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Math 240
Name:
Worksheet Week 4
Feb 2017
TA: Amy Huang
Section 1.6 Problem 19 : Determine whether each of these arguments is valid. If an
argument is correct, what rule of inference is being used? If it is not, what logical error
occurs?
1. If n is a real number such that n > 1, then n2 > 1.Suppose that n2 > 1. Then n > 1.
2. If n is a real number with n > 3, then n2 > 9. Suppose that n2 ≤ 9. Then n ≤ 3.
3. If n is a real number with n > 2, then n2 > 4. Suppose that n ≤ 2. Then n2 ≤ 4.
Solution:
1. Not Valid. Fallancy of affirming the conclusion.
2. Valid. Modus tollens.
3. Not valid. Fallancy of denying the hypothesis.
Section 1.6 Problem 29: Use rules of inference to show that if ∀x(P (x)∨Q(x)), ∀x(¬Q(x)∨
S(x)), ∀x(R(x) → ¬S(x)), and ∃x¬P (x) are true, then ∃x¬R(x) is true.
Solution:
1. ∃x¬P (x)
2. ¬P (c)
3. ∀x(P (x) ∨ Q(x))
4. P (c) ∨ Q(c)
5. Q(c)
6. ∀x(¬Q(x) ∨ S(x))
7. ¬Q(c) ∨ S(c)
8. S(c)
9. ∀x(R(x) → ¬S(x))
10. R(c) → ¬S(c)
11. ¬R(c)
12. ∃x¬R(x)
Premise
Existential instantiation
Premise
Universal instantiation
From (2) and (4), disjunctive syllogism
Premise
Universal instantiation
From (5) and (7), disjunctive syllogism
Premise
Premise
From (8) and (10), modus tollens
Existential generalization
Section 1.7 Problem 17:
using
Show that if n is an integer and n3 + 5 is odd, then n is even
1. a. a proof by contraposition.
2. b. a proof by contradiction.
Solution:
1. a. We must prove the contrapositive: If n is odd, then n3 + 5 is even.
Assume that n is odd. By definition, n can be written as n = 2k + 1, where k is an
integer.
We will substitute n = 2k + 1 in n3 + 5 to get (2k + 1)3 + 5, which simplifies to 8k 3 +
12k 2 + 6k + 1 + 5.
The above expression can be written as 2(4k 3 + 6k 2 + 3k + 3), since 4k 3 + 6k 2 + 3k + 3
is an integer, 2(4k 3 + 6k 2 + 3k + 3) is even by definition.
Hence n3 + 5 is even.
2. b.
Method 1 :
Suppose that n3 + 5 is odd and n is odd.
We will use two facts without proof: the product of two odd numbers is odd and the
sum of two odd numbers is even.
Since n is odd, so is n2 and n3 , as the product of two odd numbers is odd.
Hence, since 5 is odd, this implies that n3 + 5 is even.
This is a contradiction, hence the supposition is false.
Method 2
Suppose that n3 + 5 is odd and n is odd.
We will use two facts without proof: the product of two odd numbers is odd and the
difference of two odd numbers is even.
Since n is odd, so is n2 and n3 , as the product of two odd numbers is odd.
Hence, since the difference of two odd numbers is even, n3 − (n3 − 5) = 5 is even.
Since 5 is odd, this is a contradiction, hence the supposition is false.
Section 1.7 Problem 24: Show that at least three of any 25 days chosen must fall in the
same month of the day.
Solution: Let p be the proposition ”At least three of any 25 days chosen must fall in the same
month of the day.” Suppose that ¬p is true. This means that at most two of the 25 days fall
in the same month of the day. Because there are 12 months of the year, this implies that at
most 24 days could have been chosen, as for each of the month of the year, at most two of
the chosen days could fall in that month. This contradicts the premise that we have 25 days
under consideration.
Section 1.8 Problem 25: Write the numbers 1, 2, . . . , 2n on a blackboard, where n is an
odd integer. Pick any two of the numbers, j and k, write |j − k| on the board and erase j and
k. Continue this process until only one integer is written on the board. Prove that this integer
must be odd.
Solution: We show by showing that the number of odd integers at any point must be odd.
Initially, we have n odd numbers and n even numbers. Since we know n is odd, the number
of odd integers is odd. We then observe that at each step, we have one of the following three
scenario:
1. Both j and k are even, so |j − k| is even. In this case the number of odd integer is
unchanged, and the number of even integer is reduced by 1.
2. Both of j and k are odd, so |j − k| is even. In this case the number of odd integer is
reduced by 2, and the number of even integer is increased by 1.
3. One of j and k is odd and the other is even, so |j − k| is odd. In this case the number
of even integer is reduced by 1 and the number of odd integer is unchanged.
So we notice that no matter how we select j and k, the parity of the number of odd integers
is unchanged. Therefore at the end, there must be an odd number of odd integers remaining.
So the last integer must be odd.
Section 1.8 Problem 35: Prove that between every two rational numbers there is an
irrational number.
Solution: We use a constructional proof. Assume x and y are two different rational numbers,
√
2 is an irrational
without loss of generality, assume x < y, since we √have already seen that
√
2
2
− y|. Now n is still an irrational
number, we can choose n large enough such that n < |x
√
√
2
number because√ 2 is. In addition, the number x + n is in between x and y and also
irrational since n2 is irrational and x is rational. Hence we have successfully construct an
irrational number in between x and y.