used to prove the stronger inequality 1 2
3 4 2n 1 2n < 1 3n + 1 for all
integers greater than 1, which, together
with a verification for the case where n =
1, establishes the weaker inequality we
originally tried to prove using
mathematical induction. 7
integers n, because (1) tells us P (1) is
true, completing the basis step and (2)
tells us that P (1) P (2) P (k)
implies P (k + 1), completing the inductive
step. Example 1 illustrates how strong
induction can help us prove a result that
cannot easily be
of a golf course with an infinite number of
holes and that if this golfer plays one hole,
then the golfer goes on to play the next
hole. Prove that this golfer plays every
hole on the course. Use mathematical
induction in Exercises 317 to prove
summation
and the last two lines must meet in a
common point p2. But in this case, p1 and
p2 do not have to be the same, because
only the second line is common to both
sets of lines. Here is where the inductive
step fails. Guidelines for Proofs by
Mathematical Indu
Assume that whenever max(x, y) = k and x
and y are positive integers, then x = y.
Now let max(x, y) = k + 1, where x and y
are positive integers. Then max(x 1, y
1) = k, so by the inductive hypothesis, x
1 = y 1. It follows that x = y, completing
the in
this stable assignment. Use strong
induction to show that the deferred
acceptance algorithm produces a stable
assignment that is optimal for suitors. 25.
Suppose that P (n) is a propositional
function. Determine for which positive
integers n the statement
If k + 1 is prime, we immediately see that
P (k + 1) is true. Otherwise, k + 1 is
composite and can be written as the
product of two positive integers a and b
with 2 a b 3. Consider the first three
elements of this cycle, p1, p2, and p3.
There are two pos
whenever n is an integer greater than or
equal to 3. In Exercises 47 and 48 we
consider the problem of placing towers
along a straight road, so that every
building on the road receives cellular
service. Assume that a building receives
cellular service if
nonnegative integer. 6. Prove that 1 1! +
2 2!+ n n! = (n + 1)! 1 whenever n
is a positive integer. 7. Prove that 3 + 3 5
+ 3 52+ 3 5n=3(5n+1 1)/4
whenever n is a nonnegative integer. 8.
Prove that 2 2 7 + 2 72 + 2(7)n =
(1 (7)n+1)/4 whenever n is a
nonne
property can be used to show that there is
a unique greatest common divisor of two
positive integers. Let a and b be positive
integers, and let S be the set of positive
integers of the form as + bt, where s and t
are integers. a) Show that S is nonempty.
INDUCTIVE STEP: We show that the
conditional statement [P (1) P (2)
P (k)] P (k + 1) is true for all positive
integers k. Note that when we use strong
induction to prove that P (n) is true for all
positive integers n, our inductive
hypothesis is the assu
into (n2 + n + 2)/2 regions if no two of
these lines are parallel and no three pass
through a common point. 63. Let a1,
a2,.,an be positive real numbers. The
arithmetic mean of these numbers is
defined by A = (a1 + a2 + an)/n, and
the geometric mean of th
recursively defined set unless it is in the
initial collection specified in the basis step
or can be generated using the recursive
step one or more times. Later we will see
how we can use a technique known as
structural induction to prove results
about re
k, then P (k + 1) is true. That is, for the
inductive hypothesis we assume that P (j )
is true for j = 1, 2,.,k. The validity of both
mathematical induction and strong
induction follow from the wellordering
property in Appendix 1. In fact,
mathematical in
string containing no symbols).
RECURSIVE STEP: If w and x , then
wx . GABRIEL LAM (17951870)
Gabriel Lam entered the cole
Polytechnique in 1813, graduating in
1817. He continued his education at the
cole des Mines, graduating in 1820. In
1820 Lam went to
of the ladder. However, there is no obvious
way to complete this inductive step
because we do not know from the given
information that we can reach the (k +
1)st rung from the kth rung. After all, we
only know that if we can reach a rung we
can reach the
why these steps show that this inequality
is true whenever n is an integer greater
than 1. 20. Prove that 3n< n! if n is an
integer greater than 6. 21. Prove that 2n >
n2 if n is an integer greater than 4. 22. For
which nonnegative integers n is n2 n!?
Pr
attempt to prove P (n) for all integers
nwith n 3 using strong induction, the
inductive step does not go through. b)
Show that we can prove that P (n) is true
for all integers n with n 3 by proving by
strong induction the stronger assertion
Q(n), for n 4,
must successively make to break the bar
into n separate squares. Use strong
induction to prove your answer. 11.
Consider this variation of the game of
Nim. The game begins with n matches.
Two players take turns removing
matches, one, two, or three at a ti
triangles in the triangulation have two
sides that border the exterior of the
polygon. 18. Use strong induction to
show that when a simple polygon P with
consecutive vertices v1, v2, ., vn is
triangulated into n 2 triangles, the n 2
triangles can be numbe
b. 5. State what needs to be proved under
the assumption that the inductive
hypothesis is true. That is, write out what
P (k + 1) says. 6. Prove the statement P (k
+ 1) making use the assumption P (k). Be
sure that your proof is valid for all
integers k w
k i=1 pi = 1.] 54. Use mathematical
induction to show that given a set of n + 1
positive integers, none exceeding 2n,
there is at least one integer in this set that
divides another integer in the set. 55. A
knight on a chessboard can move one
space horizo
(k + 1) follows from P (k) for every
positive integer k, then it also follows that
P (k + 1) follows from all the statements P
(1), P (2), . . . , P (k), because we are
assuming that not only P (k) is true, but
also more, namely, that the k 1
statements P
Lam was the foremost French
mathematician of his time. However,
French mathematicians considered him
too practical, whereas French scientists
considered him too theoretical. P1: 1
CH05-7R Rosen-2311T MHIA017-Rosenv5.cls May 13, 2011 10:25 350 5 /
Inductio
the recursive step. To prove results about
recursively defined sets we use a method
called structural induction. Recursively
Defined Functions We use two steps to
define a function with the set of
nonnegative integers as its domain: BASIS
STEP: Specify th
, where a and b are real numbers. Show
that An = an 0 0 bn
for every positive integer n. 57. (Requires
calculus) Use mathematical induction to
prove that the derivative of f (x) = xn
equals nxn1 whenever n is a positive
integer. (For the inductive step, u
2, because rn < rn1. This implies that
rn 1 = f2, rn1 2rn 2f2 = f3, rn2
rn1 + rn f3 + f2 = f4, r2 r3 + r4
fn1 + fn2 = fn, b = r1 r2 + r3 fn +
fn1 = fn+1. It follows that if n divisions
are used by the Euclidean algorithm to
find gcd(a, b) with a b, the
a strong induction proof that P (n) is true
for n 8. a) Show that the statements P
(8), P (9), and P (10) are true, completing
the basis step of the proof. b) What is the
inductive hypothesis of the proof? c)
What do you need to prove in the
inductive ste