sol4

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.


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.

We get the same result using the general formula for the sum of a geometric series:

Section 4.2.

7.Prove by induction: 1+5+9+…+(4n-3) = n(2n-1) , n1, n

Proof:


Let P(k) be the statement 1+5+9+…+(4k-3) = k(2k-1)

Basis Step: Show true for n=1.LHS : 1RHS : n(2n-1) = 1*(2*1-1) = 1*1 = 1Since 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: nZ, n2, 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.


20. 5+10+15+20+…+300= 5(1+2+3+4+…+60) by generalized distributive law

= 5( ) = 5*((60*61)/2) = 9150. by “sum of first n integers”

Section 4.3

10. Prove by induction: n3-7n+3 is divisible by 3, nZ, n0Proof: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)! , nZ, n0

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)!

sol4

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