Category Archives: Real numbers

The real numbers have an order relationship and fulfill the axiom of completeness: every set bounded superiorly possesses supreme

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

Numbers, Real numbers, Infinitesimal calculus

Irrational numbers

Irrational numbers

Irrational numbers are numbers that cannot be expressed, such as a fraction \frac{m}{n}, where m, n \in\mathbb{Z} and n\not = 0

This property is fulfilled by the real numbers that are not rational

An infinite decimal place (i.e. with infinite figures) aperiodic, as \sqrt{7} = 2,645751311064591 can not be represented as a rational number

Such numbers are called irrational numbers

This designation means the impossibility of represent that number as a ratio of two whole numbers

Are denoted as \mathbb{I}

Irrational numbers most well-known

  • Pi number:
    reason between the length of a circumference and its diameter, \pi\approx 3,14159\cdots
  • Euler Number:
    e=\lim\limits_{n\to\infty}\left(1+{\frac {1}{n}}\right)^{n}\approx 2,7182\cdots
  • Aureum number:
    \Phi={\frac {1+{\sqrt {5}}}{2}}\approx 1,6180\cdots

Demonstration: Root of 2 is irrational

The value that verifies d^2=2 it is denoted by \sqrt{2} and it is a real number

Consider the set: S=\{x\in\mathbb{R}|x\geq 0, x^2 \leq 2\}

Set S is non empty (1 \in S) and is superiorly bound, as x\in S, x^2\leq 2 < 4 = 2^2, then x < 2

As a non empty set dimensioned above, we will have for the axiom of completeness, which S possesses supreme, that supreme we will denote by v. It cannot happen that v^2 > 2 or v^2 < 2 and, therefore, you have to v^2 = 2; that is, it will be the value we have denoted as \sqrt{2}

Suppose that v^2 > 2, then taking h=min\{v,\frac{(v^2 - 2)}{2\cdot v}\} it would have h > 0, v-h \geq 0 and (v-h)^2=v^2+2\cdot h \cdot v + h^2 \leq v^2 +2\cdot h \cdot v +h \cdot v = v^2 + 3 \cdot h \cdot v \leq v^2 + (2-v^2)=2, that is, v+h\in S

But this is not possible, because v+h > v and change \forall x \in S you have to x \leq v. So, \sqrt{2} it is an irrational number

Gauss's motto

Gauss's motto was published in article 42 of the Disquisitiones Arithmeticae (1801) and says:

Each real root of a polynomial monico with integral coefficients is integer or irrational

Demonstration of Gauss's motto

Be r a real root of the monican polynomial: P(x)=x^n+c_{n-1}\cdot x^{n-1}+\cdots + c_0 where n is a positive integer, and c_0,\cdots,c_{n-1} are integers and suppose that r is rational but not integer

Then there exists a unique integer q such that q < r < q+1. Since r is rational, so will r^2, \cdots, r^{n-1}

Therefore, the set: M=\{m>0| m, m\cdot r, m\cdot r^2, \cdots, m\cdot r^{n-1}\text{ they are integers}\} possesses an element and is not empty

Considering that r is the root of P(x), the identity will be fulfilled: r^n=-(c_{n-1}\cdot r^{n-1}+\cdots +c_0) and in addition \forall m \in M, m(c_{n-1}\cdot r^{n-1}+\cdots +c_0) is an integer, and therefore, m\cdot r^n it is also an integer

Let's look for a contradiction with the principle of good sorting of natural numbers, thus proving that set M has no minimum element; I mean: \forall m \in M, \exists m'\in M \Rightarrow 0 < m' < m

We consider m\in M and we take m'=(r-q)\cdot m. So we have to for every i=0,1,2,\cdots, n-1 we have to m'\cdot r^i=m\cdot r^{i+1}-q\cdot m^i, then it's an integer and it's true that m'\in M and 0 < m' < m because 0 < r-q < 1

Then M can not have an item minimum, and the root r must be integer or irrational

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}