Homework 1

Throughout the course, we take \mathbb{N}=\{1, 2, 3, \ldots\} as the set of natural numbers.

Exercise. For each subset of \mathbb{R}, give its maximum, minimum, supremum and infimum, if they exist:\{0, 8\}, (3, \infty), \left\{(-1)^n\cdot \left(1+\frac{1}{n}\right): n\in\mathbb{N}\right\}, \left\{ \frac{n}{n+1} : n\in\mathbb{N}\right\}

Proof.

Set	Maximum	Minimum	Supremum	Infimum
\{0, 8\}	8	0	8	0
(3, \infty)	–	–	–	3
\left\{(-1)^n\cdot \left(1+\frac{1}{n}\right): n\in\mathbb{N}\right\}	3/2	-2	3/2	-2
\left\{ \frac{n}{n+1} : n\in\mathbb{N}\right\}	–	1/2	1	1/2

Exercise: Prove that for any real number x, there exists an integer m such that m-1\leq x<m.

Proof (sketch). Let A=\left\{ m\in\mathbb{Z} : x<m\right\}. Use Archimedean property to prove that A is not empty and bounded below. Take m, the infimum of A and prove that m satisfies the property.

Exercise: Chapter 1, Ex 1

Proof (sketch). Prove by contradiction.

Exercise: Chapter 1, Ex 2

Proof. Suppose there exist two non-zero rational numbers p, q such that12=\left(\frac{p}{q}\right)^2.We have12\cdot q^2=p^2.By the fundamental theorem of arithmetic, the left-hand side has odd number of factors of 3, while the right-hand side even. Contradiction.

Exercise: Chapter 1, Ex 3

Proof. First, notice that (1/x)x=x(1/x)=1. For (a), the axioms (B) give y=1y=((1/x)x)y=(1/x)(xy)=(1/x)(xz)=((1/x)x)z=1z=z.
For (b), as xy=x=1x=x1. By (a), we have y=1.
For (c), as xy=1=x(1/x). By (a), we have y=1/x.
Subquestion (d) is a little bit tricky. One might start to argue as the following.
Since (1/x)x=x(1/x)=1, by (c) we have x=1/(1/x).
But to use (c), one need to verify that 1/x\neq 0. Also 1/(1/x) only makes sense when 1/x\neq 0.
We prove by contradiction. Suppose 1/x = 0. Then 1=x(1/x)=x0=0x=0. Contradiction. This finishes the proof for (d).

Exercise: Chapter 1, Ex 4

Proof (sketch). For E is nonempty, pick any element in it and compare it with \alpha, \beta.

Exercise: Chapter 1, Ex 5

Proof (sketch). Denote \alpha=\inf A. Prove that -\alpha is an upper bound of -A and it is the least one.

Exercise: Chapter 1, Ex 8

Proof. Suppose an order can be defined in the complex field to turn it into an ordered field. Then 0<i^2=-1<0. Contradiction.

Exercise: Chapter 1, Ex 9

Proof (sketch). Verify the lexicographic relation is really an order first. Now consider the y-axis and prove that it is non-empty and bounded above but it does not have the least upper bound. Thus under the dictionary order, the set of complex numbers does not have the least-upper-bound property. (Update: By y-axis, I mean all of the complex numbers whose real part is zero. And under lexicographic order, they are bounded by 1 from the top.)

Exercise: Chapter 1, Ex 13

Proof (sketch). View the complex numbers as the two dimensional Euclidean space. Then let z=0 in theorem 1.37(f) to get \lvert x\rvert\leq\lvert x-y\rvert+\lvert y\rvert. Switch x and y, we have \lvert y\rvert\leq\lvert y-x\rvert+\lvert x\rvert. Finally, combine the two inequalities together.