MATH 310 Spring 2007 Exam Solutions: Proofs and Properties of Real Numbers - Prof. Andrew , Exams of Mathematics

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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Pre 2010

Uploaded on 03/10/2009

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MATH 310 SPRING 2007
EXAM I SOLUTIONS
1. (9 points) Negate each of the following statements. (Feel free to use the symbols ,, “s.t.”,
etc.)
(a) For all real x,x6= 0 x2>0.
Negation: There exists an xRsuch that x6= 0 but x20.
(b) For every real M, there exists a real number Nsuch that, for every n>N,|f(n)|> M.
Negation: There exists a real number Msuch that, for every real N, there exists an
n>N such that |f(x)| M.
(c) For all P, there is a δPsuch that, for all x, y R,
|xy|< δ |f(x)f(y)|< .
Negation: There exists an Psuch that, for every δP, there exist x, y Rsuch
that |xy|< δ but |f(x)f(y)| .
2. (10 points) We proved in class the following:
FACT:
n
X
i=1
i=n(n+ 1)
2for all nN.
Use this fact and induction to prove that
n
X
i=1
i3= n
X
i=1
i!2
for all nN.
Proof. We prove this statement by induction on n.
Basis Step: If n= 1,
n
X
i=1
i3= 1 and n
X
i=1
i!2
= 1. X
Induction Step: Suppose that
k
X
i=1
i3= k
X
i=1
i!2
for some kN. We want to show that
k+1
X
i=1
i3= k+1
X
i=1
i!2
.
1
pf3

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MATH 310 — SPRING 2007

EXAM I SOLUTIONS

  1. (9 points) Negate each of the following statements. (Feel free to use the symbols ∀, ∃, “s.t.”, etc.)

(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)| ≥ .

  1. (10 points) We proved in class the following:

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

= 1. X

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 + 1)^3

( (^) k ∑

i− 1

i

  • (k + 1)^3 (by the induction hypothesis)

[

k(k + 1) 2

] 2

  • (k + 1)^3 (by the FACT)

= (k + 1)^2

[

k^2 4

  • (k + 1)

]

= (k + 1)^2

[

k^2 + 4k + 4 4

]

[

(k + 1)(k + 2) 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.

  1. (6 points) Let A, B, and C be sets. Prove that

(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).

  1. (5 points) Let X = { 1 , 2 , 3 } and Y = { 1 , 2 }. Give an example of a bijective function f from X to Y or explain, in two or three sentences, why no such function exists. ANSWER: There is no such function f : X → Y. A function must match each element of X with an element of Y. There are three elements of X and only two elements of Y. So any function from X into Y must take at least two elements of X to the same element of Y and, thus, f cannot be one-to-one and therefore cannot be bijective.