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)