Download CS 173: Discrete Structures, Fall 2011 Homework 3

CS 173: Discrete Structures, Fall 2011
Homework 3 - Solution
This homework contains 4 problems worth a total of 40 points. It is due on Friday, September
16th at 4pm.
When writing your proofs, be sure to use the definitions of key concepts (e.g. divisible) as
presented in class. Also one goal of this problem set is to practice certain proof techniques.
So be sure to use the proof technique specified by the problem instructions, even if there
might be other ways to prove the claim.
1. Proof by cases [8 points]
Problem 5 from Homework 2 defined the “extended real numbers.” An extended real
number has the form a + bǫ, where ǫ is a special new positive number whose square is
zero. To compare the size of two extended real numbers, we use the definition:
a + bǫ < c + dǫ whenever either a < c, or a = c and b < d.
Using this definition and proof by cases, prove the following claim:
Claim: For any extended real numbers a+bǫ, c+dǫ, and p+qǫ, if a+bǫ < c+dǫ
and c + dǫ < p + qǫ, then a + bǫ < p + qǫ.
Solution: Suppose that a + bǫ, c + dǫ and p + qǫ are extended real numbers such that
a + bǫ < c + dǫ and c + dǫ < p + qǫ.
Since a + bǫ < c + dǫ, a < c or a = c and b < d. Since c + dǫ < p + qǫ, either c < p or
c = p and d < q. So we have four possible cases:
Case 1: a < c and c < p. Thus a < p which implies a + bǫ < p + qǫ.
Case 2: a = c and b < d and c < p. Thus a < p which implies a + bǫ < p + qǫ.
Case 3: a = c and b < d and c = p d < q. Thus a = p and b < q which implies
a + bǫ < p + qǫ.
Case 4: a < c and c = p d < q. Thus a < p which implies a + bǫ < p + qǫ.
In all four cases, a + bǫ < p + qǫ. Therefore our claim is proved.
2. Direct Proof [14 points] Use direct proof to prove the following claims.
(a) (7 points) For every odd integer n there is an integer k such that n2 = 8k + 1.
Solution:
Lemma: Notice that if m is an even integer, then m has the form 2k where k is
an integer. So, for any integer s, ms = 2ks must also be even.
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Now suppose that n is an odd integer. According to the definition of odd there
exists some integer p s.t. n = 2p + 1. n2 = (2p + 1)2 = 4p2 + 4p + 1 = 4p(p + 1) + 1.
We know that either p or p + 1 is even. So by the lemma above, p(p + 1) must be
even. Thus there exists some integer k s.t. p(p + 1) = 2k. Then n2 = 4(2k) + 1 =
8k + 1. This is what we needed to show.
(b) (7 points) If a > 0 and b ≤ −2 show that a2 + ab + b2 ≥ 0.
Solution: Suppose that a and b are integers such that a > 0 and b ≤ −2. Then
ab must be negative since it’s the product of a positive number and a negative
one. So ab < 0 and thus −ab > 0.
Now, (a + b)2 ≥ 0 since the square of a real number can’t be negative. Adding
together these two inequalities, we find that (a + b)2 − ab ≥ 0. But (a + b)2 − ab =
(a2 + 2ab + b2 ) − ab = a2 + ab + b2 . So a2 + ab + b2 ≥ 0.
3. Proof by Contrapositive [10 points]
Consider the following claim:
Claim: For all integers x and y, if 3x + 5y = 47 then at least one of x and y
is not divisible by 7.
(a) (2 points) State the converse of the claim.
Solution: For all integers x and y if at least one of x and y is not divisible by 7
then 3x + 5y = 47.
(b) (2 points) State the contrapositive of the claim.
Solution: For all integers x and y if both x and y are divisible by 7 then 3x+5y 6=
47.
(c) (6 points) Prove the (original) claim using proof by contrapositive.
Solution: Instead of proving original claim using direct proof we can take contrapositive of the claim and prove it. That is, let x and y be integers and suppose
that x and y are both divisible by 7. We need to prove that 3x + 5y 6= 47.
Since x and y are both divisible by 7 then there exists two integers m and n s.t.
x = 7m and y = 7n.
Now 3x + 5y = 3(7m) + 5(7n) = 7(3m + 5n). So 3x + 5y is an integer divisible by
7 but 47 is not divisible by 7. Therefore 3x + 5y 6= 47.
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4. Proof by Contradiction [8 points]
Prove the following claim using proof by contradiction. Be sure to start your proof by
writing out the negation of the claim explicitly.
For any real numbers a, b, and c, show that at least one of these three
numbers is greater than or equal to the average of the three numbers.
Hint 1: choose a variable name for the average and turn the key parts of the claim into
equations which you can manipulate using algebra.
Hint 2: the average of x, y, and z is
x+y+z
.
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Solution: Suppose not. That is, suppose that there are real numbers a, b, and c such
that all three numbers are less than the average of the three numbers.
= m. Then our assumption states that (a < m) and
Let m be the average a+b+c
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(b < m) and (c < m). By adding all the inequalities we get a + b + c < 3m. But m is
defined to be a+b+c
, so a + b + c = 3m.
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But now we have that 3m = a + b + c < 3m. So 3m < 3m which is an obvious
contradiction. Thus our claim is true.
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