Download Math 300, Section C, Summer 2011, Solutions to Midterm Exam

Math 300, Section C, Summer 2011, Solutions to Midterm Exam
1. For the following statements decide if the statement is True or False. If True, write a proof. If False,
give a counter example.
(a) ∀x ∈ R ∀y ∈ R ∃z ∈ R xz = y
False. The true statement is its negation: ∃x ∈ R ∃y ∈ R ∀z ∈ R xz 6= y.
Let x = 0 and y = 1. Then for any z, xz = 0 6= 1.
(b) y 2 − 4y + 3 < 0 ∧ (y ∈ Z) =⇒ y = 2
True. Assume y 2 −4y+3 < 0 and y ∈ Z. Then, from the first equation we get that (y−1)(y−3) < 0.
Since the product is negative, one of the factors must be negative and the other must be positive.
So, 1 < y < 3. But, y is an integer so y must be 2.
(c) Let x and y be integers. Then, xy is odd if and only if x and y are odd.
True. This is an iff statement, so we should prove both way of the implication.
Assume x and y are odd. Then, there are integers k and m so that x = 2k + 1 and y = 2m + 1.
Then
xy = (2k + 1)(2m + 1) = 2(2km + k + m) + 1
is an odd number. So, if x and y are odd, so is xy.
We will prove the converse by counterpositive. Assume x or y is even.
If x is even, then x = 2m for some m. Then, xy = 2my is even. If y is even, then y = 2k for some
k. Then, xy = 2kx is even. In any case, the product is not odd.
2. Answer the following.
(a) Translate the following stament into logic and math.
Every rational number is the product of two irrational numbers.
Define and use sets, equations, quantifiers and logic symbols. Then, negate your logic statement.
Which one is true? The statement or its negation? Prove the true statement.
Let Q be the set of rational numbers.
∀x ∈ Q ∃y ∈
/ Q ∃z ∈
/ Q x = yz.
The stament is false. Its negation
∃x ∈ Q ∀y ∈
/ Q ∀z ∈
/ Q x 6= yz.
is true.
Let x = 0. Let y, z 6= Q be arbitrary.
If yz = x = 0, then y = 0 or z = 0 but 0 ∈ Q contradicting y, z 6= Q. So, yz 6= x.
(b) Translate the following stament into English. Do not use variable names, logic symbols and
equations. (You should be able to text your statement on a mobile phone.)
∀x ∈ R [(x ∈ Z ∧ ¬ (y ∈ Z ∧ x = 7y)) =⇒ (∃z ∈ R (z ∈ Z ∧ x = 2z))]
Then, negate the statement. Which one is true? The original or its negation? Prove the true
statement.
If an integer is not a multiple of 7, then it is even. This is false as 9 is not a multiple of 7 and is not
even. The negation of the statement is
∃x ∈ R [(x ∈ Z ∧ ¬ (y ∈ Z ∧ x = 7y)) ∧ (∀z ∈ R (z ∈
/ Z ∨ x 6= 2z))]
Let x = 9. Then if y ∈ Z then x 6= 7y. So y ∈
/ Z or x 6= 7y. And, for any z ∈ R − Z, 9 6= 2z.
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3. The Fibonacci sequence defined by the recursive formula
F1 = 1, F2 = 1
and when n ≥ 3
Fn = Fn−1 + Fn−2 .
Use strong induction to prove the formula
Fn =
When n = 1,
(1 +
√
(1 +
√
5)n − (1 −
√
2n 5
5) − (1 −
√
2 5
√
5)
√
5)n
.
√
2 5
= √ = 1,
2 5
when n = 2
(1 +
√
5)2 − (1 −
√
22 5
√
5)2
√
√
√
(1 + 2 5 + 5) − (1 − 2 5 + 5)
4 5
√
=
= √ = 1.
4 5
4 5
Now, assume the formula is true for all k < n. Then,
Fn = Fn−1 + Fn−2
√
√
(1 + 5)n−2 − (1 − 5)n−2
(1 + 5)n−1 − (1 − 5)n−1
√
√
+
by induction hypothesis
=
2n−1 5
2n−2 5
h i
√
√
√
√
1
√
=
2 (1 + 5)n−1 − (1 − 5)n−1 + 4 (1 + 5)n−2 − (1 − 5)n−2
2n 5
h
i
√
√
√
√
1
√
=
2(1 + 5)n−1 + 4(1 + 5)n−2 − 2(1 − 5)n−1 + 4(1 − 5)n−2
2n 5
h
i
√
√
√
√
1
√
=
(1 + 5)n−2 2(1 + 5) + 4 − (1 + 5)n−2 2(1 − 5) + 4 .
2n 5
√
√
Now since
(1 +
and
(1 −
we have
Fn =
1
√
2n 5
√
√
√
5)2 = 1 + 2 5 + 5 = 2(1 + 5) + 4
√
√
√
5)2 = 1 − 2 5 + 5 = 2(1 − 5) + 4
h
√
√
√ i
√
(1 + 5)n−2 (1 + 5)2 − (1 + 5)n−2 (1 − 5)2
=
(1 +
√
5)n − (1 −
√
2n 5
2
√
5)n
.
4. Let
f:
(a) Prove that f is injective.
Assume f (x) = f (y). Then
R − {−1}
x
→
→
R
x
1+x .
y
x
=
1+x
1+y
so
x(1 + y) = y(1 + x)
x + xy = y + yx
x = y.
Therefore, f is injective.
(b) Prove that f is not surjective. What is the range of f ?
x
The range of f is the set of all y’s such that 1+x
= y. Then solving for x we get
x = y(1 + x) = y + yx
x(1 − y) = y.
If y 6= 1 we can solve for x to get
x=
y
.
1−y
So the range of f is all numbers except for y = 1. Since the range is not all numbers, the function
is not surjective. Nothing gets mapped to 1.
(c) Prove that f is increasing. Do not use calculus.
y
x
To prove this stament, you assume x < y and then you have to compare 1+x
and 1+y
. But when
you try to cross multiply to get rid of the denominators, you have to be careful. a < b =⇒ ac < bc
when c > 0. So, you have to check whether 1 + x and 1 + y are positive or not. The proof will
fail.
This function is not increasing. We have −2 < 2 but f (−2) = 2 ≥ f (2) = 32 .
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