Case 3.ATay-Sachs Disease
A 6-month-old infant girl presented to the emergency room with her first seizure. Her parents
reported that she had not had any fever, nor had she had any vomiting, diarrhea, or signs of
viral illness. The parents were also conce

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

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

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

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.

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

(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

divides 4n+1 + 52n1 whenever n is a
positive integer. 37. Prove that if n is a
positive integer, then 133 divides 11n+1 +
122n1. Use mathematical induction in
Exercises 3846 to prove results about
sets. 38. Prove that if A1, A2,.,An and B1,
B2,.,Bn are se

an ear if the line segment connecting the
two vertices adjacent to vi is an interior
diagonal of the simple polygon. Two ears
vi and vj are called nonoverlapping if the
interiors of the triangles with vertices vi
and its two adjacent vertices and vj and
i

(described in Section 8.1) appeared.
Fibonacci also wrote books on geometry
and trigonometry and on Diophantine
equations, which involve finding integer
solutions to equations. P1: 1 CH05-7R
Rosen-2311T MHIA017-Rosen-v5.cls May
13, 2011 10:25 5.3 Recursiv