .
n
18.
Given a sequence of numbers a(1), a(2),..., the number
 a(i) is defined
i 1
inductively by
1
(i)
 a(i)  a(1) , and
i 1
 k

a
(
i
)


  a(i)  a(k  1) for k  1
i 1
 i 1

2n
n
i 1
1 x
Prove that  (1  x 2 ) 
for x  1 . What happens if x  1 ?
1 x
i 1
k 1
(ii)
20.
Prove that, for a positive integer n, a 2n  2n square grid with any one square
removed can be covered using L-shaped tiles of the following shape:
.
21.
Suppose that x is a real number such that x  x 1 is an integer. Prove by induction
on n that x n  x  n is an integer for all positive integers n.
25.
Let un be the nth Fibonacci number. Prove, by induction on n, that
um n  um1un  umun 1 .
12.
Suppose 3 divides 4n  5 for some integer n. Then q  , 4n  5  3q . By
multiplication by 4 and distributivity, 4n 1  20  12q . By subtraction,
4n1  5  12q  15 . Distributivity yields 4n1  5  3(4q  5) . Since is closed under
multiplication and addition, r  , r  4q  5 , so for some integer r, 4n 1  5  3r ; 3
divides 4 n1  5 . Thus, for the inductive step, (3 divides 4n  5 )  (3 divides 4 n1  5 ).
But the proposition is true when n  0 since 3 divides 40  5  6  3(2) . By
induction, 3 divides 4n  5 for all integers n  0 . ☺
13.
Suppose n !  2 n for some integer n  4 . Then by multiplication by the positive
integer n  1, (n  1)!  2n (n  1) . But n  4  n  1  n  1  2  2n (n  1)  2n1 . Thus,
since (n  1)!  2n (n  1) and 2n (n  1)  2n1 , by transitivity, (n  1)!  2n 1 . Then the
inductive step is valid; n !  2 n  (n  1)!  2n 1 .
The proposition holds for n  4 since 4!  24 , 2 4  16 , and 24  16 . By
induction, n !  2 n for all integers n  4 . ♣
Suppose that x  | x  1 , n  | n  0 and (1  x)n  1  nx . Then since
x  1  x  1  0 , multiplication by 1  x yields (1  x)n (1  x)  (1  nx)(1  x) . By
distributivity and the rule of exponents, (1  x)n1  1  (n  1) x  nx 2 . Since n is nonnegative, 0  nx 2 (appeal to previous homework notion that x  , 0  x2 ). By
14.
addition, then 1  (n  1) x  1  (n  1) x  nx 2 . By transitivity, then (1  x)n1  1  (n  1) x .
The inductive step holds; (1  x)n  1  nx  (1  x)n1  1  (n  1) x .
And the proposition holds for n  0 since (1  x)0  1  0 x 1  1 . By induction,
for all integers n  0 and real numbers x  1 , (1  x)n  1  nx . ☼
n
16.
Suppose that for some positive integer n,
1
n
 i(i  1)  n  1 .
i 1
n
1
n
n
1
1
n
1
 i(i  1)  n  1   i(i  1)  (n  1)(n  2)  (n  1)  (n  1)(n  2)
i 1
i 1
n 1
n 1
1
n(n  2)  1
1
n2  2n  1
1
(n  1) 2






(n  1)(n  2)
(n  1)(n  2)
(n  1)(n  2)
i 1 i (i  1)
i 1 i (i  1)
i 1 i (i  1)
n 1
n
1
n 1
1
n




. Then the inductive step is valid, 
n2
n 1
i 1 i (i  1)
i 1 i (i  1)
n 1
1
n 1
1
1
1 1


  .
. And the proposition holds for n  1 since

n2
1(1  1) 1  1 2 2
i 1 i (i  1)
n
1
n

Then by induction, for all positive integers n, 
.◙
n 1
i 1 i (i  1)
n 1
17.
(i) Suppose that for some integer n  1, an  0 . Then since 6  0 , 6an  0 , and
then 6an  5  5 , and then since 5  0 , by transitivity, 6an  5  0 . For the same reason,
(an  2) 2
1
 0 . So by
 0 , but by division, this implies that
an  2  0 . And so
an  2
(an  2)
6a  5
 0 . Thus an  0  an1  0 . The inductive step holds.
multiplication, n
an  2
Since a1  1, a1  0 . By induction, an  0 for all integers n  1. ‫ﭿ‬
(ii) Suppose that for some integer n  1, an  5 . Then 6an  5an  10  5 , then
6a  5
 5 . Thus
6an  5  5an  10 , and then 6an  5  5(an  2) , and so n
an  2
an  5  an1  5 .
Since a1  1, a1  5 , and by induction, an  5 for all integers n  1. Җ
n
18.
Suppose that for some integer n ,
n 1
(1  x )(1  x )
, then
)
1 x
 (1  x
i 1
 (1  x
2i 1
i 1
n 1
 (1  x
i 1
2i1
2n
)
2n
n 1
 (1  x
1  x2
. Then
)
1 x
n
2i 1
1  x 22
, and then, as desired,
)
1 x
n
2i1
i 1
2n1
1 x
.
1 x
The proposition holds for n  1 since 1  x1 
(1  x)(1  x) 1  x 2
. By induction,

(1  x)
1 x
1  x2
for all positive integers n,  (1  x ) 
.¶
1 x
i 1
If x  1 , the formula fails because of a simplification of the actual iterated
n
n
n
product. In this case,
2i 1
n
 (1  x 2 )   2  2n .
i 1
i 1
i 1
20.
Suppose that for some positive integer n, we may tile a 2n  2n square grid with
any 1 square removed using the L-shaped tiles. Any 2n1  2n1 grid with 1 square
removed consists of four 2n  2n grids, where one of them has 1 square removed. By the
inductive hypothesis, the 2n  2n grid which is missing a square may be tiled. The three
grids which remain are always connected in the same L-shape. Since, by the inductive
hypothesis, we can tile all three of these provided we remove any one square from each,
we choose to remove the square which is closest to the center of the L-joint. This leaves
room for precisely one additional L-shaped tile, which we place in the void. Then the
entire 2n1  2n1 grid is tiled except for any one square. The inductive step holds.
In the case of n  1 , rotation of one tile shows that a 21  21 grid missing one
square may always be tiled. By induction, for all positive integers n, any 2n  2n grid
may be tiled with the L-shaped tiles. £
This is an illustration of the inductive step:
21.
Suppose that x is a real number such that x  x 1 is an integer, and n is an integer
such that x n  x  n and x n 1  x  ( n 1) are integers. Then since is closed under
multiplication, ( x n  x  n )( x  x 1 ) is an integer. But then so is
is closed under addition and subtraction as
x n1  x ( n1)  ( x n1  x ( n1) ) , and since
n 1
 ( n 1)
well, then so is x  x
, as desired. The inductive step holds.
1
We let x  x be an integer, so the proposition holds for n  1 . This implies that
it holds for n  2 since ( x  x 1 )2  2  x 2  x 2  2  2  x 2  x 2 is an integer. By strong
induction, x n  x  n is an integer for all positive integers n. ¥
25.
Suppose that for all positive integers m, the proposition holds for two consecutive
integers n 1 and n; that is, um n1  um1un1  umun and um n  um1un  umun 1 . By the
Fibonacci definition, um n1 : um n1  um n . Substitution yields
umn1  um1un1  umun  um1un  umun 1 , and then um n1  um1 (un1  un )  um (un  un1 ) .
But the Fibonacci definition guarantees that um n1  um1un1  umun 2 and so
um  ( n 1)  um1u( n 1)  umu( n 1) 1 . The inductive step holds.
The case of n  1 holds since um1  um1u1  umu2 is guaranteed by the definition
of the Fibonacci numbers and the fact that u1  1 and u2  1 . The case of n  2 holds as
well since um 2 : um1  um , and then um 2  um1  um  um , so um 2  um1 (1)  um (2) , and
then um 2  um1u2  umu3 , as desired. Since these are consecutive cases, by strong
induction, for all positive integers m and n, um n  um1un  umun 1 .۞
Suppose um divides umn for all m and some n. By the previous result,
um ( n 1)  umn  m  umn 1um  umn um 1 . By distributivity, since um divides umn and um divides
umn1um , um divides um ( n 1) . Since um divides u m (1) intrinsically, by induction, um divides
umn for all positive integers m and n. ۩
This is mildly puzzling; since m and n are free, doesn’t commutativity permit
um n  um1un  umun1  um1un  umun1  un (um1  um1 )  um (un1  un1 )  0 ? But then
un (um  um1  um1 )  um (un  un1  un1 )  0 , and so unum  umun  0 . Oh. Dur.
So every Fibonacci number is the difference of its neighbors. This is like the
exponential function. To expose the Fibonacci nature of the exponential function, we
should show that there must exist some positive b such that
a b
ea 
 e dx  e
x
a b
 ea b  ea (eb  eb ) . Strangely, for that b, eb  eb  1 , so that eb
a b
satisfies the quadratic equation (eb )2  (eb )  1  0 , whose roots are the golden ratios. So
the exponential function behaves precisely like the Fibonacci sequence provided we
consider the neighbors of a point located at x to be located at x  ln( ) .
So why is the Fibonacci sequence different from the exponential function at all?
Because no two exponential neighbors x and x  ln( ) are equal. The numbers
u1  u2  1 are the generators of the standard Fibonacci sequence. The embedded
Fibonacci sequence of the exponential function has the generators a1  1 and a2   ,
producing a different sequence with the same basic sequential property.