Section 3

$\mathbb{Q}$

$\mathbb{Q}$ is a field

The set $\mathbb{Q}$ can be thought of as the set of fractions, and is not only a set of $\frac{m}{n}$ where $n\ne 0$ as two fractions can represent the same element in $\mathbb{Q}$. Think $\frac{1}{2}$ and $\frac{3}{6}$

Formally here are the conditions met in $Q$ (A and M)

1. Addition
   1. Commutativity
   2. Associativity
   3. Identity (with the $0$) element
   4. Additive inverses
2. Multiplicative
   1. Associativity
   2. commutativity
   3. identity
   4. inverses
3. Distributive property Notice that these fit the definition for $\mathbb{Q}$ to be a field according to Fields.

Order structure in $\mathbb{Q}$ (O)

Order structure laws in $\mathbb{Q}$. Kinda obvious but to state them:

1. O1: For any $a,b$ either $a\le b$ or $b\le a$
2. O2: If $a\le b$ and $b\le a$ then $b=a$
3. O3: If $a\le b$ and $b\le c$ then $a\le c$ (transitive law)
4. O4: If $a\le b$ then $a+c\le b+c$
5. O5: If $a\le b$ and $0\le c$ then $ac\le bc$

These properties make $\mathbb{Q}$ and ordered field.

$\mathbb{R}$

Should be mostly familiar with this so I will just note the important parts:

1. Includes all rational numbers, and algebraic elements ($\pi ,e,\dots$)
2. No gaps on the number line, unlike $\mathbb{Q}$ or $\mathbb{Z}$. Every point corresponds to a real number and vice versa
3. Unlike group theory, the main purpose of this course is not to prove the structure of $\mathbb{R}$ , but instead to accept it as a mathematical system and to study some important functions and properties on $\mathbb{R}$. From the axioms given above, we know $\mathbb{R}$ is an ordered field. The following proofs, although completed in $\mathbb{R}$ are valid for any ordered field (such as $\mathbb{Q}$)

Some proofs

The following proofs are very elementary proofs from group theory, so I am only going to reattempt some of them:

1. i.
   1. $a+c=b+c⟹a+c+(-c)=b+c+(-c)⟹a=b$
2. ii.
   1. $a\cdot 0=0⟹a0=a\cdot (0+0)=a0+a0$
   2. Thus we have $a0=a0+a0$
   3. We can represent this as $a0+0=a0+a0$
   4. By i. we can conclude $0=a0$
3. iii.
   1. $(-a)b=-ab$
   2. Notice $(a+(-a))b=ab+((-a)b)=0b=0=ab+(-ab)$
   3. Thus, look closer at the $ab+((-a)b)=ab+(-ab)$ and again by i. we know $(-a)b=-ab$
4. iV. and V. todo left as 3.3
5. Vi.
   1. Assume $ab=0$
   2. then $0=b^{-1}\cdot 0=0\cdot b^{-1}=(ab)b^{-1}=a(b{b}^{-1})=a(1)=a$

Here are some more proofs that show how u can use the above axioms on hw, I only did some the rest can be found in Abstract Algebra Home:

1. i.
   1. Notice axiom O4 stating $a\le b⟹a+c\le b+c$
   2. Lets use this by letting $c=(-a)+(-b)$
   3. Then $a\le b⟹a+[(-a)+(-b)]\le b+[(-a)+(-b)]$
   4. Using additive associativity and commutativity properties, we can show $-b\le -a$
2. ii.
   1. First take $a\le b$
   2. From O5 we know $az\le bz$ for some $z\ge 0$
   3. From i. we also know that $-bz\le -az$
   4. Utilizing associativity and iii. from the previous set, we can show $(-z)b\le (-z)a$ were $-z\le 0$ as we assumed $c$ to be non-negitive. Let $c=-z$ /

Abs value

Formal def: $|a|=a$ for $a\ge 0$, $|a|=-a$ for $a\le 0$.

The distance then is just the abs value of the difference, i.e, dist$(a,b)=|a-b|$

Properties

These have mostly been seen before

1. $|a|>0$ (duh)
2. $|ab|=|a|\cdot |b|$ (proof by cases/exhaustion)
3. $|a+b|\le |a|+|b|$ (again proof by cases/exhaustion; triangle inequality)

From above, we have the implication that

1. dist$(a,c)$ $\le$ dist$(a,b)$ + dist$(b,c)$
   1. What is being said here is $|a-c|\le |a-b|+|b-c|$
   2. From the triangle inequality $|[a-b]+[b-c]|\le |a-b|+|b-c|$
   3. This implies $|a[-b+b]-c|=|a-c|\le |a-b|+|b-c|$

This is used so commonly in this book I find it beneficial to state it: $|a|=|a-b+b|\le |a-b|+|b|$ Often, $|a-b|$ is bounded by something, i.e. $|a-b|<1$ This lets us say $|a|<1+|b|$

Triangle inequality is said as such because no side of a trangle is larger than the sum of the other two:

An informal way of thinking about it is going directly from point a to b is better than taking some shortcut.

Exercises

3.1

1. For $\mathbb{N}$ we have it so A4 fails as there are no additive inverses. Also M4 fails as there are no multiplicative inverses. Additive identity fails as there is no 0.
2. For $\mathbb{Z}$ we have it so M4 fails due to lack of multiplicative inverses.

3.2

1. For ref here is the proof The statement for i. is $a+c=b+c⟹a=b$. The statement was in the form $c+a = c+b Thus we used the commutative property of addition to get it into the right form.
2. Similar result as 1.