Homework Problems for Intermediate/Advanced Calculus - Week 1: Real Number Axioms, Assignments of Calculus

A set of homework problems for the first week of intermediate/advanced calculus course, focusing on the field and order axioms of real numbers. It includes proof-based exercises related to the distributivity, additive inverse, multiplicative inverse, and order axioms.

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

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MAT 370/371 Intermediate/Advanced Calculus Aug 25, 2004
Homework for week 1
Problems for section 1.2: Real number axioms, part I - The field axioms
Remember to assume nothing beyond the field axioms and the results that have been proved using
them. Refer to the hand-outs sections on the WWW for the real number axioms.
1. Let x,y, and zbe real numbers. Prove that x(y+z) = xy +xz.
(Note: The textbook has the order reversed due to a different order in the distributivity axiom.)
3. Prove that x= (1) ·xfor every real number x.
5. d. Suppose x, y Rare both nonzero. Prove that
1
x
y
=y
x.
7. If x6= 0 and y6= 0 show that (xy)1exists and that (xy)1=x1y1.
8. Let A={E, O }be the set containing two elements: The sets of the even and odd integers,
respectively. Define two binary operations on Aas follows:
E O
E E O
O O E
E O
E E E
O E O
Prove that with these operations Asatisfies the field axioms.
Problems for section 1.3: Real number axioms, part II - The order axioms
Use the following Order axiom (i.e. do not use the axioms as stated on the handout on the WWW.)
There exists a subset R+of Rsuch that
R+is closed under addition and multiplication, i.e. if x, y R+then also x+yR+and
xy R+.
For every aRexactly one of the following holds: aR+,a= 0, or aR+.
1. Prove that if aRand bR+then a < a +b.
3. Let a, b R. Prove that if a < 0 and b < 0 then a+b < 0 and a·b > 0.
5. Prove that 1
2<1.
7. Prove the law of trichotomy: For every pair of real numbers aand bexactly one of the following is
true: a < b or a=bor a > b.
9. b. Suppose aRand 0 < a < 1. Prove a2< a.
9. d. Suppose a, b Rhave the same sign. Prove that if abthen 1
a1
b.
11. Prove that there is no rational number whose square is 2.
Hint: Proceed by contradiction. Assume that there is some rational number m
nwhose square is 2
and derive a contradiction.

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MAT 370/371 Intermediate/Advanced Calculus Aug 25, 2004

Homework for week 1

Problems for section 1.2: Real number axioms, part I - The field axioms

Remember to assume nothing beyond the field axioms and the results that have been proved using them. Refer to the hand-outs sections on the WWW for the real number axioms.

  1. Let x, y, and z be real numbers. Prove that x(y + z) = xy + xz. (Note: The textbook has the order reversed – due to a different order in the distributivity axiom.)
  2. Prove that −x = (−1) · x for every real number x.
  3. d. Suppose x, y ∈ R are both nonzero. Prove that

1 x y

y x

  1. If x 6 = 0 and y 6 = 0 show that (xy)−^1 exists and that (xy)−^1 = x−^1 y−^1.
  2. Let A = {E, O} be the set containing two elements: The sets of the even and odd integers, respectively. Define two binary operations on A as follows: ⊕ E O E E O O O E

⊗ E O

E E E

O E O

Prove that with these operations A satisfies the field axioms.

Problems for section 1.3: Real number axioms, part II - The order axioms

Use the following Order axiom (i.e. do not use the axioms as stated on the handout on the WWW.)

There exists a subset R+^ of R such that

  • R+^ is closed under addition and multiplication, i.e. if x, y ∈ R+^ then also x + y ∈ R+^ and xy ∈ R+.
  • For every a ∈ R exactly one of the following holds: a ∈ R+, a = 0, or a ∈ R+.
  1. Prove that if a ∈ R and b ∈ R+^ then a < a + b.
  2. Let a, b ∈ R. Prove that if a < 0 and b < 0 then a + b < 0 and a · b > 0.
  3. Prove that 12 < 1.
  4. Prove the law of trichotomy: For every pair of real numbers a and b exactly one of the following is true: a < b or a = b or a > b.
  5. b. Suppose a ∈ R and 0 < a < 1. Prove a^2 < a.
  6. d. Suppose a, b ∈ R have the same sign. Prove that if a ≤ b then (^1) a ≥ (^1) b.
  7. Prove that there is no rational number whose square is 2. Hint: Proceed by contradiction. Assume that there is some rational number mn whose square is 2 and derive a contradiction.