New way to introduce exhibitors

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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 .


We proceed by induction.

For  The statement is obvious: .

Let be  for .



Lemma 2 is proved.

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


Lemma 3. If  and , then , .


Indeed, by induction .

Let it be proved that.



Lemma 3 is proved.

Lemma 3 implies

Lemma 4. If  and , then .

Lemmas 3 and 4 imply an important inequality: if then


In particular, ifthen . Note that the inequality  true to all .

Lemma 5. For any natural fair inequality .


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 .


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 .


Because then . If athere is  such that . Therefore, for any  and  right


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 .


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 .


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


Ifthen from  will get


Note that t. To.then


Finally from, , will get


It is clear that


So, it is established that


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


Usingfrom we will receive:


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


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

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