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Solutions to exam i for math 310 - spring 2007. It includes proofs and properties of real numbers such as negating statements, using facts and induction, set operations, and bijective functions.
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(a) For all real x, x 6 = 0 ⇒ x^2 > 0. Negation: There exists an x ∈ R such that x 6 = 0 but x^2 ≤ 0. (b) For every real M , there exists a real number N such that, for every n > N , |f (n)| > M. Negation: There exists a real number M such that, for every real N , there exists an n > N such that |f (x)| ≤ M. (c) For all ∈ P , there is a δ ∈ P such that, for all x, y ∈ R,
|x − y| < δ ⇒ |f (x) − f (y)| < .
Negation: There exists an ∈ P such that, for every δ ∈ P , there exist x, y ∈ R such that |x − y| < δ but |f (x) − f (y)| ≥ .
FACT:
∑^ n
i=
i =
n(n + 1) 2
for all n ∈ N.
Use this fact and induction to prove that
∑^ n
i=
i^3 =
( (^) n ∑
i=
i
for all n ∈ N.
Proof. We prove this statement by induction on n.
Basis Step: If n = 1,
∑^ n
i=
i^3 = 1 and
( (^) n ∑
i=
i
Induction Step: Suppose that
∑^ k
i=
i^3 =
( (^) k ∑
i=
i
for some k ∈ N. We want to show that
k∑+
i=
i^3 =
(k+ ∑
i=
i
k∑+
i=
i^3 =
( (^) k ∑
i=
i^3
( (^) k ∑
i− 1
i
k(k + 1) 2
= (k + 1)^2
k^2 4
= (k + 1)^2
k^2 + 4k + 4 4
(k + 1)(k + 2) 2
(k+ ∑
i=
i
(again by the FACT)
Thus, by induction on n,
∑^ n
i=
i^3 =
( (^) n ∑
i=
i
for all n ∈ N.
(A ∪ B) − (B ∩ C) ⊆ (A ∩ (B ∩ C)c) ∪ (B − C).
Proof. Choose an arbitrary x ∈ (A ∪ B) − (B ∩ C). Then x ∈ A ∪ B but x /∈ B ∩ C. So, x ∈ A or x ∈ B but x /∈ B ∩ C. If x ∈ A, then x ∈ A ∩ (B ∩ C)c, which is contained in (A ∩ (B ∩ C)c) ∪ (B − C). If x /∈ A, then x ∈ B. If x ∈ B but x /∈ B ∩ C, then x must be in C. So, x ∈ B − C, which again is contained in (A ∩ (B ∩ C)c) ∪ (B − C). Thus, x ∈ (A ∩ (B ∩ C)c) ∪ (B − C). Since this is true for any arbitrary element of (A ∪ B) − (B ∩ C), we have shown that
(A ∪ B) − (B ∩ C) ⊆ (A ∩ (B ∩ C)c) ∪ (B − C).