Discrete Mathematics
Summer 03
Assignment # 4 : Solutions
Section 3.8
13. Use Euclidean algorithm to hand calculate gcd (544, 1001).
1. Divide 1001 by 544 to get 1001 = 544 * 1+ 457.
Hence gcd (1001, 544) = gcd (544, 457)
2. Divide 544 by 457 to get 544 = 457 *1 + 87.
Hence gcd (544, 457) = gcd (457, 87)
3. Divide 457 by 87 to get 457 = 87*5+22.
Hence gcd (457, 87) = gcd (87, 22)
4. Divide 87 by 22 to get 87 = 22*3+21.
Hence gcd(87, 22) = gcd(22, 21).
5. Divide 22 by 21 to get 22= 21*1 + 1.
Hence gcd(22, 21) = gcd(21, 1)
6. Divide 21 by 1 to get 21 = 21*0+1.
Hence gcd (21, 1) = gcd (1, 0) = 1.
Then we have: gcd (1001, 544) = gcd (544, 457) = gcd (457, 87) = gcd (87, 22)
= gcd(22, 21) = gcd(21, 1) = gcd (0, 1) = 1
26. Definition. We say that the positive integer c is the least common multiple of the
nonzero integers a,b if:
(1) c is a multiple of both a and b (i.e., a "common multiple")
(2) c divides _every_ common multiple of a and b
More formally, we say that lcm(a,b)=c if:
(1) a | c and b | c.
(2) m  , if a | m and b | m then c | m.
Claim: a,b +, a | b iff lcm (a, b) = b.
Since this is an “if and only if” statement there are two steps.
Part 1: lcm (a, b) = b  a | b.
Discrete Mathematics
Summer 03
Proof 1:
By the first part of the definition of lcm(a, b) = c, we see that lcm(a, b) = b
states that a | b and b | b. Therefore, the definition itself includes the statement that had to be
shown, namely that a | b.
Part 2: a | b  lcm (a, b) = b.
Proof 2:
Suppose a | b.
We need to show that
(1) a | b and b | b.
(2) a | m and b | m  b | m.
(These two statements are obtained by instatiating the definition of lcm(a, b) =
c to the special case where c=b.)
a|b, by our assumption, and b|b for all b. Hence (1) is proven.
(2) is a logical tautology, and so no further proof is required.
Section 4.1
16. Explicit formula for the sequence with initial terms 2, 6, 12, 20, 30…
ak = k(k+1) OR ak = ak-1 +2k
21.

4
m 0
1
1
1
1
1
1 16  8  4  2  1 31
 0  1 2  3 4 

m
16
16
2
2
2
2
2
2
We get the same result using the general formula for the sum of a geometric series:
1
31
5
1
31
4

m0 2 m  21  32
1 16
1
2
2
1
Section 4.2.
7.Prove by induction:
1+5+9+…+(4n-3) = n(2n-1) , n1, n
Proof:
Discrete Mathematics
Summer 03
Let P(k) be the statement 1+5+9+…+(4k-3) = k(2k-1)
Basis Step: Show true for n=1.
LHS : 1
RHS : n(2n-1) = 1*(2*1-1) = 1*1 = 1
Since RHS = LHS, we’ve proved the basis step.
Inductive Step: Show that P(k)  P(k+1)
Suppose P(k): 1+5+9+…+(4k-3) = k(2k-1)
Show: 1+5+9+…+(4k-3)+(4(k+1)-3) = (k+1)*(2(k+1)-1).
LHS :
[1+5+9+…+4(k-3)]+4(k+1)-3
= k(2k-1) + 4(k+1)-3 (by induction hypothesis P(k))
= 2k2-k+4k+4-3
= 2k2+3k+1
RHS:
(k+1)*(2(k+1)-1)
= (k+1) * (2k+1)
= 2k2+3k+1
Since RHS = LSH, we’ve proved the inductive step.
By induction, it follows that P(k) is true for all k>=1. That concludes the proof.
17. Given: Distributive law:
c, a1, a2 R , c(a1+a2) = ca1+ca2.
Prove: Generalized distributive law:
nZ, n2, if c, a1, a2,…,an R , c(a1+a2+…+an) = ca1+ca2+…+can.
Proof:
Basis Step: Show true for n=2.
c(a1+a2) = ca1+ca2. (from distributive law).
Inductive Step: Suppose true for n=k, then show true for n=k+1.
Suppose: c(a1+a2+…+ak) = ca1+ca2+…+cak
Show:
c(a1+a2+…+ak+ak+1) = ca1+ca2+…+cak+ cak+1
LHS:
c(a1+a2+…+ak+ak+1)
= c((a1+a2+…+ak)+(ak+1))
= c(a1+a2+…+ak) + cak+1 (by distributive law for binary expressions)
= ca1+ca2+…+cak+ cak+1 (by induction hypothesis)
Since LHS= RHS, we’ve proved the induction step.
Discrete Mathematics
20.
5+10+15+20+…+300
= 5(1+2+3+4+…+60)
Summer 03
by generalized distributive law
60
= 5(  k ) = 5*((60*61)/2) = 9150. by “sum of first n integers”
k 1
Section 4.3
10. Prove by induction: n3-7n+3 is divisible by 3, nZ, n0
Proof:
Basis Step: When n=0, n3-7n+3 = 0 + 0 + 3 = 3, which is clearly divisible by 3.
Inductive Step:
Suppose: (k3-7k+3) is divisible by 3.
Show: ((k+1)3-7(k+1)+3) is divisible by 3.
(k+1)3-7(k+1)+3
= (k2+2k+1)(k+1) – 7k-7+3
= (k3+k2+2k2+2k+k+1) – 7k-4
= k3+3k2-4k-3
rewrite as:
=(k3-7k+3) + (3k2+3k-6)
=(k3-7k+3) + 3*(k2+k-2)
By the induction hypothesis, the first term is divisible by 3. The second term is
evidently divisible by 3. Hence the sum is divisible by 3. QED.
17. Prove by induction: 2n < (n+2)! , nZ, n0
Proof:
Basis Step: When n=0, what is to be shown is that 20<(0+2)!, i.e. that 1<2. Which is true.
Inductive Step:
Suppose that 2k < (k+2)!
Then 2k+1 = 2*2 k < 2*(k+2)! < (k+3)*(k+2)! = (k+3)!