New way to introduce exhibitors

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

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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 $$.

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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