# 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. Ifthen .

### 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, ifthen . 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 thatthere 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 functionby putting  and . For any non-empty subsets,  sets  real numbers set where .

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

### Evidence

Because then . If athere 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 thenthat 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 letwhere , . 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. Ifthen .

### 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, sincethen 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)

Ifthen 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)

Usingfrom  we will receive:

(11)

T. K., then from  follows that i.e.  increases by . Furtherso 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