## Real Analysis – Basics

— Introduction —

Physics is the name of the game we’ll be playing in this blog. And for one to know and understand Physics, one has to know and understand some Mathematics. We’ll start things off with Calculus (Real Analysis) because it is the cornerstone of Physics.

Topics that will be dealt with will be the Axioms of the Real numbers, the notion of limit and limits per se, Differential Calculus, Integral Calculus, and Series.

At this point I still don’t know the precise order this topics will be dealt with (the main problem is where should we study Series) but all of these things will appear.

Please see this link because mathematical symbols will be used a lot on this blog and I want you to be able to follow things through. At first a more verbal approach will be used but gradually mathematical stenography will appear more and more.

— 1. Field Axioms —

First of all let us assume the existence of a set of Real numbers, which we denote by the symbol ${ \mathbb{R}}$ in which two operations are defined. These operations are the addition operation, and the multiplication operation.

Let us also suppose that there exists a subset of ${ \mathbb{R}}$ which we’ll call the set of the positive numbers, and denote by ${ \mathbb{R}^+}$. These four terms will be taken as primitive concepts (also called undefined terms).

 Axiom 1 Addition and multiplication are commutative That is, for every ${ x}$, and ${ y}$ belonging to ${ \mathbb{R}}$ it is ${ x+y=y+x}$ and ${ x\cdot y=y\cdot x}$. Where ${ +}$ denotes the addition operation and ${\cdot}$ denotes the multiplication operation.
 Axiom 2 Addition and multiplication are associative That is, for every ${ x}$, ${ y}$, and ${ z}$ belonging to ${ \mathbb{R}}$ it is ${ \left( x+y\right)+z=x+\left(y+z\right)}$ and ${ \left( x\cdot y\right)\cdot z=x\cdot\left(y\cdot z\right)}$
 Axiom 3 Multiplication is distributive in relation to addition For every ${ x}$, ${ y}$, and ${ z}$ belonging to ${ \mathbb{R}}$ it is ${ x\cdot\left(y+z\right)=x\cdot y+x\cdot z}$
 Axiom 4 Both the addition operation and the multiplication operation have a neutral element, and the neutral elements regarding both operations are distinct from each other. That is, for the addition operation there exists an element, ${ a}$, that for every element ${ x\in\mathbb{R}}$ it is ${ x+a=a+x=x}$. And for the multiplication operation there exists an element ${ b}$, that for every element ${ x\in\mathbb{R}}$ it is ${ x\cdot b=b\cdot x=x}$.

A few remarks are now in order. First of all let us note that because of Axiom 1 we can prove that each one of these neutral elements is unique in ${ \mathbb{R}}$. Let us do it in the case of the multiplication (the addition case has a perfectly analogous proof) operation.

What we are trying to prove is: if we assume that we have two neutral elements, ${ b}$ and ${ c}$ for the multiplication operation they are one and the same. So here it goes:

${ x\cdot b=x}$ ${ \forall x\in\mathbb{R}}$ (so in particular the equation is valid for ${ x=c}$ and so ${ c\cdot b=c}$) and ${ x\cdot c=x}$ ${ \forall x \in \mathbb{R}}$ (so in particular this equation is valid for ${ x=b}$ and so ${ b\cdot c=b}$).

Thus ${ c\cdot b=c}$ and ${ b\cdot c=b}$. But considering Axiom 1 multiplication is commutative and so ${ c\cdot b=b\cdot c}$. Hence ${ c=b}$.

Please note the crucial point that Axiom 1 had in this proof. Thinking a little bit more we can see that whenever addition and multiplication are commutative in a given field we have an unique neutral element. For non-commuting fields things need not to be so.

After proving the unicity of both neutral elements and taking into account Axiom 4 (which states that the two neutral elements are distinct) we can now make some new definitions.

 Definition 1 The neutral element of addition will be called zero and denoted by the symbol ${ 0 }$. To the neutral element of multiplication we will call one and denote it by the symbol ${ 1}$.

We will now introduce a further axiom and some more results.

 Axiom 5 Following definition 1 ${ \forall x \in \mathbb{R} \, \exists\,y \in \mathbb{R} : x+y=0}$. ${ \forall x \in \mathbb{R} \setminus \{0\} \, \exists\,y \in \mathbb{R} : x\cdot y=1}$

From the previous axiom follows:

1. The element ${ y}$ is called the additive inverse of ${ x}$ ( or opposite ) and it is easy to prove its unicity.

Let us suppose that we have ${ y}$ and ${ y^\prime}$ for which ${ x+y=0}$ and ${ x+y^\prime=0}$ are valid equalities.

But

{\begin{aligned} y^\prime &= y^\prime+0 \\ &= y^\prime+(x+y)\\ &= (y^\prime+x)+y\\ &= 0+y\\ &= y \end{aligned}}

Where Axiom 4 ; definition 1; Axiom 2; Axiom 1, Axiom 4 and definition 1; Axiom 2 and Axiom 4 were used respectively.

Thus there is only one opposite to every real number.

After proving the unicity of the opposite in the addition operation it is possible for us to denote the symmetric of ${ x}$ by the symbol ${ -x}$ so that it is ${ x+(-x)=0}$.

• ${y}$ is called the multiplicative inverse, or the reciprocal, of ${ x}$. In an analogous manner that we proved that the opposite is unique we can also prove that the reciprocal is unique. Thus it makes sense for us to denote it by a symbol and the symbol is ${ x^{-1}}$ or ${ \dfrac{1}{x}=1/x}$.
 Definition 2 A field is defined to be a set where ${ \cdot}$ and ${ +}$ are well defined operations and all five previous axioms are verified.

— 1.1. Consequences of the Field Axioms —

Now we will state (and in some cases prove ) some familiar results from the building blocks we achieved so far.

 Theorem 1 ${\forall x,y,z \in \mathbb{R}}$ if ${x+y=x+z}$ then ${y=z}$. Proof: In this case a proof won’t be given because I feel that the result is much more important than the proof. $\Box$
 Theorem 2 (Possibility and unicity of subtraction) There exists an operation that is inverse to the addition operation. $\displaystyle \forall x,y \in \mathbb{R} \,\exists \,z : x=y+z \ \ \ \ \ (1)$   Proof: First we tackle the existence part of the theorem: {\begin{aligned} y+[x+(-y)] &= (y+x)+(-y)\\ &= (x+y)+(-y)\\ &= x+(y-y)\\ &= x+0\\ &= x \end{aligned}} Hence by taking ${ z=x+(-y)}$ we indeed have ${ x=y+z}$. Now we tackle the unicity part of the theorem: ${ y+z=x}$ and ${ y+z^\prime=x}$. By Theorem 1 it follows ${ z=z^\prime}$. $\Box$

${ z}$ Is called the difference between ${ x}$ and ${ y}$ and is denoted by the ${ x-y}$

 Theorem 3 (Possibility and unicity of division) There exists an operation that is inverse to the multiplication operation. $\displaystyle \forall x,y \in \mathbb{R} \setminus\{0\} \, \exists z \in \mathbb{R} : x=y\cdot z \ \ \ \ \ (2)$

Proof: It is analogous to Theorem 2 $\Box$

${ z}$ is called the quotient between ${ x}$ and ${ y}$ and denoted by ${ \dfrac{x}{y}}$ or ${ xy^{-1}}$

 Theorem 4 ${\forall x \in \mathbb{R} \quad x\cdot 0=0}$ Proof: {\begin{aligned} 0+0 &= 0 \\ x\cdot (0+0) &= x\cdot 0 \\ x\cdot 0+x\cdot 0 &= x\cdot 0 \\ x\cdot 0+x\cdot 0 &= x\cdot 0+0 \\ x\cdot 0 &= 0 \end{aligned}} $\Box$
 Theorem 5 ${\forall x,y \in \mathbb{R} x\cdot y=0 \Rightarrow x=0 \vee y=0}$ Proof: Left as an exercise for the reader. $\Box$

All of this may seem tautological to people who aren’t used to proper mathematical reasoning but it isn’t. The point of all of these definitions, axioms and theorems is that Mathematics is built is being built from the ground up so that no doubts can be cast to the results we state.