Category Archives: Natural numbers

The natural numbers are the ordered set of numbers that mankind has used to tell

Natural numbers, 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, 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, Natural numbers, Integer numbers, Rational numbers, Real numbers, Infinitesimal calculus

Inequalities

Inequalities

Below you are going to list some inequalities that are useful or that have not been mentioned

Up to now we have shown some properties that are verified in the various sets of numbers

Considering that each numerical set we have defined above contains the previous one and the set of real ones contains them all, all of them (except the principle of good sorting of natural numbers) are checked for the actual numbers

For example, using some of the inequalities mentioned above, it is possible to prove that:

  • 0\leq a \leq b \Rightarrow a^2 \leq b^2
  • 0 < a \leq b \Rightarrow \frac{1}{b} \leq \frac{1}{a}

Absolute value of a real number

It is defined for a real number in the same way that it has already been made for integers, but it also meets some inequalities that deserve to be pointed out

  • -\|a\| \leq a \leq \|a\|
  • \|a\| \leq b \Leftrightarrow -b \leq a \leq b
  • \|a\| \geq b \Leftrightarrow \begin{cases} a \geq b \\ a \leq -b \end{cases}
  • \|a\cdot b\| = \|a\|\cdot \|b\|
  • a^2 \leq b^2 = \|a\| \leq \|b\|

Keep in mind that in the equality \sqrt{a^2}=\|a\|, will only be true \sqrt{a^2}=a if a \geq 0

Distance

Dice a, b \in \mathbb{R}, is called distance between a and b to the actual non negative number \|a-b\|

This notation is fundamental to interpret inequalities of the form \|x-a\| \leq b, such as the distance of x to a is less than or equal to b

Triangular inequality

We've seen it before, but because it's an important inequality we remember it for real numbers

Dice a, b \in \mathbb{R} is true that \|a+b\| \leq \|a\|+\|b\|

Demonstration: triangular inequality

We take \begin{cases} -\|a\|\leq a \leq \|a\| \\ -\|b\|\leq b \leq \|b\| \end{cases}

We add both inequalities and we obtain -(\|a\|+\|b\|) \leq a+b \leq \|a\|+\|b\|

And therefore \|a+b\| \leq \|a\|+\|b\|

Inequality triangular reverse

Dice a, b \in \mathbb{R} is true that \|a\|-\|b\| \leq \|a-b\|

Demonstration: reverse triangular inequality

The inequality triangular inverse is equivalent to prove that -(a-b) \leq \|a\|-\|b\| \leq \|a-b\|

By the inequality triangle has to

\|a\|=\|a-b+b\| \leq \|a-b\|+\|b\|
\|a\|-\|b\| \leq \|a-b\|

with what the right-hand side of the inequality is proved

By the inequality triangle has to

\|b\|=\|b-a+a\| \leq \|b-a\|+\|a\|
\|b\|-\|a\| \leq \|b-a\|
-(a-b) \leq \|a\|-\|b\|

with what the left-hand side of the inequality is proved

Inequality between arithmetic and geometric mean

One of the most useful and popular inequalities is the inequality between the arithmetic and geometric mean (sometimes referred to as AM – GM). Which is defined as follows:

Dice a_1, a_2, \cdots, a_n \in \mathbb{R^+}

We define the arithmetic mean as M_{n, 1}=\frac{a_1, a_2, \cdots, a_n}{n}

It defines the average geomética as M_{n, 0}=\sqrt[n]{a_1, a_2, \cdots, a_n}

And inequality is defined as M_{n , 0} \leq M_{n, 1}

Demonstration: inequality between arithmetic and geometric mean

This demonstration was published in the magazine Mathematical Intelligencer in 2007, vol. 29, number 4 by M. D. Hirschhorn. It is simple to understand and is based on an induction on n

If n=1 then M_{1, 0}=M_{1, 1}

Suppose that it is true for n

Let's use the following observation, seemingly unrelated, to achieve the objective pursued:

x^{n+1}-(n+1)\cdot x + n \geq 0\text{, if }x > 0

The demonstration of this fact is evident by using the identity

x^{n+1}-(n+1)\cdot x +n=(x-1)^2\cdot(x^{n-1}+2\cdot x^{n-2}+\cdots +(n-1)\cdot x+ n)

It can also be tested by induction. If n=1 is inferred from the identity x^2-2\cdot x+1=(x-1)^2

Suppose that it is true for n

(x-1)^2\cdot (x^n+2\cdot x^{n-1}+\cdots + n\cdot x + n +1)= =(x-1)^2\cdot [x\cdot (x^{n-1}+2\cdot x^{n-2}+\cdots + (n-1)\cdot x + n) + n +1]= =x\cdot [(x-1)^2\cdot (x^{n-1}+2\cdot x^{n-2}+\cdots + (n-1)\cdot x + n)] + (x-1)^2\cdot (n +1)= =x\cdot (x^{n+1}-(n+1)\cdot x + n) + (x^2-2\cdot x + 1)\cdot (n +1)=

=x^{n+2}-(n+2)\cdot x + n+1

Now we take \begin{cases}a=\frac{a_1+a_2+\cdots+a_n+a_{n+1}}{n+1} \\ b=\frac{a_1+a_2+\cdots+a_n}{n}\end{cases}, using in the chosen identity x=\frac{a}{b} we will have to \begin{cases}(\frac{a}{b})^{n+1}-(n+1)\cdot\frac{a}{b}+n\geq 0 \\ a^{n+1}\geq ((n+1)\cdot a - n \cdot b)\cdot b^n \end{cases}

Which can be rewritten as

\begin{cases}(\frac{a_1+a_2+\cdots+a_n+a_{n+1}}{n+1})^{n+1}\geq a_{n+1}\cdot (\frac{a_1+a_2+\cdots +a_n}{n})^n \\ (M_{n+1, 1})^{n+1} \geq a_{n+1}\cdot (M_{n,1})^n \end{cases}

since M_{n,0}\geq M_{n,1}

you have to (M_{n+1,1})^{n+1})\leq a_{n+1}\cdot (M_{n,0})^n=a_{n+1}\cdot a_n \cdots a_1

that is equivalent to M_{n+1, 0} \leq M_{n+1,1}

So at the end of the demonstration met the argument of induction

Notes

It is interesting to observe that the equality M_{n,0}=M_{n,1} is true only if and only if a_1=a_2=\cdots=a_n. This fact follows in view of the fact that equality x^{n+1}-(n+1)\cdot x+n=0, for x > 0, is only true if x=1 and has been chosen as the argument of induction suitable

The arithmetic mean and the geometric mean are only two particular cases of a much wider class of means. \forall s \in \mathbb{R}, the mean order s of the positive actual values is defined
a_1,a_2,\cdots,a_n\text{ as }M_{n,s}=(\frac{a_{n}^{s}+\cdots+a_{1}^{s}}{n})^\frac{1}{s}\text{, }s\not=0
M_{n,0} as has already been done. Limit cases can also be considered \begin{cases}M_{n, -\infty}=min\{a_1,a_2\cdots,a_n\} \\ M_{n, +\infty}=max\{a_1,a_2\cdots,a_n\} \end{cases}

Inequality between arithmetic and geometric mean is, in turn, a particular case of a more general chain of inequalities: M_{n, s} \leq M_{n, r}\text{, if }s < r

The average M_{n, -1} is called the harmonic mean and can be inferred from elementary inequality M_{n, 0}\leq M_{n,1} that M_{n, -1}\leq M_{n,0}