Category Archives: Numbers

Description of the numbers

Numbers, Natural numbers, Infinitesimal calculus

A model of the natural

A model of the natural numbers

We're going to see a model of the natural

A theory set of axioms is consistent when there appears some kind of contradiction that makes the axioms do not make sense

Peano's axioms introduce natural numbers from set theory. After exposing the axioms, some additional definitions (the sum, the product and the principle of good management) have been given. But are Peano's axioms consistent? That is, is there a model based on the assembly model that meets Peano's axioms? Let's see how to build a model of natural numbers from set theory

Note if we had a model, we would be reducing the consistency of numbers to that of set theory, and the set theory is consistent? At some point we have to stop descending, and you have to assume something as true without the possibility that it will be demonstrated. In 1930, He tested his famous theorems of incomprehensibility, which in a simplified way state the following:

  • First theorem of incompleteness
    In any consistent formalization of mathematics that is strong enough to define the concept of natural numbers, an affirmation can be constructed that cannot be demonstrated or refuting within that system
  • Second theorem of incompleteness
    Any consistent set may be used to prove itself

These results were very negative for the philosophical approach to mathematics, proposed by David Hilbert, known as Hilbert's formalization program. He had proposed that the consistency of more complex systems, such as actual analysis, had to be tested in terms of simpler systems. Finally, the consistency of all mathematics could be reduced to basic arithmetic

With his first theorem, Godel demonstrated that arithmetic is incomplete, and in this way it will be impossible for it to be used to demonstrate the consistency of any axiom system. The paradise devised by Hilbert did not exist, it was a utopia

Let's stop the logic and go back to the natural numbers. A model of Peano's axioms would be a triplet (\mathbb{N},0,s) with \mathbb{N} a set, 0\in \mathbb{N}, s|\mathbb{N}\rightarrow\mathbb{N} an application that would satisfy these axioms. It is not difficult to see that if we have two models, they should be the same, since given two models (\mathbb{N}_A, 0_A, s_A) and (\mathbb{N}_B, 0_B, s_B) Peano's axioms, the application f|\mathbb{N}_A\rightarrow \mathbb{N}_B defined by:

\begin{cases}f(0_A)=0_B \\ f(s_A(n))=s_B(n) \end{cases}

it's a bijection. Since we have to, there can only be one model of Peano's axioms, but we haven't defined any yet!

Logicians Gottlob Fregge and Bertrand Russell built models of Peano's axioms based on the intuitive idea that a natural number is the cardinal of an ensemble. That is, each natural number is the "essence" shared by the bijective (and finite) sets whose cardinal number is that cardinal number. Then a natural number would be defined as the equivalence case of those sets; in addition, zero corresponds to the empty set (actually, its equivalence class) and s applied to an assembly is to add an element that is not in the set (for each set of the equivalence class). However, this idea has serious logical difficulties, because, according to the laws of Zermelo-Fraenkel's axiomatics on which the current mathematics is based, put informally: "the whole of all sets is not a whole", the equivalence classes of bijective sets also do not form a set of

The model commonly used is that of the Hungarian nationalized American mathematician, János von Neumann, who not only made important contributions in many fields of mathematics, but in computer science with von Neumann's architecture, currently used in computers and in the manufacture of the first atomic bomb

The model devised by von Neumann begins with the definition of 0 as the empty set (\emptyset), and also defines an operator s acting on the sets by s(A)=A\cup(A)

Defines the set of natural numbers \mathbb{N}_0, such as the intersection of all closed sets under the action of s (that is, of all C sets such that s(C)\subset C) containing the empty set. Each natural number (understood as a set), is the set of natural numbers less than him:

\scriptsize\begin{cases}0=\emptyset \\ 1=s(0)=\emptyset\cup\{\emptyset\}=\{\emptyset\}=\{0\} \\ 2=s(1)=\{\emptyset\}\cup\{\{\emptyset\}\}=\{\emptyset,\{\emptyset\}\}=\{0,1\} \\ 3=s(2)=\{\emptyset,\{\emptyset\}\}\cup\{\{\emptyset,\{\emptyset\}\}\}=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}=\{0,1,2\} \\ \cdots \\ n=s(n-1)=\cdots=\{1,2,3,\cdots,n\} \end{cases}

The set \mathbb{N}_0, along with 0 and the "next" function s|\mathbb{N}_0\rightarrow\mathbb{N}_0, satisfies Peano's axioms

Numbers, Natural numbers, Infinitesimal calculus

Operations of the natural

Operations of the natural numbers

We're going to see the operations of the natural

With the axioms of Peano and with the zero as the first element of \mathbb{N}_0, we introduce the notation of others:

\begin{cases}s(0)\text{ we call 1} \\ s(1)\text{ we call 2} \\ \cdots \\ s(n-1)\text{ we call n}\end{cases}

Operation sum

The sum or (addition) in \mathbb{N}_0 it is an operation that +|\mathbb{N}_0\rightarrow \mathbb{N}_0 which is recursively defined as:

\begin{cases}a+0=a \\ a+s(b)=s(a+b) \end{cases}

Properties of the sum

Dice a, b, c\in\mathbb{N}_0 comply with:

  • Associative property for addition
    a+(b+c)-(a+b)+c (as a result of the previous property, there is no need to enter parentheses and a+b+c can be written)
  • Commutative property for addition
    a+b=b+a
  • Neutral element of the sum
    0\in\mathbb{N}_0,\forall a\in\mathbb{N}_0|a+0=a; \forall a\in\mathbb{N}_0, \exists b\in\mathbb{N}_0|a+b=0
  • Cancellation property (simplification) in the sum
    \text{If }a+c=b+c\rightarrow a=b

Operation product

The product or (multiplication) in \mathbb{N}_0 it is an operation that \cdot|\mathbb{N}_0\rightarrow \mathbb{N}_0 which is recursively defined as:

\begin{cases}a\cdot 0=0 \\ a\cdot s(b)=a+a\cdot b \end{cases}

Properties of the product

Dice a, b, c\in\mathbb{N}_0 comply with:

  • Associative property for the product
    a\cdot(b\cdot c)=(a\cdot b)\cdot c (as a result of the above property, there is no need to indicate parentheses and can be written a\cdot b\cdot c)
  • Commutative property for the product
    a\cdot b=b\cdot a
  • Neutral element of the product
    1\in\mathbb{N}_0,\forall a\in\mathbb{N}_0|a\cdot 1=a; \forall a\in\mathbb{N}_0,a\not=0, \exists b\in\mathbb{N}_0|a\cdot b=1
  • Cancellation property (simplification) on the product
    \text{If }a\cdot c=b\cdot c\text{ con }c\not= 0\rightarrow a=b

Operation sum and product

In addition, sum and product share the following property:

Property of the sum and the product

  • Distributive property of the product regarding the sum
    \text{If }a\cdot (b+c)=(a\cdot b)+(a\cdot c)

Note if we had chosen to represent the natural ones without zero, in the axioms of Peano it would be enough to change \mathbb{N}_0 by \mathbb{N} and 0 by 1. Since the first axiom only serves to ensure that \mathbb{N}_0 is not the empty set, and the name given to the first element is not relevant in its definition. When it becomes relevant, it is when the sum and product operations are defined. If the natural ones are built starting at 1, the definition of the sum is started with a+1 s(a), and that of the product by a\cdot 1=a

Numbers, Natural numbers, Integer numbers, Infinitesimal calculus

Integer numbers

Integer numbers

With natural ones we can add numbers, but we can't always subtract. We will use the whole numbers to be able to subtract any pair of natural

Although natural numbers seem obvious, mathematicians had a hard time treating negative numbers equally to positive numbers. Historically, the use of negative numbers is far after the use of fractions or even irrational positives

An example of complications from negative numbers for both Greek Diophanto and European Renaissance algebrists, an equation of the x^2+b\cdot x+c=0 (in which for us parameters b and c can be both positive and negative and the resolution method remains the same) was not unique, but should be analyzed in four cases \begin{cases}x^2+b\cdot x+c=0 \\ x^2+b\cdot x=c \\ x^2+c=b\cdot x \\ x^2=b\cdot x+c \end{cases} with b and c always positive (even more cases if we allow the possibility that b or c are worth zero) and each case had its own method of resolution

Note: the rating x^n has usual sense x^n=\underbrace{x\cdots x}_{\text{n times}}, \forall n \in\mathbb{N} and x belongs to any of the sets of numbers we consider (natural, integer, rational, and real). Also when no. 0 we have to x^0=1

It was the Dutch mathematician Simon Stevin, at the end of the 16th century, who first recognized the validity of negative numbers by accepting them as a result of the problems he worked with. In addition, it recognized the equality between subtracting a positive number and adding a negative number (i.e., a-b-a+(-b), with a, b > 0). For this reason, just as Brahmagupta is considered to be the father of zero, Stevin is regarded as the father of negative numbers (in fact, Stevin made many more contributions to the world of numbers, in the field of real numbers)

In \mathbb{N}_0 can add numbers, but we can't always subtract. That is why the need arises to create a new set of numbers with which to subtract any pair of natural numbers; this new set is that of integers and is denoted as \mathbb{Z}

It is totally elementary to see that this relationship is of equivalence, and that allows us to define whole numbers as the quotient set \mathbb{Z}=\mathbb{N}_0\times\mathbb{N}_0/\sim

Numbers, Natural numbers, Integer numbers, Infinitesimal calculus

Natural differences and integers

Natural differences and integers

Differences between natural and whole:

The set of integers it is not a set well-ordered (since the subset of negative integers has no minimum)

Any subset not empty of \mathbb{Z} lower bound has minimum (and multiplied by (-1), if bound above, has maximum)

To multiply integers we have to distinguish between positive and negative, applying the so called rule of the signs:

\begin{cases} \text{positive }\cdot\text{positive = positive} \\ \text{positive }\cdot\text{negative = negative} \\ \text{negative }\cdot\text{positive = negative} \\ \text{negative }\cdot\text{negative = positive} \end{cases}

The order of \mathbb{Z} is a total order, but it should be noted that any negative number is less than any positive. In addition, these properties must be taken into account for operations:

\tiny\begin{cases} a \leq b \Rightarrow a + c \leq b + c \\ a \leq b, c \geq 0 \Rightarrow a \cdot c \leq b \cdot c \\ a \leq b, c < 0 \Rightarrow a \cdot c \geq b \cdot c & \text{if multiplied by a negative number} \end{cases}

We have at our disposal the absolute value function (or module):

\|a\|=\begin{cases} a & \text{if } a \geq 0 \\ (-a) & \text{if } a < 0 \end{cases}

From the absolute value function and thanks to its geometric implications, we also have at our disposal the Triangular inequality:

\|a + b\| \leq \|a\| + \|b\|
Numbers, Natural numbers, Integer numbers, Infinitesimal calculus

Operations of integers

Inclusion of the natural numbers in the integers

We're going to see the operations of integers

\mathbb{N}_0 includes in \mathbb{Z} associating to each a\in\mathbb{N}_0 the equivalence class of (a, 0)

The natural number 0 in \mathbb{Z} is (0, 0), which is the kind of equivalence given by \{(a, a) | a \in\mathbb{N}_0\}, which is the element of the sum in \mathbb{Z}

The natural 1 in \mathbb{Z} is (1, 0), and its kind of equivalence, which is the neutral element of the product in \mathbb{Z}

Unlike in \mathbb{N}_0, in \mathbb{Z}, given a number, you can always find another one that adds to the first give zero (the neutral element of the sum). If we have (a, b), just take (b, a) and it is fulfilled:

(a, b)+(b, a)=(a + b, a + b)\sim (0, 0)

This number (b, a), which is the opposite (a, b), and we denote it as -(a, b). Which allows us to define the subtraction:

(a, b) - (c, d)=(a + b) + (-(c, d))=(a, b)+(d, c)=(a + d, b + c)

Or equivalently:

(a, b) - (c, d)=(r + s) \Longleftrightarrow (a, b)=(r, s) + (c, d)

Operation sum

Given that (a, b) it is the way that we have in \mathbb{Z} to indicate the subtraction a-b and in the same way, (c, d) represents c-d, there's nothing more to think about (a-b)+(c-d)=(a+c)-(b+d) to give a proper definition of the sum:

(a, b)+(c, d)=(a+c, b+d)

Operation product

Similarly, to define the product we have to (a-b)\cdot(c-d)=(a\cdot c + b\cdot d)-(a\cdot d+b\cdot c), therefore the appropriate definition for the product:

(a,b)\cdot (c,d)=(a\cdot c+b\cdot d, a\cdot d+ b\cdot c)

Relationship of equivalence

For the definitions in \mathbb{N}_0\times\mathbb{N}_0 given above are valid in \mathbb{Z}=\mathbb{N}_0\times\mathbb{N}_0/\sim, we need to prove that they are compatible with the relationship of equivalence, that is, if we have (a_1,b_1)\sim(a_2, b_2)\text{ y }(c_1,d_1)\sim(c_2, d_2) it is fulfilled that:

\begin{cases} (a_1,b_1)+(c_1,d_1)\sim(a_2,b_2)+(c_2,d_2) \\ (a_1,a_1)\cdot(c_1,d_1)\sim(a_2,b_2)\cdot(c_2,d_2) \end{cases}

Order relationship

There is also to define the order relationship in \mathbb{Z}. To define when (a,b)\leq(c,d), let's think once again about (a,b) as in \displaystyle a-b and in (c,d) as c-d. Then just look at that a-b\leq c-d is equivalent to a+d\leq b+c to realize that the definition that we seek has to be (a,b)\leq (c,d)\Leftrightarrow a+d\leq b+c

As in the sum and the product, we must check the compatibility of that definition with the equivalence ratio, that is, if we have (a_1,b_1)\sim(a_2,b_2)\text{ y }(c_1,d_1)\sim(c_2,d_2), is fulfilled that:

(a_1,b_1)+(c_1,d_1)\sim(a_2,b_2)+(c_2,d_2)
Numbers, Natural numbers, Rational numbers, Infinitesimal calculus

Rational numbers

Rational numbers

The extent of the integers, \mathbb{Z}, to rational numbers \mathbb{Q}, has a clear parallel with the extent of \mathbb{N}_0 to \mathbb{Z}. As we could not subtract in \mathbb{N}_0, invented a new type of numbers to achieve it. Now we find the problem that we can't always divide into \mathbb{Z}, and invented a new type of numbers to achieve it

To define \mathbb{Z} we took pairs of natural numbers, and applied an equivalence ratio. The definition of \mathbb{Q} follow the same steps, but one in an even clearer way: rational numbers will be pairs of integers, which correspond to the numerator and denominator of each fraction; equivalence classes must also be taken to identify fractions representing the same number

As the denominator of a fraction cannot be null, instead of taking \mathbb{Z}x\mathbb{Z}, the set is taken:

\mathbb{Z}x(\mathbb{Z}\backslash\left\{0\right\})=\left\{(a, b) | a, b \in \mathbb{Z} \quad b \not= 0\right\}

and in it we define the relationship \sim given by:

(a, b)\sim (c, d) \Longleftrightarrow a\cdot d = b\cdot c

that can be easily demonstrated, that it is of equivalence

(a, b) and (c, d) will end up being, respectively, the rational \frac{a}{b} and \frac{c}{d}; you still can't talk about equality \frac{a}{b}=\frac{c}{d} (for these fractions do not yet exist), but if it made sense it would be equivalent to saying that a\cdot d = b\cdot c, which is what we're using to define the equivalence ratio

Now we define \mathbb{Q} as the quotient set:

\mathbb{Q}=\mathbb{Z}x(\mathbb{Z}\backslash \left\{0\right\})/\sim
Numbers, Rational numbers, Infinitesimal calculus

Operations of the rational

Inclusion of the integers in the rational

We're going to see the operations of the rational

\mathbb{Z} includes in \mathbb{Q} associating to each a\in\mathbb{Z} the equivalence class of (a, 1)

Being the sum and the product in \mathbb{Q} extensions of \mathbb{Z}

In addition, since (-a, -b)\sim(a, b), and as by definition b\not= 0, as long as we take (a, b) in \mathbb{Q}, we can assume that b > 0 (i.e. denominators of fractions can be imposed as positive)

Operation sum

It is defined by the rules of fractions, when it makes sense:

\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b \cdot c}{b \cdot d}

and to achieve this the formal definition must be:

(a, b) + (c, d) = (a \cdot d + b\cdot c, b\cdot d)

Operation product

It is defined by the rules of fractions, when it makes sense:

\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b \cdot d}

and to achieve this the formal definition must be:

(a, b) \cdot (c, d) = (a \cdot c, b\cdot d)

Relationship of equivalence

For the definitions in \mathbb{Z}\times(\mathbb{Z}\backslash\left\{0\right\}) given above are valid in \mathbb{Q}=\mathbb{Z}\times(\mathbb{Z}\backslash \left\{0\right\})/\sim, we have to prove that they are compatible with the equivalence ratio, that is, if we have (a_1,b_1)\sim(a_2, b_2) and (c_1,d_1)\sim(c_2, d_2) it is fulfilled that:

\begin{cases} (a_1,b_1)+(c_1,d_1)\sim(a_2,b_2)+(c_2,d_2) \\ (a_1,a_1)\cdot(c_1,d_1)\sim(a_2,b_2)\cdot(c_2,d_2) \end{cases}

Order relationship

There is also to define the order relationship in \mathbb{Q}. To do this we have to express the still non existent inequality \frac{a}{b}\leq \frac{c}{d} in terms of inequalities with integers, but as we can assume that b > 0 and d > 0, is equivalent to a \cdot d \leq c \cdot b (this inequality is obtained from which we wanted to obtain by multiplying b \cdot d and because we have assumed that denominators are positive, it makes inequality not change direction). So we define:

(a, b) \leq (c, d) \Longleftrightarrow a \cdot d \leq c\cdot b \quad (b, d > 0)

It can be shown that this relationship is of order, which is compatible with the equivalence ratio (we have used it when defining \leq for positive denominators only) and extends that of \mathbb{Z}. As in \mathbb{Z}, the order relationship in \mathbb{Q} is total, and we also have positive and negative numbers

Note Once we have already made a good definition of rational numbers, we can detach ourselves from auxiliary notation (a, b) to use from now on, the usual \frac{a}{b} or a / b, where a is called numerator and b denominator (as mentioned above, it can be assumed that b is always positive). Also, instead of using \frac{a}{1}, we can just use a. And instead of using the equivalence ratio as a (a, b)\sim (c, d) we will use \frac{a}{b}=\frac{c}{d}

Note In each class of equivalent fractions of \mathbb{Q}, there is a fraction that is usually taken as a representative. It is the irreducible so-called, in which the maximum common numerator divider and denominator is worth 1. Although it is elementary, we should have developed something more arithmetic in \mathbb{N}_0 to be able to insert it here

Operation division

The properties of the sum and the product in \mathbb{Z} (associativity, commutivity, neutral elements, distributivity, etc.) extend to \mathbb{Q}, in addition, order is stable with operations in the same sense as in \mathbb{Z}. But we have a novelty regarding \mathbb{Z}, we can divide

For any \frac{a}{b}\in \mathbb{Q} not null, we can find a reverse with respect to the product. Since:

\frac{a}{b}\cdot \frac{b}{a}=\frac{a\cdot b}{b \cdot a}=\frac{1}{1}=1

The notation for the inverse of \frac{a}{b} it is (\frac{a}{b})^{-1}=\frac{b}{a}

Thanks to the reverse we can divide by a rational not null. Divide \frac{a}{b} between \frac{c}{d} is to find a rational number such that \frac{a}{b} multiplied by that rational \frac{c}{d}; this operation is immediate, just multiply \frac{a}{b} by the inverse of \frac{c}{d}:

\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot(\frac{c}{d})^{-1}=\frac{a}{b}\cdot\frac{d}{c}=\frac{a\cdot d}{b\cdot c}

The existence of inverse in the product (and therefore division) makes the algebraic structure of \mathbb{Z} (which is a ring) fall short, and that \mathbb{Q} with the sum and the product becomes a body. In addition, the order structure is stable with trades, i.e. dice x, y, t \in \mathbb{Q} it is fulfilled that:

\begin{cases}x\leq y \Rightarrow x+t \leq y + t \\ x\leq y, t \geq 0 \Rightarrow t\cdot x \leq t\cdot y \end{cases}
Numbers, Natural numbers, Rational numbers, Infinitesimal calculus

Archimediedian property

Archimediedian property

Archedian property:

It x \in \mathbb{Q}, x > 0. So for any y \in \mathbb{Q} there is n \in \mathbb{N}_0 such that n\cdot x > y

Demonstration: Archimedian property

If y \leq 0 the result is trivial, because it is enough to take n = 1. We assume that y > 0. We want to prove that it exists n \in \mathbb{N}_0 that meets n > \frac{y}{x}

The quotient of rational numbers \frac{y}{x} it will be a rational number \frac{a}{b} with positive integers a and b. So, n > \frac{y}{x}=\frac{a}{b} is equivalent to say n \cdot b > a And that is accomplished by taking n = a + 1 since:

(a + 1)\cdot b = a\cdot b + b \geq a + b \geq a + 1 > a

that is just what we wanted

Numbers, Natural numbers, Rational numbers, Real numbers, Infinitesimal calculus

Real numbers

Real numbers

The real numbers complete what already made the rational numbers

The rational numbers \mathbb{Q}, offer us many possibilities already. But you can immediately realize that they are not enough for any of the tasks one would like to entrust to numbers: measuring units. In fact, it was the Pythagorics in the 4th century BC. who observed this significant lack of rational numbers, which was a huge crisis in the conception of the mathematics that they had

The idea of Pyrtagics was that everything had to be reduced to numerical proportions; and such proportions would amount to our fractions. Possibly, his deep disappointment was the result of his discovery that the diagonal of the square was immeasurable with length, that is, that the diagonal cannot be expressed as a rational number of times the length of the side

The diagonal length of a square can be obtained by the Pythagorean theorem: in a right triangle, the sum of the squares of the catetos is equal to the square of the hypotenuse. And its formula:

c^2=a^2+b^2

where a and b are the legs and c is your hypotenuse

There are many demonstrations of the theorem of Pythagoras and of very different nature because of the type of techniques that they use

One of the most modern is the one based on an area argument, published in the journal The New England Journal of Education and due to the twentieth president of the united States James A. Garfield

There are three methods commonly used to define real numbers \mathbb{R}, based on rational \mathbb{Q}, and all three have their origins in the nineteenth century, which is when mathematical analysis reached the rigor that is required today. These methods are as follows:

  • Cuts of Dedekind
  • Equivalence classes of successions of Cauchy of rational
  • Equivalence classes of pairs of probate monotone convergent

The real numbers \mathbb{R}, are a body with a relationship of order verifying the completeness axiom (also called the supreme's axiom): all set bounded superiorly possesses supreme

That is, the actual numbers are a set with two operations, sum and product, and an order relationship fulfilling exactly the same properties as in the case of rational ones. The only difference between real and rational is that for the former the axiom of completeness is fulfilled

A real number that is not rational what we refer to as the irrational

This definition of real numbers simply by axioms is misleading, because without formal construction from previous concepts, nothing guarantees that such an axiomatic definition model exists. We'll try to build it using Dedekind's cuts and uniqueness. Some of the other two methods will also be discussed

Propositions

For the properties involved \mathbb{Z} and \mathbb{Q} we're not going to give all the details of the demonstrations, just some indication of how to put them forward. The one that will be developed is that of the Arquimedian property, given its great importance and usefulness

Proposition 1: Archimedian Property

It x\in\mathbb{R}, x > 0. Then \forall y\in\mathbb{R}, \exists n\in\mathbb{N}_0\Rightarrow n\cdot x > y

Demonstration of Proposition 1

If y \leq 0, if we take n = 1 there's nothing to prove. Yes y > 0 we will prove by reduction to absurdity

Suppose that n\cdot x \leq y for all n\in\mathbb{N}_0 and we consider the set S=\{n\cdot x | n \in\mathbb{N}_0\}. Set S, which is nonnempty, is fully bound (by y), then by the axiom of completeness possesses supreme

It s=\text{ sup }S, as x > 0 then s - x < s, by the definition of supreme s - x cannot be the top dimension of S. Therefore there will be some element of S of the way m \cdot x, with m\in\mathbb{N}_0 such that s - x < m \cdot x

But this implies that s < (m+1) \cdot x and obviously (m+1) \cdot x \in S, with what s is not the upper level of s, so we come to a contradiction with the fact that it is the supreme of S, thus proving the Archimedian property by reducing the absurdity

Proposition 2

Taking x = 1 we have to \mathbb{N} is not tied up at all. And it can be easily tested with the Archedian property

Proposition 3

\forall \epsilon\in\mathbb{R}, \epsilon > 0, \exists n\in\mathbb{N}\Rightarrow \frac{1}{n} < \epsilon

Considering y=1 and x=\epsilon > 0 we can prove the proposition using the property arquimediana and it will be of great use when we study limits of successions

Proposition 4

If \alpha, \beta \in \mathbb{R} that meet \beta - \alpha > 1 then \exists k\in\mathbb{Z} that meets \alpha < k < \beta

To test this proposition it will suffice to take \alpha as the largest integer that satisfies \alpha\leq\alpha and consider k=a+1, which will be fulfilled \alpha < k < \beta

Given \alpha\in\mathbb{R} the largest integer such that a \leq \alpha < a+1, is called an entire part of \alpha and is denoted by \mid\alpha\mid

To be absolutely rigorous, we must prove the existence and uniqueness of that value

Proposition 5

Given \alpha\in\mathbb{R}, \exists a \in\mathbb{N} unique such that a\leq\alpha < a+1

It can also be demonstrated on the basis of the property arquimediana

Proposition 6

The following is true:

  • Are \alpha, \beta \in\mathbb{R} with \alpha < \beta. Then \exists r\in\mathbb{Q}\Rightarrow\alpha < r < \beta
  • Are r, s \in\mathbb{Q} with r < s. Then \exists \alpha\in\mathbb{R} \backslash \mathbb{Q}\Rightarrow r < \alpha < s
  • Are \alpha, \beta \in\mathbb{R} with \alpha < \beta. Then \exists \Upsilon\in\mathbb{R} \backslash \mathbb{Q}\Rightarrow \alpha < \Upsilon < \beta

The first part is inferred by Proposition 3: \exists n\in\mathbb{N} with \frac{1}{n} < \beta - \alpha where n\cdot\beta - n\cdot\beta > 1 and by Proposition 4: \exists k \in \mathbb{Z} with n\cdot\alpha < k < n\cdot \beta. Take r=\frac{k}{n}

The second part is enough to use that there is an irrational between 0 and 1 (or between any other fixed pair of rationals), displace and scale

The third part is a consequence of the previous two (between \alpha and \beta we can interleave rationals twice)

Numbers, Real numbers, Infinitesimal calculus

Pythagorean theorem

Pythagorean theorem

Pythagorean theorem: in a right triangle, the sum of the squares of the catetos is equal to the square of the hypotenuse

James A. Garfield

Garfield was the twentieth president of the United States, was an amateur mathematician and published this demonstration in the magazine The New England Journal of Education (vol. 3, p. 161) in 1876, five years before his arrival at the White House and his death, as he died in September 1881, the year of his appointment, as a result of injuries sustained in an attack in July of that year

Demonstration: Pythagorean Theorem

Based on the following scheme:

Pythagorean theorem Demonstration of Garfield

the demonstration is based on the observation that:

\text{area}(T_1) + \text{area}(T_2) + \text{area}(T_3) = \text{area}(T_1 \cup T_2 \cup T_3)

where T_1 is the triangle on the left, T_2 the triangle on the right and T_3 the central triangle

so we have to:

\text{area}(T_1)=\text{area}(T_2)=\frac{1}{2}\cdot a \cdot b
\text{area}(T_3)=\frac{1}{2}\cdot c^2

as T_1 \cup T_2 \cup T_3 is a trapezoid of bases a and b and height a + b you have to:

\text{area}(T_1 \cup T_2 \cup T_3)=\frac{1}{2} \cdot (a + b)^2

therefore if we replace in the initial formula we have:

2\cdot \frac{1}{2} \cdot a \cdot b + \frac{1}{2} \cdot c^2 = \frac{1}{2} \cdot (a + b)^2
2\cdot \frac{1}{2} \cdot a \cdot b + \frac{1}{2} \cdot c^2 = \frac{1}{2} \cdot (a^2 + a \cdot b + a \cdot b + b^2)
2\cdot \frac{1}{2} \cdot a \cdot b + \frac{1}{2} \cdot c^2 = \frac{1}{2} \cdot (a^2 + 2 \cdot a \cdot b + b^2)
2\cdot \frac{1}{2} \cdot a \cdot b + \frac{1}{2} \cdot c^2 = \frac{1}{2} \cdot (a^2 + b^2) + \frac{1}{2} \cdot 2 \cdot a \cdot b
\frac{1}{2} \cdot c^2 = \frac{1}{2} \cdot (a^2 + b^2) + \frac{1}{2} \cdot 2 \cdot a \cdot b - 2\cdot \frac{1}{2} \cdot a \cdot b
\frac{1}{2} \cdot c^2 = \frac{1}{2} \cdot (a^2 + b^2)
2 \cdot \frac{1}{2} \cdot c^2 = 2 \cdot \frac{1}{2} \cdot (a^2 + b^2)
c^2 =a^2 + b^2

with what the demonstration is concluded