Download Discrete Maths hw6 solutions

DT6248 Discrete Maths
Homework Solutions.
proofs and Mathematical Induction
1) Prove that for all integers m and n, if m and n are odd, then mn is odd.
Answers: Let m and n be odd integers. Then there exist k1 and k2 such that m =
2k1 + 1 and n = 2k2 +1. Now mn = (2k1 + 1) (2k2 +1) = 4k1k2 +2k1 +2k2 + 1 =
2(2k1k2 + k1 +k2) +1. Therefore, mn is odd.
2) Show, by giving a proof by contradiction, that if 40 coins are distributed among
nine bags so that each bag contains at least one coin, at least two bags contain the
same number of coins.
Answers: Suppose that the conclusion is false, that is, that no two bags contains
the same number of coins. Suppose that we rearrange the bags in increasing order
of the number of coins that they contain. Then the first bag contains at least one
coin; the second bag contains at least two coins, and so on. Thus, the total
number of coins is at least 1 + 2 +3 + …..+ 9 = 45. This contradicts the
hypothesis that there are 40 coins. The proof by contradiction is complete.
3) Use proof by contradiction to show that 1  2 is not rational.
Answer:
Suppose 1  2 is rational, i.e. 1  2  m / n where m and n are integers in
lowest terms (no common factors).
1 2  m / n


2  m / n 1
mn
2
note: if m / n is in lowest terms, then (m / n)  1
n
is also in lowest terms.
let t  m  n  2 
t
n
We've already seen that 2 cannot be expressed as a fraction in lowest terms.
therefore, we have a contradiction and 1+ 2 is not rational.
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4) Use proof by contradiction to show that  real numbers x  y  2 , then either
x  1 or y  1.
Answer:
Assume neither
x  1 or y  1.
 x  1 and x  1.
 x  y  1  1  2.
contradiction: x  y  2.
Therefore, if x  y  2 , then either x  1 or y  1.
5)
Using induction, verify that each equation is true for every positive integer n.
a) 1 + 3 + 5 + ….+ (2n – 1) = n2
Answers:
1  12
Assume true for n.
1
 (2n  1)  (2n  1)  n 2  2n  1  (n  1) 2
1
1
1
1
n
b)


 

13 35 5 7
 2n 1 2n  1 2n  1
Answers:
1
 1/ 3
(1 3)
Assume true for n.
1

13

1

1
 2n 1 2n  1  2n  1 2n  3

n
1
n 1


2n  1  2n  1 2n  3 2n  3
6) Use the geometric sum to prove that
1
r 0  r1   r n 
for all n  0 and 0  r  1.
1 r
Answers: r 0  r1 
rn 
1  r n1
1

1 r
1 r
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7) Use induction to prove that 7n – 1 is divisible by 6 , for all n  1.
Answers:
71  1  6 is divisible by 6.
Suppose that 6 divides 7 n  1.
Now, 7 n 1  1  7 7 n  1  7 n  1  6 7 n.
Since divides both 7 n  1 and 6 7 n , it divides their sum,
which is 7 n 1  1.
8) Prove that the sum of the squares of the first n positive integers is
n(n  1)(2n  1)
6
Answer:
For n  1: 1 
1(2)(3)
1
6
Assume true for n ie ( 1  4  9  ...  n 2 
n(n  1)(2n  1)
6
Prove true for n  1:
Show
1 4  9 
n
2
 n 2  (n  1) 2 
 3n  2   2n  3
 n  1 n  2  2(n  1)  1
6
2n3  12n 2  13n  6
6
6
n(n  1)(2n  1)
1  4  9   n 2  (n  1) 2 
 (n  1) 2 
6
2
3
n(n  1)(2n  1) 6(n  1)
2n  3n 2  n  6n 2  12n  6 2n3  12n 2  13n  6



6
6
6
6


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9) Prove that, for all positive integers n, 4
2 n 1
 3n2 is divisible by 13.
Answer:
n  1: 43  33  91  13  7.
Assume true for n.
Prove true for n  1.
42 n 1  3n  2  13m
and 42 n 1  13m  3n  2
42( n 1) 1  3n 1 2  42 n 3  3n 3  42 n 1  42  3n  2  3
 13m  3n  2   42  3n  2  3  13 16m   16  3n  2  3n  2  3
 13 16m   3n  2  3  16   13 16m   13(3n  2 )
Which is clearly divisible by 13.
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