Download Sample Chapter

2
Proof Techniques
31
C H A P T ER
In this chapter, a number of statements are given and the different techniques to prove such statements
are dealt with. Proof is an art of convincing the reader that the given statement is true. The proof
techniques are chosen according to the statement that is to be proved. Some of the proof techniques
start with the given statement and some of them start with the opposite of the given statement and
some statements are proved by constructing a model. This fundamental idea of the proving techniques
is required to prove the theorems that are stated in the later chapters.
Proof of a mathematical statement is an art of convincing the reader that the given statement is correct.
There are different methods to prove a statement based on the way the proof starts and proceeds.
Direct proof technique is used to prove implication statements which have two parts, an “if-part”
known as Premises and a “then part” known as Conclusions. In this proof technique one starts with
the premise and proceeds directly to conclusions with a chain of implications that use known facts,
laws and formulas.
The proof technique is explained in detail in the following illustration.
ILLUSTRATION 2.1: Prove that “If n is an odd integer then n2 is odd”.
Given: n is an odd integer
To Prove: n2 is odd
PROOF : The if-part of the given statement is “n is an odd integer” and then-part of the given statement
is “n 2 is odd”.
Proof Techniques
33
To prove: 0 < x < 1
PROOF : Here the premises p is x 3 + x 2 – 2x < 0 and the conclusion c is 0 < x < 1. It is assumed that
the conclusion is false.
Therefore, for x ≥ 1, it is to be shown that x3 + x2 – 2x ≥ 0
x3 + x2 – 2x = x (x2 + x – 2)
= x(x – 1) (x + 2)
(2.3)
Since it is assumed that x ≥ 1, (x – 1) and (x + 2) will be positive numbers in equation (2.3). When
a set of positive numbers are multiplied, result will be a positive number, hence x 3 + x2 – 2x > 0.
The assumption ‘that the conclusion is false’ leads to negation of premises. The given statement
is true by contra positive method.
Proof by contradiction is a technique which can be used for all the type of statements. In this method
it is assumed the statement that is to be proved is false and showed that this assumption leads to
contradiction of some well-known fact or contradicts some other assumption made earlier in the proof.
Hence, it is concluded that the statement cannot be false.
ILLUSTRATION 2.4: Prove the statement S “ 2 is irrational” using proof by contradiction technique.
A real number r is rational if only if it can be written as r = a/b where a and b are integers and b π 0.
To prove:
PROOF :
2 is irrational
It is assumed that the statement is false (i.e.) assumed that
2 is rational.
A rational number can be written as division of two integers therefore 2 can be written as a/b
where ‘a’ and ‘b’ are some integers. After cancelling the common factors of a and b
2 = p/q
(2.4)
Since p and q do not have any common factors they are said to be relatively prime. Squaring equation
(2.4) on both sides yields
p2
=2
q2
p 2 = 2q 2
(2.5)
(2.6)
It may be inferred from equation (2.6) that p2 is even since it is a multiple of 2. It is already known
that p is even whenever p2 is even. Therefore, p can be written as 2 k where k is an integer. Replace
p in equation (2.6)
(2k) 2 = 2q2
Reducing the above equation leads to q 2 = 2k2, which implies q is even. Since both p and q are even
they have a common factor 2 which contradicts the earlier assumption that p and q do not have
common factors. Hence, “ 2 is irrational” is proved by contradiction method.
34
Theory of Computation
Proof by contradiction technique can be applied to implication statements also. When proof by
contradiction technique is used for implication statement, it is assumed that the premise is true and
conclusion is false and show that this assumption leads to conclude that something already known or
assumed is false.
ILLUSTRATION 2.5:
If ‘a’ is a real number and greater than zero, then
1
is also greater than zero.
a
Given: ‘a’ is a real number and greater than zero
To Prove:
1
is greater than zero
a
PROOF : The given statement is an implication statement, hence, it is assumed that the premise is true
and the conclusion is false. The assumption is
a > 0 and
Since
1
£0
a
1
is less than zero there exists some real number b > 0 such that
a
1
+b=0
a
Multiplying both sides by a, the following equation is obtained
1 + ab = 0
(2.7)
As per the assumption made earlier, the numbers a and b are greater than zero hence multiplication
of a and b is also greater than zero. From equation (2.7) one may infer that
1<0
which is contradiction to the well known fact that 1 > 0. Therefore, the assumption made earlier is
1
wrong. Hence, the statement “If ‘a’ is a real number and greater than zero, then is also greater than
a
zero” is true.
In this method of proof the properties in the statement given are demonstrated with an existing object
(or) if such an object does not exist then a method (algorithm with steps) to create the object is
provided.
ILLUSTRATION 2.6: If a person can add any two numbers then he can add n numbers.
Given: a person can add any two numbers
To Prove: he can add n numbers
PROOF :
The statement is proved by proof by construction and the steps are as follows:
Proof Techniques
35
To add the n numbers a1, a2, a3, …, a n:
1.
2.
3.
4.
Assign Sum = a1 and j = 2.
Sum = Sum + aj.
j = j + 1, go to Step 2 till j £ n.
Sum contains the result of addition of n numbers.
Thus, the statement is proved.
This is different from the proving methods seen above, this method of proof is used to prove that the
given statement is false. Using this method it is easy to prove that generalized statements are false. It
is sufficient to choose one sample and say the statement does not hold good.
ILLUSTRATION 2.7:
Consider the following statement “All primes are odd” and prove it is false.
PROOF : The statement given above is a generalized statement and it generalizes that the set of all prime
numbers are odd. It can be proved false by considering one prime and show that the prime number
is even. The number 2 is considered, 2 is a prime number but it is even hence the given statement is
false.
ILLUSTRATION 2.8: Prove the statement “There is no pair of positive integers a and b such that a mod
b = b mod a” is false.
To Prove: There is a pair of integers a and b such that a mod b = b mod a
PROOF : Since this is a generalized statement, the statement is proved to be false by showing that there
is a pair of integers for which a mod b = b mod a.
When number a is equal to number b, a mod b = b mod a = 0. Hence, the statement is proved false
using proof by counter example technique.
If and only If statements are special type of if statements which implies that when the premise is true
the conclusion is true and when the conclusion is true premise is also true. These types of statements
are special type of implication statements similar to English statements that have ‘vice versa’ as part
of their statements. ‘If and only If’ statements may be shortly referred using ‘iff’.
To prove ‘iff’ statements the statement is divided into two implication statements and both the
statements are proved using any of the proof techniques seen above. The two statements obtained after
splitting the ‘if and only if’ statements are called ‘if-part’ and ‘only-if part’. Consider a ‘iff’ statement
‘A if and only if B’, this statement is splitted as follows
1. If-part: if B then A.
2. Only-if part: if A then B.
36
Theory of Computation
ILLUSTRATION 2.9: Ceil of a real number x is equal to its floor iff x is an integer.
Ceil of a real number is the least integer greater than the given number and floor of a real number
is the greatest integer less than that of the given number. Ceil of a number 3.6 represented as È3.6˘
is 4 and floor of the same number 3.6 represented as Î3.6˚ is 3.
PROOF:
To prove this statement split it into if-part and only if-part as follows:
If-part: If x is an integer, then its ceil is equal to its floor.
Only-if part: If ceil of a number is equal to its floor, then the number is an integer.
This part of the statement is proved using direct proof technique. By the definition of ceil and floor,
when x is an integer, its ceil is equal to its floor and it is x itself. Hence, the statement is proved.
This part of the statement is proved using proof by contradiction. It is assumed that Îx˚ = Èx˘ and x
is not an integer. By the definition of ceil and floor for any x, Èx˘ >= x and Îx˚ <= x, since it is assumed
that Îx˚ = Èx˘, it may be concluded that Îx˚ = Èx˘ = x.
Again by considering the definition of ceil and floor it is noted that x must be an integer which
contradicts the assumption that x is not an integer. Hence, the statement ‘If ceil of a number is equal
to its floor, then the number is an integer’ is true.
Different proving techniques may be used to prove the two parts of the iff statements as in
Illustration 2.9.
ILLUSTRATION 2.10: Prove the statement ‘a mod b is equal to b mod a if and only if a is equal to
b where a and b are positive integers’.
PROOF: The first step in proving “iff” statements is to split the given statement into two parts if-part
and only-if part. The given statement is split into “if-part” and “only-if part” as given below:
If-part: if a = b then a mod b = b mod a
Only-if part: if a mod b = b mod a then a = b.
This part is proved using direct proof technique. By the definition of modulus, x mod x = 0. Therefore,
when a = b, a mod b = b mod a = 0.
This part of the statement is proved using proof by contradiction. Therefore, it is assumed that “a mod
b is equal to b mod a”, and “a is not equal to b”.
When ‘a’ is not equal to ‘b’, either a > b or a < b.
Let r = a mod b and r1 = b mod a
Proof Techniques
37
When a > b, r < b, as per the definition a mod b returns only values from 0 to b – 1, and r1 = b.
The assumption “a mod b is equal to b mod a” leads to conclude that b is less than itself. By
mathematical theory, a number cannot be less than itself hence the assumption made earlier has led
to a contradiction.
2.7
PROOF BY INDUCTION
Proof by induction is a technique used to prove statements about objects that can be defined using
recursive functions.
2.7.1
Recursive functions
A function f is called a recursive function when the same set of operations are to be performed on the
result of f (n) to obtain f (n + 1), irrespective of the value of n. Recursive functions have two conditions
i. Base Condition: The result of f (n0) is explicitly given, where n0 is the lowest value that the
variable in the function can take, and this is the condition that will be used in evaluation before
it terminates.
ii. Recursive Condition: Set of operations to be performed on the result of f (n) to find f (n + 1).
ILLUSTRATION 2.11:
A recursive definition is given for the function g(n) which finds the sum of
integers 1 to n. (i.e.) g (n) = 1 + 2 + 3 + … + n =
Âi = 1 i .
n
i. Base Condition: The variable in the given statement is n and the lowest value that can be
assigned for n is 1 therefore give the value of g(1) explicitly in the base condition, g(1) = 1.
ii. Recursive Condition: Here the set of operations to be performed are specified on the result
of f (n) to obtain f (n + 1). In this illustration f (n + 1) can be obtained by adding (n + 1) to f (n).
Using the recursive definition of g (n) the value of g (5) is found as follows. Here the value
of n is 5. Only the value of g (1), can be directly found using the base condition of the function.
For other values of n, the recursive condition is applied repeatedly till g (1) as shown below:
g(5) =
=
=
=
=
=
g(4) + 5
g(3) + 4 + 5
g(2) + 3 + 4 + 5
g(1) + 2 + 3 + 4 + 5
1+2+3+4+5
15.
ILLUSTRATION 2.12: Give the recursive definition for a function f (n), that finds factorial of n.
Factorial of a number n is found by multiplying numbers 1 to n (i.e.) n! = 1 ¥ 2 ¥ 3 ¥ … ¥ n
i. Base Condition: Generally, factorial is found for numbers greater than one. f(1) is defined as
f(1) = 1
ii. Recursive Condition: f (n +1) can be found by multiplying (n + 1) to the result of f (n).
38
Theory of Computation
The principle of induction states that to prove a statement S is true for all values of n ≥ n0, it is sufficient
to do the following steps:
i. Base Step: S is proved to be true for n0 using some facts or laws already known.
ii. Induction Hypothesis: S is assumed to be true for some value of k, where k is greater than
n0.
iii. Induction Step: S is proved to be true for (k + 1) using the induction hypothesis, well-known
facts and formulas. The same concept is given as flow diagram in Fig. 2.1.
Prove that S is true
for minimum value n0
Assume that statement is
true for some value k > n0
Prove that S is true for
(k + 1) using the
assumption made in
induction hypothesis
Fig. 2.1
Flow diagram for principle of induction
The above concepts are illustrated with the following examples:
ILLUSTRATION 2.13: The statement “Summation of numbers from 1 to n is equal to n(n + 1)/2” is
proved using the principle of mathematical induction.
n
The given statement can be written as
Â
=
i =1
n (n + 1)
.
2
i. Base Step: In the base step the veracity of the statement for the minimum value of n is analyzed.
In the given statement the minimum value that n can take is 1.
Summation of numbers from 1 to n, when n = 1 is 1.
and
n(n +1)/2 for n = 1 is 1*(1 + 1)/2 = 1.
Hence, the statement is true for the base condition.
ii. Induction Hypothesis: It is assumed that the statement is true for some value k > 1, which
k
is written as
Â
= k (k + 1)/2.
i =1
iii. Induction Step: It is to be proved that the statement is true for the value (k + 1) using the
k +1
induction hypothesis (i.e.) prove that
Â
i =1
= (k + 1) ((k + 1) + 1)/2.
40
Theory of Computation
k 2 ≥ 2k + 1
(k2 – 1) ≥ 2k
(k + 1) (k – 1) ≥ 2k
(2.12)
Dividing on both sides by 2, and using the idea that the minimum value that can be assigned for
k is 4 equation (2.12) can be written as
(k +1) *m ≥ k
Where m ≥ 1.5, since (k – 1) will be at least 3 and (k – 1)/2 will be 1.5 and any number (k + 1)
multiplied by 1.5 will result in a value greater than k. Hence, the relation is proved to be true
using proof by induction.
This principle is similar to the principle of mathematical induction; and provides three steps to be
followed to prove a statement. The basis and the induction step are same as that of principle of
induction and a slight modification is being made in the induction hypothesis. Instead of assuming a
statement is true for some value k > n0, here it is assumed that the statement is true for all values greater
than n0 and less than k, for k > n0 and n0 is the minimum value for which the statement is true.
The flow diagram for the steps in strong principle of induction is given in the following Fig. 2.2.
Prove that S is true for
minimum value n0
Assume that the statement is
true for all values greater
than n0 and less than k
Prove that S is true for (k + 1)
using the assumption made in
Fig. 2.2
Flow diagram for steps in strong principle of induction
ILLUSTRATION 2.15: Prove the statement “If n ≥ 2, then n is either a prime number or n can be written
as product two or more prime numbers”.
The value of n0, the minimum value of integer for which the statement is true is 2.
Given: n ≥ 2
To Prove: n is either a prime number or n can be written as product two or more prime numbers.
i. Basis Step: The statement is verified for n = 2. Since 2 is a prime number S is true for basis
step.
Proof Techniques
41
ii. Induction Hypothesis: It is assumed that S is true for all values greater than 2 and less than
k. Therefore, numbers between 2 and k are either prime or can be written as product of two
or more prime numbers.
iii. Induction Step: It is to be proved that (k +1) is prime or can be written as product of two or
product of two or more prime numbers. There are only two possibilities for the number (k + 1),
either (k + 1) can be prime number or not. If (k + 1) is prime, then the statement S is true trivially.
Otherwise (k + 1) can be written as product of some two numbers p and q
(k + 1) = p * q
p and q are less than k and by induction hypothesis all numbers less than k may be written as
product of two or more prime numbers. Hence, if the number (k + 1) is not prime, then it can
be written as product of two or more prime numbers.
ILLUSTRATION 2.16: Consider the statement “If n ≥ 8 then n can be written as sum of 3¢s and 5¢s”
and prove using the strong principle of induction.
i. Basis Step: The minimum value for which the statement is given to be true is 8, and 8 can be
written as 8 = 3 + 5 hence S is true for base case.
ii. Induction Hypothesis: Assume S is true for all values greater than 8 and less than some value k.
iii. Induction Step: It is to be proved that S is true for (k + 1). Further, (k + 1) can be written
as
(k + 1) – 3 + 3 = (k – 2) + 3
By induction hypothesis (k – 2) can be written as sum of 3¢s and 5¢s hence (k + 1) can be written
as sum of 3¢s and 5¢s.
In structural induction, the principle of induction discussed above is used to prove statements about
recursively defined structures used in mathematics and computer science. Here the non-linear
structure tree is considered to explain structural induction. Tree is a non-linear data structure widely
used in the field of computer science.
Trees can be constructed from a given finite set of elements by first partitioning the set into two
subsets S1 and S2. S1 contains only one element and that element is the root of the tree to be
constructed. The elements of other subset S2 can be either nodes or trees or combination of nodes
and trees. Elements of S2 are attached to the root through a link generally referred as edges of the
tree.
When two nodes X1 and X 2 are joined through an edge, the node X1 in the top of the edge is known
as the parent node and X2 (generally referred as child) is at the bottom of the edge. Nodes in the last
level of the tree are known as leafs of the tree; leaves of the tree do not have any children. The recursive
definition may be formally given for construction of trees as follows:
i. Base Condition: A single node is a tree and the tree contains root as well as leaf.
Proof Techniques
43
The following examples illustrate the concept of structural induction for the recursively defined
structure, tree.
Illustration 2.18: Consider the following statement and prove it using structural induction “for
every tree the number of nodes is the number of edges plus one”.
i. Basis Step: A tree can have minimum of one node in it and the number of edges is zero when
the number of nodes is one hence the statement is true for the base case.
ii. Induction Hypothesis: It is assumed that S is true for some value k (i.e.) if a tree contains k
edges then it contains (k + 1) nodes.
iii. Induction Step: It is to be proved that if a tree contains (k + 1) edges then it contains
(k + 2) nodes. To construct a tree with (k + 2) nodes a node has to be added to the tree with
(k + 1) edges through an edge and by induction hypothesis there are k nodes in a tree with
(k + 1) edges; hence there will be (k + 2) for a tree with (k + 1) edges.
In this chapter, the different techniques of proof are dealt. The techniques to be used are chosen
according to the type of statement that is to be proved. The key points given in this chapter are recalled
here:
∑ Proof is an art of convincing the reader that the given statement is true.
∑ Direct proof technique is used to prove implication statements which have two parts, an “if part”
known as Premises and a “then part” known as Conclusions.
∑ Proof by Contra positive technique, is used to prove implication statements when direct proof
is little bit difficult.
∑ Contra positive of the statement “if p then c” is “if not c then not p”.
∑ Proof by Contradiction is a technique which can be used for all the type of statements and in
this method it is assumed the statement to be proved is false and show that this assumption leads
to contradiction of some well-known fact or assumption made.
∑ In Proof by Construction, proof of the properties in the given statement is demonstrated with
an existing object (or) if such an object does not exist then a method (algorithm with steps) to
create the object is provided.
∑ Proof by Counter Example technique is used to prove a statement is false.
∑ “If and only if ” statements have two parts if-part and only-if part.
∑ To prove an “if and only if ” statement both the parts of the statements are proved.
∑ Proof by induction is a technique used to prove statements about objects that can be defined
using recursive functions.
∑ A function f is called recursive function when the same set of operations is to be performed on
the result of f(n) to obtain f(n + 1) irrespective of the value of n. Recursive functions have two
conditions.
∑ Tree is a non-linear data structure, which can be constructed using recursive definition.
Proof Techniques
45
x + y = 2*m + 2*n + 1
= 2* (m + n) + 1
From the sum of x and y it may be inferred that it is odd. Hence, the statement is proved using
contra positive technique.
2.5 Show that the following statement is true “If x2 – y2 = 1 then x and y are not positive integers”.
Solution: This statement is proved using Proof by Contradiction. Assume that x 2 – y2 = 1
where x and y are positive integers (i.e.) integers greater than 0.
x2 – y2 = (x + y) (x – y)
=1
2
(2.10)
2
Since x – y = 1, either (x + y) and (x – y) must be 1 or both (x – y) and (x + y) must be –1.
If (x + y) = 1 and (x – y) = 1 then the possible values that can be assigned are x = 1 and y = 0
which contradicts the assumption made earlier. When x – y = –1 and x + y = –1 the values of
x may be –1 and 0 respectively, which also contradicts the assumption made earlier. Therefore,
the given statement is true.
2.6 Prove that “If n is a positive integer such that n mod 4 is 2 or 3 then n is not a perfect square.”
Solution: Using the basic idea of modulus, (n mod m) ≥ 0 and < m it may be concluded that
value of n mod 4 should be 0, 1, 2 or 3.
The given statement is proved by Proof of Contra positive technique. Contra positive of the
given statement is “If n is a perfect square, then n mod 4 is either 0 or 1”. Since n is a perfect
square, there exists some integer such that k = n .
Considering all the four cases of remainders for k mod 4
1. When k mod (4) = 0, k can be written as 4q, where q is any integer, then
n = (4q)2
= 4(4q2)
Hence, n is divisible by 4, and n mod 4 = 0.
2. When k mod (4) = 1, k can be written as 4q + 1, then
n=
=
=
=
k2
(4q + 1)2
16q2 + 8q + 1
4(4q2+ 2q) +1.
Therefore, n mod 4 will be 1.
3. When k mod 4 = 2, k can be written as 4q + 2, and
n=
=
=
=
Therefore, n mod 4 will be 0.
k2
(4q + 2)2
16q2 + 16q + 4
4(4q2 + 4q + 1) + 0
46
Theory of Computation
4. When k mod (4) = 3, k can be written as 4q + 3 and
n=
=
=
=
k2
(4q + 3) 2
16q 2 + 24q + 9
4(4q 2 + 8q + 2 ) + 1.
Therefore n mod 4 will be 1. Hence, the statement is proved by Proof by Contra positive
technique.
2.7 Prove that “If a is an integer, then a is not exactly divisible by 3 if and only if a2 – 1 is exactly
divisible by 3.”
Solution: Whenever a if and only if statement is proved, the statement is divided into two parts
known as if-part and only-if part. The given statement is divided as follows:
If-part: If (a2 – 1) is exactly divisible by 3, then ‘a’ is not exactly divisible by 3.
Only-if part: If a is not exactly divisible by 3, then a2 – 1 is exactly divisible by 3.
PROOF : The if-part is proved first.
It is assumed that (a2 – 1) is exactly divisible by 3 and proved that a is not exactly divisible
by 3.
(a2 – 1) can be written as (a + 1) ¥ (a – 1) if (a2 – 1) is exactly divisible, then its factors either
(a + 1) or (a – 1) is divisible by 3.
In both the cases the number a is not divisible by 3. Hence, the if-part of the statement is
proved.
Only-if part of the statement is proved as follows. It is assumed that number ‘a’ is not divisible
by 3 and prove that a2 – 1 is divisible by 3.
Since it is assumed that, ‘a’ is not divisible by 3, ‘a’ can be written as 3q + r, where r is the
remainder and r can take the values 0, 1, 2. Value of r cannot be zero since it is assumed that
a is not exactly divisible by 3.
When
When
r
(a – 1)
(a2 – 1)
r
=
=
=
=
1, a = 3q + 1
3q, which means (a – 1) is exactly divisible by 3
(a + 1) (a – 1), hence, (a2 – 1) is divisible by 3.
2, a = 3q + 2,
(a – 2) = 3q which means (a – 2) is divisible by 3. The next number that is exactly divisible
by 3 can be got by adding 3 to this number.
Hence, (a – 2) + 3 = (a + 1), it is inferred that (a + 1) is divisible by 3. Hence, (a2 – 1) which
is (a + 1) (a – 1) is also exactly divisible by 3. Thus, the statement is proved.
2.8 Prove that for every n > 0, (5 n – 2 n) is divisible by 3.
Solution:
The statement is proved using Proof by Induction method.
i. Basic step: When n = 0, (5 n – 2 n) is 0, which is divisible by 3.
ii. Induction Hypothesis: It is assumed that the given statement is true for some k i.e
5 k – 2 k = 3P, where P is some integer.
48
Theory of Computation
Recursive Definition has two conditions known as base and recursive conditions and they are
i. Base Condition: In the base condition it is defined that S 0, S0 = {L}
ii. Recursive Condition: S k+1 can be formed by concatenating S to S k.
S k +1 = S ◊ Sk, then union of S 0, S 1,… S n where n is infinity is found to determine S*.
2.12 Given S = {a, b} find S 3 using the recursive definition given above.
Solution: To determine S 3, the recursive condition in definition is used, S3 is written as S ◊ S2.
Recursive condition is repeatedly used till base condition is reached.
Therefore,
S3 =
=
=
=
=
=
=
=
=
S3 =
S ◊ S2
S ◊ S ◊ S1
S ◊ S ◊ S ◊ S0
S ◊ S ◊ S ◊ {L}
S ◊ S ◊ {a, b} ◊ {L}
S ◊ S ◊ {a, b}
S ◊ {a, b} ◊ {a, b}
S ◊ {a2, ba, ab, b2}
{a, b} ◊ {a2, ba, ab, b2}
{a3, aba, aab, ab2, ba2, b2a, bab, b3}
1. In direct proof technique the proof is started by assuming _________.
(a) If part is true
(c) Both are true
(b) Then part is true
(d) Both are false.
2. Contra positive of the statement – If not P then not C is _________.
(a) If C then P
(c) If C then not P
(b) If not C then P
(d) If not P then C
3. When the proof of a statement is started by assuming the statement is false the type of proof
is called _________.
(a) proof by contra positive
(c) proof by contradiction
(b) proof by induction
(d) proof by deduction
4. In the proof of counter example technique, a statement is proved to be _________.
(a) false
(c) neither true nor false
(b) true
(d) none of the above.
5. In proof by construction method a statement S is proved true by _________.
(a) assuming S as false
(b) assuming S as true
50
Theory of Computation
13. Give a recursive definition to find the length of a string.
14. Give a recursive definition for multiplying two numbers by repeated addition.
1. a
6. d
2. a
7. d
3. c
8. a
4. a
9. b
5. c
10. c