Download MATH 1700 Midterm 1 February 16, 2017

MATH 1700
Midterm 1
NAME_________________________
1) Construct a truth table for (𝑝 →∼ 𝑞) → (∼ 𝑝 → 𝑞). Is the statement a tautology, a contradiction
or neither? (4 pts)
𝑝 𝑞
∼𝑝 ∼𝑞
𝑝 →∼ 𝑞
∼𝑝→𝑞
(𝑝 →∼ 𝑞) → (∼ 𝑝 → 𝑞)
T
T
F
F
F
T
T
T
F
F
T
T
T
T
F
T
T
F
T
T
T
F
F
T
T
T
F
F
So, it is neither one.
2) Explain why the following argument is invalid: (3 pts)
𝑝 →∼ 𝑞
𝑞 →∼ 𝑝
∴𝑝∨𝑞
By the last row of the following table
𝑝
𝑇
𝑇
𝐹
𝐹
𝑞
𝑇
𝐹
𝑇
𝐹
∼𝑝 ∼𝑞
𝐹
𝐹
𝐹
𝑇
𝑇
𝐹
𝑇
𝑇
𝑝 →∼ 𝑞
𝐹
𝑇
𝑇
𝑇
𝑞 →∼ 𝑝 𝑝 ∨ 𝑞
𝐹
𝑇
𝑇
𝑇
𝑇
𝑇
𝑇
𝐹
the argument is invalid, because if 𝑝 and 𝑞 are both 𝐹, then the premises are 𝑇 while the
conclusion is 𝐹.
3) Write the negation of the following statement: (3 pts)
For all real numbers 𝑟, there exists a real number 𝑠 such that 𝑟𝑠 = 10.
There is a real number 𝑟 such that for all real numbers 𝑠, 𝑟𝑠 ≠ 10.
4) Prove the following statement using the contrapositive: (3 pts)
For all integers 𝑛, if 𝑛2 is not a multiple of 5, then 𝑛 is not a multiple of 5.
The contrapositive of the statement is the following:
For all integers 𝑛, if 𝑛 is a multiple of 5, then 𝑛2 is a multiple of 5.
Assume 𝑛 is a multiple of 5. So, for an integer 𝑘, we have 𝑛 = 5𝑘 and hence
𝑛2 = (5𝑘)2 = 25𝑘 2 = 5(5𝑘 2 )
2
which means 𝑛 is a multiple of 5. Therefore, the contrapositive and hence the statement itself
are correct.
5) Is the following statement true or false? If false find a counterexample and if true prove it. (2 pts)
For all real numbers 𝑥 and 𝑦, ⌈𝑥 ⋅ 𝑦⌉ = ⌈𝑥⌉ ⋅ ⌊𝑦⌋.
For 𝑥 = 𝑦 = 0.1, we have ⌈𝑥 ⋅ 𝑦⌉ = ⌈0.1 × 0.1⌉ = ⌈0.01⌉ = 1 while ⌈𝑥⌉ = ⌈0.1⌉ = 1 and ⌊𝑦⌋ =
⌊0.1⌋ = 0 and hence 1 = ⌈𝑥 ⋅ 𝑦⌉ ≠ ⌈𝑥⌉ ⋅ ⌊𝑦⌋ = 1 × 0. Therefore, the statement is false.
February 16, 2017
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MATH 1700
Midterm 1
6) Find each of the following. (3 pts)
a) 43 div 8 = 5
b) ⌈−2.185⌉ = −2
c) 1013 mod 5 = 3
7) Indicate whether the statements below are true or false. (3 pts)
a) ∀ real numbers 𝑥, ∃ real number 𝑦 such that 𝑥 < 𝑦. T
b) ∃ real number 𝑦 such that ∀ real numbers 𝑥, 𝑥 < 𝑦. F
c) ∀ real numbers 𝑥, and ∀ real numbers 𝑦, 𝑥 2 + 𝑦 2 ≥ 3𝑥𝑦. F
8) Prove the following statement by contradiction: (4 pts)
There is no largest rational number.
Suppose not. That is, assume there is the largest rational number and call it 𝑁. Since 1 is a rational
number and we know that the sum of two rational numbers is a rational number, 𝑁 + 1 is also a
rational number. But 𝑁 + 1 is strictly larger than 𝑁, which is in contradiction with 𝑁 being the
largest rational number. So, the original assumption is false and hence there is no largest rational
number.
9) Use mathematical induction to show that for all integers 𝑛 ≥ 0
8𝑛 − 1 is divisible by 7.
Let 𝑃(𝑛): 8𝑛 − 1 is divisible by 7.
a) State what you need to prove in the base step and carry out the proof. (1pt)
b) State what you need to prove in the inductive step and carry out the proof. (4 pts)
a) We need to prove 𝑃(0): 80 − 1 is divisible by 7. Since 80 − 1 = 1 − 1 = 0 it is divisible by 7.
b) We need to assume 𝑃(𝑘): 8𝑘 − 1 is divisible by 7 and prove 𝑃(𝑘 + 1): 8𝑘+1 − 1 is divisible
by 7.
By induction assumption we have 8𝑘 − 1 = 7𝑘, for some integer 𝑘. Therefore,
8𝑘+1 − 1 = 8 ⋅ 8𝑘 − 1 = (7 + 1) ⋅ 8𝑘 − 1 = 7 ⋅ 8𝑘 + 8⏟𝑘 − 1 = 7 ⋅ 8𝑘 + 7𝑘 = 7 ⋅ (8𝑘 + 𝑘)
=7𝑘
and hence 8𝑘+1 − 1 is divisible by 7.
February 16, 2017
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