# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 4

    Operations
We consider the mathematical operations addition and multiplication, within the real numbers, as mappings
    $\mathbb {R} \times \mathbb {R} \longrightarrow \mathbb {R} ,$
so we assign to each pair
    $(x,y)\in \mathbb {R} \times \mathbb {R} \,$
the real number $x+y$ (and $x\cdot y$ respectively). Such a mapping is called an operation.

## Definition

An operation (or binary operation) $\circ$ on a set $M$ is a mapping
    $\circ \colon M\times M\longrightarrow M,(x,y)\longmapsto \circ (x,y)=x\circ y.$
The domain of an operation is thus the product set of $M$ with itself, and the range is also $M$. Addition, multiplication and subtraction (on $\mathbb {Z}$, on $\mathbb {Q}$, or on $\mathbb {R}$) are operations. On $\mathbb {Q}$ and $\mathbb {R}$, the division is not an operation, since it is not defined in case the second component equals $0$. However, the division is an operation on $\mathbb {R} \setminus \{0\}$.
    Axioms
In mathematics, we encounter many structures which occur again and again. E.g., the rational numbers and the real numbers share many common properties, but they also differ in regard with certain other properties. This observation is the fundament for the axiomatic approach to mathematics. In this approach, one puts several structural properties which appear in a certain context consistently into a new concept. The goal of this is to infer in a logical way further properties from basic properties. The argumentation does then not take place on the level of specific familiar examples like the real numbers, but on the level of the properties themselves in a logical-deductive manner. The gain of this method is that one has to do the mathematical conclusions only once on the abstract level of the properties, and these conclusions hold then for all models which fulfill the basic properties. At the same time, one recognizes logical dependencies and hierarchies between the properties. Basic properties of mathematical structures are called axioms.
In the axiomatic approach, the properties (their principles and rules) are in the center. Mathematical objects which obey these rules are then examples or models for these concepts. In particular, one chooses properties which, on one hand, are easy to formulate, and which, on the other hand, allow strong conclusions. The advantages of this approach are the following items.
As disadvantages, one can consider the following items.
    Fields
We are going to describe the properties of the real numbers within an axiomatic framework. The axioms for the real numbers fall into three kinds: algebraic axioms, axioms for the ordering and the completeness axiom. The algebraic axioms are put together in the concept of a field. By algebraic properties, we mean properties which refer to the operations of computations, like addition, subtraction, multiplication and division. These operations assign to two elements of a given set $M$ (like the set of real numbers) another element of this set, they are binary operations. The following definition uses only two operations, namely addition and multiplication, subtraction and division will be treated as derived operations.

## Definition

A set $K$ is called a field if there are two binary operations (called addition and multiplication)
    $+:K\times K\longrightarrow K{\text{ and }}\cdot :K\times K\longrightarrow K$
and two different elements $0,1\in K$, which fulfill the following properties.
* Axioms for the addition:
** Law of associativity: $(a+b)+c=a+(b+c)$ holds for all $a,b,c\in K$.
** Law of commutativity: $a+b=b+a$ holds for all $a,b\in K$.
** $0$ is the neutral element of the addition, i.e. $a+0=a$ holds for all $a\in K$.
** Existence of the negative: For every $a\in K$, there exists an element $b\in K$ with $a+b=0$.
* Axioms of the multiplication:
** Law of associativity: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all $a,b,c\in K$.
** Law of commutativity: $a\cdot b=b\cdot a$ holds for all $a,b\in K$.
** $1$ is the neutral element for the multiplication, i.e. $a\cdot 1=a$ holds for all $a\in K$.
** Existence of the inverse: For every $a\in K$ with $a\neq 0$, there exists an element $c\in K$ such that $a\cdot c=1$.
* Law of distributivity: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all $a,b,c\in K$.
It is known from school that all these axioms hold for the real numbers (and for the rational numbers) together with the natural operations.
In a field, we use the convention that multiplication connects stronger than addition. Hence, we write $a\cdot b+c\cdot d$ instead of $(a\cdot b)+(c\cdot d)$. To further simplify the notation, the product sign is usually omitted. The special elements $0$ and $1$ in a field are called (the) zero and (the) one. By definition, they have to be different.
For us, the most important examples for a field are the field of rational numbers, the field of real numbers and the field of complex numbers (to be introduced later).

## Lemma

Suppose $K$ is a field. Then for every element $x\in K$ the element $y$ fulfilling $x+y=0$ is uniquely determined. For $x\neq 0$ the element $z$ fulfilling $xz=1$

### Proof

Let $x$ be given and suppose that $y$ and $y'$ are elements fulfilling $x+y=0=x+y'$. Then
    $y=y+0=y+(x+y')=(y+x)+y'=(x+y)+y'=0+y'=y'\,,$
which means altogether $y=y'$. For the second part see Exercise 4.3 .
$\Box$

## Example

We are trying to find a structure of a field on the set $\{0,1\}$. If $0$ is supposed to be the neutral element of the addition and $1$ the neutral element of the multiplication, then everything is already determined: The equation $1+1=0$ must hold, since $1$ has an inverse element with respect to the addition, and since $0\cdot 0=0$ holds, due to Lemma 5.5 . Hence the operation tables look like
and
With some tedious computations, one can check that this is indeed a field.

## Lemma

Let $K$ be a field,
$a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$
$K$
* $a0=0$ (annulation rule).
* (rules for sign).
    $(-a)b=-ab=a(-b)\,$
    $(-a)(-b)=ab\,.$
    $a(b-c)=ab-ac\,.$
* From $a\cdot b=0$ one can deduce $a=0$ or $b=0$.
* ${\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (general law of distributivity).

### Proof

* We have $a0=a(0+0)=a0+a0$. Subtracting $a0$ (meaning addition with the negative of $a0$) on both sides gives the claim.
* See Exercise 4.4 .
* See Exercise 4.4 .
* See Exercise 4.4 .
* We prove this by contradiction, so we assume that $a$ and $b$ are both not $0$. Then there exist inverse elements $a^{-1}$ and $b^{-1}$ and hence $(ab){\left(b^{-1}a^{-1}\right)}=1$. On the other hand, we have $ab=0$ by the premise and so the annulation rule gives hence $0=1$, which contradicts the field properties.
    $(ab){\left(b^{-1}a^{-1}\right)}=0{\left(b^{-1}a^{-1}\right)}=0\,,$
* This follows with a double induction, see Exercise 4.22 .
$\Box$
    Proof by contradiction
We have just given a proof by contradiction, we want to illustrate this method by further examples.
We give two classical examples for a proof by contradiction.

## Theorem

$2$
${\sqrt {2}}$

### Proof

We make the assumption that there exists some rational number whose square equals $2$, and we have to derive a contradiction from this. Our assumption means the existence of
    $x\in \mathbb {Q} \,$
fulfilling the property
    $x^{2}=2\,.$
A rational number can be written as a fraction, where numerator and denominator are integers. Hence, the rational number $x$ has the form
    $x={\frac {a}{b}}\,.$
Moreover, we can suppose that this fraction has been reduced to its lowest terms, so that the greatest common divisor of $a$ and $b$ is $1$. In fact it is enough to suppose that at least one of $a$ and $b$ are odd (if both are even we can divide both by $2$ until one gets odd). The property
    $x^{2}=2\,$
means then
    $x^{2}={\left({\frac {a}{b}}\right)}^{2}={\frac {a^{2}}{b^{2}}}=2\,.$
Multiplication by $b^{2}$ yields
    $2b^{2}=a^{2}\,$
(this is an equation in $\mathbb {Z}$ and even in $\mathbb {N}$). This equation means that $a^{2}$ is even, since $a^{2}$ is a multiple of $2$. This implies that $a$ itself is even, because the square of an odd number is again odd. Therefore, we can write
    $a=2c\,$
with an integer $c$. Putting this into the equation, we deduce
    $2b^{2}=(2c)^{2}=2^{2}c^{2}\,.$
Dividing both sides by $2$ we obtain
    $b^{2}=2c^{2}\,.$
Hence, also $b^{2}$ is even and so $b$ is even. But this is a contradiction, as $a$ and $b$ are not both even.
$\Box$
The following theorem is called Theorem of Euclid.

## Theorem
### Proof

We assume that the set of all prime numbers is finite, say $\{p_{1},p_{2},\ldots ,p_{r}\}$ is a complete list of all prime numbers. We consider the number
    $N=p_{1}\cdot p_{2}\cdot p_{3}\cdots p_{r}+1\,.$
This number can not be divided by any of the prime numbers $p_{i}$, since the reminder of $N$ by division through $p_{i}$ is always $1$. Hence, the prime factors of $N$, which exist due to Theorem 2.6 , are not contained in the given set. This is a contradiction.
$\Box$
    The binomial theorem

## Definition

For a natural number $n$, one puts
    $n!:=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\,,$
$n$
We put $0!=1$

## Definition

Let $k$ and $n$ denote natural numbers with $k\leq n$. Then
    ${\binom {n}{k}}:={\frac {n!}{k!(n-k)!}}\,$
$n$
$k$
One can write this fraction also as
    ${\frac {n(n-1)(n-2)\cdots (n-k+2)(n-k+1)}{k(k-1)(k-2)\cdots 2\cdot 1}},$
because th factors from $(n-k)!$ are also in $n!$. In this representation, we have the same number of factors in the numerator and in the denominator. Sometimes it is useful to allow also negative $k$ or $k>n$ and define in these cases the binomial coefficients to be $0$.
From the very definition, it is not immediately clear that the binomial coefficients are natural numbers. This follows from the following relationship.

## Lemma

The binomial coefficients fulfill the recursive relationship
    ${\binom {n+1}{k}}={\binom {n}{k}}+{\binom {n}{k-1}}\,.$

### Proof

$\Box$
The following formula brings addition and multiplication together in some sense.

## Theorem

Let $a,b$ be elements of a field and let $n$ denote a natural number. Then
    $(a+b)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}$

### Proof

We do induction over $n$. For $n=0$ we have on one hand $(a+b)^{0}=1$ and on the other hand $a^{0}b^{0}=1$ as well. Suppose now that the statement is true for $n$. Then
    ${\begin{aligned}(a+b)^{n+1}&=(a+b)(a+b)^{n}\\&=(a+b){\left(\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}\right)}\\&=a{\left(\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}\right)}+b{\left(\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}\right)}\\&=\sum _{k=0}^{n}{\binom {n}{k}}a^{k+1}b^{n-k}+\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k+1}\\&=\sum _{k=1}^{n+1}{\binom {n}{k-1}}a^{k}b^{n-k+1}+\sum _{k=0}^{n+1}{\binom {n}{k}}a^{k}b^{n-k+1}\\&=\sum _{k=1}^{n+1}{\left({\binom {n}{k-1}}+{\binom {n}{k}}\right)}a^{k}b^{n+1-k}+b^{n+1}\\&=\sum _{k=1}^{n+1}{\binom {n+1}{k}}a^{k}b^{n+1-k}+b^{n+1}\\&=\sum _{k=0}^{n+1}{\binom {n+1}{k}}a^{k}b^{n+1-k}.\end{aligned}}$
$\Box$

## Remark

For the binomial coefficient
    ${\binom {n}{k}}$
there is an important interpretation. It describes the number of subsets with exactly $k$ elements inside a set with $n$ elements. For example, within a set with $49$ elements there are exactly
    ${\binom {49}{6}}={\frac {49\cdot 48\cdot 47\cdot 46\cdot 45\cdot 44}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}=13983816\,$
subsets with $6$ elements. The inverse of this number is the probability to get at the lotto all six numbers right.