# New way to introduce exhibitors

- Tutorial

*The article proposes a new, very unusual way to define an exponent and, based on this definition, its main properties are derived.*

Every positive number $$ set in correspondence $$where $$ and $$.

Lemma 1. Of $$ it follows that for each element $$ there is an item $$ such that $$.

Will write $$, if a $$ upper bound of the set $$. Similarly, we will write$$, if a $$ - lower bound of the set $$.

Lemma 2.If$$then $$.

### Evidence

We proceed by induction.

For $$ The statement is obvious: $$.

Let be $$ for $$.

Then$$

$$.

Lemma 2 is proved.

In the following, we show that each set $$limited. From Lemma 2 it follows that

$$

**(one)**

Lemma 3.If $$ and $$, $$then $$, $$.

### Evidence

Indeed, by induction $$.

Let it be proved that$$.

Then $$

$$

$$

$$.

Lemma 3 is proved.

Lemma 3 implies

Lemma 4.If $$ and $$, $$then $$.

Lemmas 3 and 4 imply an important inequality: if $$then

$$

**(2)**

In particular, if$$then $$. Note that the inequality $$ true to all $$.

Lemma 5.For any natural$$ fair inequality $$.

### Evidence

Let be $$, $$.

Evaluate the work$$. From Lemma 2 it follows that

$$ for $$.

therefore $$.

Because $$, then applying Lemma 4, we obtain $$i.e. $$.

Thus, Lemma 5 is proved.

Lemma 6.Let $$ two non-empty bounded subsets of the set of real numbers $$. If for any $$ there is an item $$ such that $$then $$.

### Evidence

It's clear that $$. Assuming that$$there is $$ such that $$. So for any $$ true inequality $$. But in $$ there is an item $$. So each $$ less of this $$, which contradicts the condition of the lemma, and the proof is complete.

## Function definition $$ (exhibitors)

We see (see Lemma 1 and Lemma 5) that for any $$ lots of $$limited. This allows you to define a function$$by putting $$ and $$. For any non-empty subsets$$, $$ sets $$ real numbers set $$where $$.

Lemma 7.If$$, $$ non-empty bounded subsets $$then $$.

### Evidence

Because $$then $$. If a$$there is $$ such that $$. Therefore, for any $$ and $$ right

$$

**(3)**

Select the sequence $$ elements of the set $$converging to $$ and sequence $$ elements of the set $$converging to $$. But then$$that contradicts $$.

Lemma 7 is proved.

Lemma 8.For$$ fair equality $$.

### Evidence

Consider the sets $$, $$ and $$. Turning on $$obviously. Prove that for any $$ there are $$ and $$ such that $$. Indeed let$$where $$, $$. Consider sets of positive numbers.$$.

It's clear that $$, $$.

Set $$, $$.

It's clear that$$, $$ and $$

$$,

That zavernaet proof of Lemma 8.

So$$. But Lemma 7 implies that$$.

We have built a valid function.$$defined on a set of positive numbers such that $$. We finish it on the whole numerical line, putting $$ and $$ for any negative number $$.

So function$$ defined on the whole number line.

Lemma 9.If$$then $$.

### Evidence

If one of the numbers $$, $$, $$ equally $$then for them the statement of the lemma is true.

For the case when $$the lemma follows from Lemma 8.

Further, if the lemma is true for numbers$$, $$, $$then it is true for numbers $$, $$, $$. Indeed, since$$then $$i.e. $$. So, it suffices to prove the lemma for the case$$. But then either $$, $$either $$, $$either $$, $$. Happening$$, $$already disassembled. For definiteness, we set$$, $$. So,$$, Consequently $$where $$, $$ and $$. So$$ or $$i.e. $$.

Lemma 9 is proved.

## About function $$

We built a function $$defined on a set of real numbers such that for any $$ right:

$$, $$

**(4)**

For$$ of $$ follows

$$

**(5)**

If$$then from $$ will get

$$

**(6)**

Note that t. To.$$then

$$

**(7)**

Finally from$$, $$, $$ will get

$$

**(8)**

It is clear that

$$.

So, it is established that

$$

**(9)**

Estimate the value of$$. Putting in inequality$$ $$get for $$, $$ such that $$ and $$:

$$

**(10)**

Using$$from $$ we will receive:

$$

**(11)**

T. K.$$, $$then from $$ follows that $$i.e. $$ increases by $$. Further$$so for $$ will get

$$

**(12)**

Of$$ it follows that on the set $$ function $$uniformly continuous. So$$ continuous everywhere on $$.

Now we estimate the value of the derivative of the function$$ at an arbitrary point $$.

Let be $$ and $$, $$ at $$. Then

$$,

T. E.$$.

Because $$ at $$and $$ at $$then

$$.

It means that$$ differentiable everywhere $$ and $$.

**Slobodnik Semen Grigorievich**,

content developer for the application “Tutor: Mathematics” (see the article on Habré ), Candidate of Physical and Mathematical Sciences, math teacher of school 179, Moscow