American Journal of Applied Science and Technology
73
https://theusajournals.com/index.php/ajast
VOLUME
Vol.05 Issue 07 2025
PAGE NO.
73-75
10.37547/ajast/Volume05Issue07-12
Ant'ye, Mantissa and Its Applications (Continued)
Pulatov Bahram Nigmatovich
Associate Professor, Oriental University,Candidate of Physical and Mathematical Sciences, Uzbekistan
Jumaniyozov Kudrat Sapoevich
Associate Professor, Candidate of Pedagogical Sciences, Oriental University, Uzbekistan
Received:
31 May 2025;
Accepted:
29 June 2025;
Published:
31 July 2025
Abstract:
This article provides theoretical and practical information as well as proofs about the antyle and mantissa,
and highlights the main points of it. It also suggests methods for solving several complex and Olympic-type problems
with the help of the antyle and mantissa.
Keywords:
Ant’e, mantissa, sequences, functions in R set, definition of integral, canonical decomposition,
bounded functions.
Introduction:
Ant'e(French. Entier) The term integer was proposed
by Legendre in 1798 and has been in use for two
hundred years. The new concept came to his mind
from the need to count how many times the prime
number p occurs in the canonical distribution of a
complex number n.
The designation of the ant’e x as [ x ] was proposed by
Gauss in 1808.
There is almost no significant historical information in
the literature about the term mantissa.
In the old days, ant'e and the mantissa Problems on
(whole and fractional parts of a number) were
considered very narrow Olympiad problems.
Recently, while easy problems of this type have been
offered in entrance exams to universities in some
foreign countries, problems with a higher level of
difficulty have begun to be offered to students in
national and international Olympiads.
So what's so interesting about the ant’e and mantissa
issues?
If we consider the ant’e and mantissa as functions,
these functions are not continuous, but rather,
piecewise continuous. The existence of such
functions
is
fundamentally
different
from
conventional solving methods.
The uniqueness of the topic "Ant’e and Mantissa" and
the complexity, or rather, the difficulty of the
problems on this topic, is that the student does not
have enough practice in this area. The solution to
most of these types of problems is based on a very
short logical analysis.
In addition, the ant’e and mantissa have properties
that require the student to perform a number of
operations. Despite these characteristics, many of the
properties can be derived while solving problems.
The article presents selected problems on ant’e and
mantissa typical of the number theory and algebra
section of the subject, and also includes some
problems recommended in national Olympiads in
foreign countries. For those interested, the article
provides complete and substantiated solutions to the
problems.
1.
[𝒇(𝒙)] = 𝒈(𝒙)
Methods for solving equations in
the form of.
The equations in question are often found in
competitive exams. Equations of this form are also
given other equations, for example, equations
involving the mantissa.
[𝒇(𝒙)] = 𝒈(𝒙)
(1.1)
The uniqueness of equations of the form is that the
American Journal of Applied Science and Technology
74
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American Journal of Applied Science and Technology (ISSN: 2771-2745)
right-hand side of the equality is an integer. From this
it follows that the equation is satisfied only by values
of the variable that take on integer values of the
function. In addition, if the value domains of the
functions are finite, equation (1.1) can be
decomposed
into
several
simpler
equations.
g(𝑥)𝑓(𝑥) 𝑣𝑎 𝑔(𝑥)
1.1.
a method of applying a function that is the
inverse of a function.
𝑔(𝑥)
Since equation (1.1) is satisfied by the values of the
variable that make the function complete,
g(x)
𝑛 = 𝑔(𝑥) ,
𝑛 ∈ 𝑍
Let us introduce the definition. Let us assume that is
a strictly monotonic (increasing) function. Then the
function is invertible and we can express it
by:
𝑔(𝑥)𝑔(𝑥)𝑥 𝑛
𝑥 = 𝑔
−1
(𝑛).
As a result, by defining the variable twice, equation
(1.1) is reduced to an equation with all variables.
[𝑓(𝑔
−1
(𝑛))] = 𝑛, [𝑓(𝑔
−1
(𝑛)) − 𝑛] = 0
(1.2)
The last equation is equivalent to the following
double inequality
0 ≤ 𝑓(𝑔
−1
(𝑛)) − 𝑛 < 1 ,
(1.3)
We can solve it by first finding the number of
solutions of (1.1) and then finding the roots of the
given equation by inverse substitution.
𝑛
The scope of application of the method considered
above to equations of the form (1.1) is somewhat
limited by the invertibility of the function and the
complexity of the inverse substitution of variables
compared to other methods.
𝑔(𝑥)
Now let's apply this method to some problems of the
form (1.1)
11
. Solve the equation.
[
𝑥+1994
81
] =
𝑥+2011
101
Solution. We can introduce the notation . Then
𝑛 =
𝑥+2011
101
𝑥 = 101𝑛 − 2011
Now we can write the double inequality
0 ≤
101𝑛 − 17
81
− 𝑛 < 1.
By solving the left side of the double inequality (which
only accepts integer values), we obtain, and by
solving the right side, we obtain. This means that the
equations have four solutions.
𝑛 ≥ 1𝑛 𝑛 ≤ 4𝑛 =
1,2,3,4
Answer: -1910, -1809, -1708, -1607
12.
Solve the equation.
[𝑥] =
√2
𝑥
Solution. We can introduce the notation . Then
𝑛 =
√2
𝑥
𝑥 𝑥 =
√2
𝑛
𝑛
We solve the double inequality with respect to the
auxiliary integer by expressing it through
𝑛
𝑛 ≤
√2
𝑛
< 𝑛 + 1.
First, let's consider the case where the values are
positive. In this case, the inequality becomes
𝑛
𝑛
2
≤ √2 < 𝑛
2
+ 𝑛.
It seems obvious that this could be the whole
solution.
𝑛 = 1
Now the inequality in the negative case is this
𝑛
𝑛
2
+ 𝑛 < √2 ≤ 𝑛
2
It turns out that this inequality, in turn, has no
solution in s: if the "right" part of the inequality in s
does not hold, then the inequality on the left side of
the inequality in s does not hold.
𝑛𝜖𝑍 < 0 𝑛 =
−1 𝑛 ≤ −2
So, the equation given in has a unique solution.
𝑛 =
1
Answer:
√2
13
. How many solutions does the equation have?
𝑥 =
{32𝑥}
Instructions: Write the equation given initially in the
form.
[32𝑥] = 31𝑥
After the definition, you can write an inequality. This
means that the given equation has 31 solutions.
𝑛 =
31𝑥0 ≤ 𝑛 < 31
Answer: 31 solutions.
1.2. Equipotent substitution method.
𝟎 ≤ 𝒇(𝒙) −
𝒈(𝒙) < 𝟏
𝑓(𝑥)
The main property of the mantissa of a
function is this
0 ≤ {𝑓(𝑥)} < 1, yoki 0 ≤ 𝑓(𝑥) − [𝑓(𝑥)] < 1
consists of relationships.
Now we can write these equally strong substitutions
for the equation
[𝑓(𝑥)] = 𝑔(𝑥)
[𝑓(𝑥)] = 𝑔(𝑥) ⇔
{
0 ≤ 𝑓(𝑥) − 𝑔(𝑥) < 1
𝑔(𝑥) 𝜖 𝑍
(1.4)
The solution of inequality (1.4) can consist of one or
more intervals, and in this case the solutions of
equation (1.1) belong to the set of solutions of
inequality (1.4). Therefore, from the above
considerations it follows that inequality (1.4) is a
consequence of equation (1.1). It also follows that if
inequality (1.4) has no solution, equation (1.1) also
has no solution.
American Journal of Applied Science and Technology
75
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American Journal of Applied Science and Technology (ISSN: 2771-2745)
Now we will solve problems using the equivalent
substitution method.
14
. (All-Russian./2010-2011) Solve the equation
𝑥
2
−
[𝑥] − 2 = 0.
Solution.
The given equation consists of an equation
of the form. We can reduce equation (1`.4) to the
form
[𝑓(𝑥)] = 𝑔(𝑥)
0 ≤ 𝑥 − 𝑥
2
+ 2 < 1.
The solution to the inequality on the left is a segment
(we will not consider the solutions that arise from the
inequality on the right, since it is a "not very
convenient discriminant"). Substituting the four
values obtained from this into the given equation, we
obtain the following.
[−1,2]𝑔[𝑥] 𝜖 {−1,0,1,2}.
𝑥
1
= −1 ( [𝑥] = −1 bo
′
lganda), 𝑥
2
= √3 ( [𝑥] = 1 bo
′
lganda ),
𝑥
3
= 2 ( [𝑥] = 2 bo
′
lganda ),,
Answer:
{−1, √3 , 2}.
15.
Solve the equation.
[2𝑥] = 𝑥
2
Solution. As a result of an equally strong substitution,
this
0 ≤ 2𝑥 − 𝑥
2
< 1,
we form the inequality. The solution to this inequality
is
𝑥 𝜖 [ 0 , 1) ∪ ( 1 , 2 ].
Now it is not difficult to choose from the generated
half-intervals those that make the value of integer.
𝑥
2
Answer: .
{ 0 , √2 , √3 , 2 }
1.3
.
Method
of
bounded
functions.
𝒇(𝒙) 𝐯𝐚 𝒈(𝒙)
If the functions are bounded, the right (left) sides of
equation (1.1) take on a finite number of integer
values. In this case
𝑓(𝑥) va 𝑔(𝑥)
Equation (1. 1) is reduced to the set of these systems
of equations
{
[𝑓(𝑥)] = 𝑛
1
,
𝑔(𝑥) = 𝑛
1
,
{
[𝑓(𝑥)] = 𝑛
2
,
𝑔(𝑥) = 𝑛
2
,
… {
[𝑓(𝑥)] = 𝑛
𝑘
,
𝑔(𝑥) = 𝑛
𝑘
,
( 1.5 )
In this case, the range of values of the function is
either the integers in the range of values or this
equality
consists
of
𝑛
1 ,
𝑛
2 ,
… , 𝑛
𝑘
lar [𝑓(𝑥)]𝑔(𝑥) funksiyaning
{𝑛
1 ,
𝑛
2 ,
… , 𝑛
𝑘
} = 𝐸 ([𝑓(𝑥)]) ∩ 𝐸(𝑔(𝑥))
∩ 𝑍 ( 1.6 )
In other words, boundedness means that one
function is bounded from above and the other from
below. The analysis of this case also leads to the set
(1.5).
𝑓(𝑥) va 𝑔(𝑥)
It is worth noting that the set (1.5) is an equally strong
permutation, only
(1.6) is satisfied. However, it is not always practical to
construct the set. In this case, the set (1.5) consists of
the result of equation (3.1).
{𝑛
1 ,
𝑛
2 ,
… , 𝑛
𝑘
}
16
. Solve the equation
[𝑙𝑜𝑔
2
𝑥 ] = 𝑐𝑜𝑠𝑥.
Solution. We use the fact that the function is bounded
and takes integer values of . We write down the set
(1.5)
𝑐𝑜𝑠𝑥 − 1, 0 va 1
{
[𝑙𝑜𝑔
2
𝑥 ] = −1,
𝑐𝑜𝑠𝑥 = −1,
{
[𝑙𝑜𝑔
2
𝑥 ] = 0,
𝑐𝑜𝑠𝑥 = 0,
… {
[𝑙𝑜𝑔
2
𝑥 ] = 1,
𝑐𝑜𝑠𝑥 = 1.
Let's look at the second system. This system gives the
only solution to the given equation. The solution to
the equation consists of a half-interval, and the
solution to the equation is in this, that is, we paid
attention to the domain of definition of the
logarithmic function. We can verify that
[𝑙𝑜𝑔
2
𝑥 ] =
0 [ 1 , 2 )𝑐𝑜𝑠𝑥 = 0 𝑥 =
𝜋
2
+ 2𝜋𝑛 , 𝑛 𝜖 𝑍 , 𝑥 >
0 [𝑙𝑜𝑔
2
𝜋
2
] = 0
The remaining two systems have no solution.
Answer:
{
𝜋
2
}
2. Examples for independent work.
2.1. Solve the equation
𝑥
2
− 3𝑥 = [3 − 2𝑥]
2.2. Solve the equation
𝑥
2
= [2𝑥]
2.3. Solve the equation
[3√𝑥 − 𝑥
2
] = 2
𝑥−
1
2
2.4. Solve the equation
[𝑥
6
− 1] = −
9(𝑥−1)
2
4
2.5. Solve the equation
[3𝑥] = 𝑙𝑜𝑔
0,5
𝑥
2.6. Solve the equation
[2 − 𝑥] = 2𝑠𝑖𝑛𝑥
2.7. Solve the equation
[2𝑥] = 𝑥
3
− 3 .
REFERENCES
ILSemyenov "Ant'ye i mantissa" Sbornil zadach
c solution". IPM im. MB Keldisha 2015g.
MA Mirzaahmedov, DA Sotiboldiev “Preparing
students for mathematical olympiads”. Tashkent,
“Teacher” 1993.
Agakhanov NX, Bagdanov II, Kajevnikov PA
Mathematics. Regional Olympiad. Class 8-11.-
Prosvesheniye, 2010.
Add T. , Andrica D. Problems for Mathematical
Contests. - GIL Publishing House, 2003.
Internet resources:
Olymp.msu.ru - the official portal of the Olympic
Games "Lomonosov".
Rroblems.ru
