History and definition of integer and fractional parts of a number

During the Middle Ages, there lived one of the greatest English scientists, the Franciscan monk William of Ockham. He was born in Ockham, the English county of Surrey, sometime between 1285 and 1300, and studied and taught at Oxford and then in Paris. Persecuted because of his teachings, Ockham found refuge at the court of Louis.IVBavarian in Munich and, wisely not leaving him, lived there until his death in 1349.

Ockham is considered one of the predecessors of the great thinkers Rene Descartes and Immanuel Kant. According to his philosophical views, reality is the existence of a concrete thing, therefore “it is in vain to do with more what can be done with less.” This statement became the basis of the principle of economy of thinking. William Ockham used it with such devastating force that it later received the now so popular name “Occam’s razor.”

For many people who are not good at mathematics, commonplace became questions like “What else can be discovered in mathematics?” Considering the mathematical preparedness of those asking, we can assume that we are talking only about school-level mathematics. Quite in the spirit of Ockham, we offer questioners, and first of all the students themselves, some tasks that vary the concepts of integer and fractional parts of a number that are well known to them. Using these problems, we will show how important it is to consider not each problem separately, but to combine them into a system, developing a general solution algorithm. This methodological technique dictates to us Ockham’s principle of economy of thinking.

Definition: the integer part of a number x is the largest integer c not exceeding x, i.e. if [x] = c,c ≤ x < c + 1.

For example: = 2;

[-1,5] = -2.

The whole part is indicated real number x with the symbol [x] or E(x).

The symbol [x] was introduced by the German mathematician K. Gauss (1771-1855) in 1808 to denote the integer part of the number x.

The function y = [x] is called the “Antje” function ( fr. entier - integer) and is denoted by E(x). This sign was proposed in 1798 by the French mathematician A. Legendre (1752-1833). Using some values ​​of the function, you can build its graph. It looks like this:

The simplest properties of the function y = [x]:

1. The domain of definition of the function y = [x] is the set of all real numbers R.

2. The range of the function y = [x] is the set of all integers Z.

3. The function y = [x] is piecewise constant.

4. The function y = [x] is non-decreasing, i.e. for any x 1 and x 2 from R such,

that x 1 ≤ x 2 ,the inequality [ x 1 ] ≤ [ x 2 ].

5. For any integer n and any real number x the following equality holds: = [x] + n.

6. If x ─ is a non-integer real number, then the following equality is valid: [-x] = -[x] - 1.

7. For any real number x the following relation is true:

[x] ≤ x< [x] + 1,причём равенство [x] = x достигается тогда и только тогда, когда х ─ integer, i.e. x Z.

The question arises: “If there is a function for the integer part of a number, maybe there is also a function for the fractional part of the number?”

Definition: the fractional part of the number (denoted by (x)) is the difference x - [x].

For example: {3,7} = 0,7

{-2,4} = 0,6.

Let's plot the function y = (x). It looks like this:

The simplest properties of the function y = (x):

1. The domain of definition of the function y = (x) is the set of all real numbers R.

2. The range of values ​​of the function y = (x) is a half-interval and y = (x) will help you complete some tasks.

TASKS:

1) Build function graphs:

A) y = [ X ] + 5;

b) y = (x) - 2;

c) y = |[ x]|.

2) What could the numbers x and y be if:

a) [x + y] = y;

b) [x - y] = x;

c) (x - y) = X;

d) (x + y) = y.

3) What can be said about the magnitude of the difference x - y if:

a) [x] = [y];

b) (x) = (y).

4) Which is greater: [a] or (a)?

2.1. The simplest equations

The simplest equations include equations of the form [x] = a.

Equations of this type are solved by definition:

a ≤ x< а +1 , где а - целое число.

If a is a fractional number, then such an equation will have no roots.

Let's look at an example solution one of these equations:

[X + 1.3] = - 5. By definition, such an equation transforms into inequality:

5 ≤ x + 1.3< - 4. Решим его. Получим -6,3 ≤ х < - 5,3.

This will be the solution to the equation.

Answer: x [-6,3;-5,3).

Let's consider another equation that belongs to the simplest category:

[x+1] + [x-2]-[x+3] = 2

To solve equations of this type, it is necessary to use the property of the integer function: If p is an integer, then the equality is true

[x ± p] = [x] ± p

Proof: x = [x] + (x)

[ [x] + (x) ± p] = [ [x] + (x)] ± p

x = k+ a, where k= [x], a = (x)

[ k + a ± p ] = [ k + a ] ± p= [x] ± p.

Let's solve the proposed equation using the proven property: We get [x] + 1 + [x] - 2 - [x] - 3 = 2. We bring similar terms and get the simplest equation [x] = 6. Its solution is the half-interval x = 1

Let's transform the equation into inequality: 1 ≤ x 2 -5x+6< 2. Двойное неравенство запишем в форме системы неравенств:

x 2 - 5x + 6< 2,

x 2 - 5x + 6 ≥ 1 and solve it;

x 2 - 5x + 4<0,

x 2 - 5x + 5>0

We get x (1;4)

X (-∞;(5 -
)/2]
[(5 +)/2; +∞),

X (1; (5 - )/2]
[(5 +)/2;4).

Answer: x (1; (5 - )/2]
[(5 +)/2;4).

SOLVE THE PROPOSED EQUATIONS BY YOURSELF:

1) = 1

2) = 0,487

3) [ x + 4] – [ x + 1] = 2

4) [x 2] = 4

5) [ x] 2 = 4

6) [ x + 1,3] = - 5

7) [x 2 – x + 4] = 2

8) = - 1

9) = 4,2

10) (x) – [x] + x = 0

11) x + (x) + [x] = 0

12) [4x – 5] = 7

2.2 Solving equations of the form [ f ( x )]= g ( x )

Equation of the form [ f(x)]= g(x) can be solved by reducing them to the equation

[ x] = a.

Let's consider example 1 .

Solve the equation

Let's replace the right side of the equation with a new variableaand let's express from herex

11 a = 16 x + 16, 16 x = 11 a – 16,

Then
=
=

Now let's solve the equation
relative to the variableA .

Let us expand the sign of the integer part by definition and write it using the system of inequalities:

From between
select all integer valuesa: 3;4;5;6;7 and perform the reverse substitution:

Answer:

Example 2.

Solve the equation:

Divide each numerator term in parentheses by the denominator:

AND

From the definition of the integer part of a number it follows that (a+1) must be an integer, which means a is an integer.Numbers a, (a+1), (a+2) - threeconsecutive numbers, which means one of them must be divisibleby 2, and one by 3. Therefore, the product of numbers is divisibleall the way to 6.

That isinteger. Means

Let's solve this equation.

a(a+1)(a+2) - 6(a+1) = 0

(a+1)(a(a+2) - 6) = 0

a + 1 = 0 or a 2 + 2a – 6 = 0

a = -1 D = 28

a= -1 ±
(are not integers).

Answer: -1.

Solve the equation:

2.3. Graphical way to solve equations

Example 1.[x] = 2(x)

Solution. Let's solve this equation graphically. Let's build graphs of the functions y = [x] and y = 2(x). Let's find the abscissas of their intersection points.

Answer: x = 0; x = 1.5.

In some cases, it is more convenient to use a graph to find the ordinates of the points of intersection of the graphs. Then substitute the resulting value into one of the equations and find the desired x values.

Solve the equations graphically:

(x) = 1 – x; 6) [|x|] = x;

(x) + 1 = [x]; 7) [|x|] = x + 4;

3x; 8) [|x|] = 3|x| - 1;

3(x) = x; 9) 2(x) – 1 = [x] + 2;

5) (x) = 5x + 2; 10) How many solutions does

equation 2(x) = 1 - .

2.4. Solving equations by introducing a new variable.

Let's look at the first example:

(X) 2 -8(x)+7 = 0

Replace (x) with a, 0 A< 1, получим простое quadratic equation

A 2 - 8a + 7 = 0, which we solve using the theorem inverse to Vieta’s theorem:The resulting roots are a = 7 and a = 1. Let's carry out the reverse replacement and gettwo new equations: (x) = 7 and (x) = 1. Both of these equations have no roots.Therefore, the equation has no solutions.

Answer: there are no solutions.

Let's consider another case solving the equation by introducing a new

variable:

3[x] 3 + 2[x] 2 + 5[x]-10 = 0

Let's make the replacement [x] = a, az. and we get a new cubic equationBehind 3 +2a 2 +5a-10=0. We find the first root of this equation by selecting:a=1 is the root of the equation. We divide our equation by (a-1). We getquadratic equation 3a 2 + 5a +10=0. This equation has a negativediscriminant, which means it has no solutions. That is, a=1 is the only oneroot of the equation. We carry out the reverse substitution: [x]=a=1. We solve the resulting equation by defining the integer part of a number: x 2 + 8[x]-9 = 0

3(x-[x]) 2 + 2([x]-x)-16 = 0

[X] 4 -14[x] 2 +25 = 0

(2 (x)+1) 3 – (2(x)-1) 3 = 2

(x-[x]) 2 = 4

5[x] 2 -7[x]-6 = 0

6(x) 2 +(x)-1 =0

1/([x]-1) - 1/([x]+1) = 3-[x]

12(x) 3 -25(x) 2 +(x)+2 = 0

10) 10[x] 3 -11[x] 2 -31[x]-10 = 0

2.5. Systems of equations.

Consider the system of equations:

2[ x] + 3[ y] = 8,

3[ x] – [ y] = 1.

It can be solved either by addition or by substitution. Let's focus on the first method.

2[ x] + 3[ y] = 8,

9[ x] – 3[ y] = 3.

After adding the two equations we get 11[x] = 11. Hence

[ x] = 1. Substitute this value into the first equation of the system and get

[ y] = 2.

[ x] = 1 and [ y] = 2 – solutions of the system. That isx= 18-y

18-x-y

3) 3[x] – 2(y) = 6

[x] 2 – 4(y) = 4

4) 3(x) – 4(y) = -6

6(x) – (y) 2 = 3.

3.1. Plotting function graphs of the form y = [ f ( x )]

Let there be a graph of the function y =f(X). To plot the function y = [f(x)], proceed as follows:

Draw straight lines y =n, nn, y =n + 1.

n, y =n+ 1 with the graph of the function y =f(X). These points belong to the graph of the function y = [f( x)], since their ordinates are integers (in the figure these are points A, B, C,D).

Let's plot the function y = [x]. For this

Draw straight lines y =n, n= 0; -1; +1; -2; +2; ... and consider one of the stripes formed by straight lines y =n, y =n + 1.

We mark the points of intersection of the lines y =n, y =n+ 1 with schedule

function y = [x]. These points belong to the graph of the function y = [x],

since their coordinates are integers.

To obtain the remaining points of the graph of the function y = [x] in the indicated strip, we project the part of the graph y = x that falls into the strip parallel to the O axis at to the straight line y =n, y =n+ 1. Since any point M of this part of the graph of the functiony = x, has the following ordinatey 0 , Whatn < y 0 < n+ 1, then [y 0 ] = n

In each other strip where there are points on the graph of the function y = x, the construction is carried out in a similar way.

TASKS FOR INDEPENDENT SOLUTION

Graph the functions:

3.2. Plotting function graphs of the form y = f ([ x ])

Let a graph of some function y = be givenf(X). Plotting a graph of the function y =f([x]) is carried out as follows:

To obtain the remaining points of the function graph y =f([x]) in the indicated band part of the graph of the function y =f(x) falling into this strip is projected parallel to the O axis at to the straight line y =f( n).

In every other strip where there are points on the graph of the function y =f(x), the construction is carried out in a similar way.

Let's consider plotting the function y = . To do this, we will draw a graph of the function y = with a dotted line. Further

numbers.

3. In each other strip, where there are points on the graph of the function y =, the construction is carried out in a similar way.

TASKS FOR INDEPENDENT SOLUTION

Graph the functions:

Let us call the following relations the main inequalities with [x] and (x): [x] > b and (x) > b. A convenient method for solving them is the graphical method. Let us explain it with two examples.

Example 1.[x] ≥ b

Solution. Let us introduce two functions y = [x] and y =band draw their graphs on the same drawing. It is clear that then two cases must be distinguished:b– whole and b– not whole.

Case 1. b– whole

y=b(bZ)

y=b (b Z)

The figure shows that the graphs coincide on [b; b + 1].

Therefore, by solving the inequality [x] ≥b there will be a beam x ≥ b.

Case 2. b– not whole.

In this case, the graphs of the functions y = [x] and y =bdo not intersect. But the part of the graph y = [x] lying above the straight line begins at the point with coordinates ([b] + 1; [ b] + 1). Thus, by solving the inequality [x] ≥b there will be a ray x ≥ [ b] + 1.

Other types of basic inequalities are studied in exactly the same way. The results of these studies are summarized in the table below.

Multiple meanings

[X]≥ b, bZ

x≥ b

[x] ≥b,

[x] >b, b- any

x≥ [b] + 1

[X]≤ b, b- any [x]< b, b- any any

X< [ b] + 1

[X]< b, bZ

X< b

{ X)≥ b, (x) >b, b≥ 1

No solutions

(X)≥ b, (x) >b, b < 0

(-∞; +∞)

(X)≥ b, (X)> b, 0 ≤ b< 1

n+b≤ x< 1+n

n+b< x< 1 + n, nZ

{ X) ≤ b, (X)< b, b≥ 1

(-∞; +∞)

(X) ≤ b, (X)< b, b< 0

No solutions

(X) ≤ b, (X)< b, 0 ≤ b<1

n≤ x≤ b+ n

n< x≤ b+ n, nZ

Let's considerexample solutions to inequality:

Let's replace [x] to the variable a, where a is an integer.

>1 ;
>0;
>0;
>0.

Using the interval method, we finda > -4 [ x] > -4

a< 1/3 [x]< 1/3.

To solve the obtained inequalities, we use the compiled table:

x ≥ -3,

X< 1. x [-3;1)

Answer:[-3;1) .

TASKS FOR INDEPENDENT SOLUTION.

1) [x]< 2

2) [x]≤ 2

3) [x] > 2.3

4) [x] 2

5) [X] 2 -5[x]-6< 0

6) [x] 2 - 7[x] + 6 0

7) 30[x] 2 -121[x] + 80< 0

8) [x] 2 + 3[x]-4 0

9) 3(x) 2 -8(x)-4< 0

10) 110[x] 2 -167[x] + 163 0

11)
> 2

12)
> 1

13)
0

14)
0

Example 1.

Prove that the number
divisible by 5 for any naturaln.

Proof: Letn– an even number, i.e.n=2 m, WheremN, Example 2. , then (years).

Voronova A.N. Inequalities with a variable under the sign of the integer part // Mathematics at school. 2002. No. 2. P.56-59.

Galkin E.V. Non-standard problems in mathematics. Algebra: Textbook. manual for students 7-11 grades. Chelyabinsk: “Vzglyad”, 2004.

Additional chapters on the 10th grade mathematics course for elective classes: A manual for students / Comp. BEHIND. Eunuch. M.: Education, 1979.

Erovenko V.A., O.V. Mikhaskova O.V. Occam's methodological principle using the example of functions of the integer and fractional parts of a number // Mathematics at school. 2003. No. 3. P.58-66.

7. Kirzimov V. Solving equations and inequalities containing an integer and

fractional part of a number // Mathematics. 2002.№30. pp. 26-28.

8. Shreiner A.A. “Tasks of regional mathematical olympiads

Novosibirsk region". Novosibirsk 2000.

9. Directory “Mathematics”, Moscow “AST-PRESS” 1997.

10. Reichmist R.B. “Graphs of functions. Tasks and exercises." Moscow.

“School – press” 1997.

11. Mordkovich A.G., Semenov P.V. and others. “Algebra and the beginnings of analysis. 10

Class. Part 2. Problem book. Profile level" Smolensk

"Mnemosyne" 2007.

Lesson objectives: introduce students to the concept of integer and fractional parts of a number; formulate and prove some properties of the integer part of a number; introduce students to a wide range of uses of the integer and fractional parts of a number; improve the ability to solve equations and systems of equations containing integer and fractional parts of a number.

Equipment: poster “Whoever does and thinks for himself from a young age later becomes more reliable, stronger, smarter” (V. Shukshin).
Projector, magnetic board, algebra reference book.

Lesson plan.

1. Organizing time.
2. Checking homework.
3. Learning new material.
4. Solving problems on the topic.
5. Lesson summary.
6. Homework.

During the classes

I. Organizational moment: lesson topic message; setting the lesson goal; message of the stages of the lesson.

II. Checking homework.

Answer students' questions about homework. Solve problems that caused difficulties when doing homework.

III. Learning new material.

In many algebra problems, you have to consider the largest integer that does not exceed a given number. Such an integer has received a special name “integer part of a number”.

1. Definition.

The integer part of a real number x is the largest integer not exceeding x. The integer part of a number x is denoted by the symbol [x] or E(x) (from the French Entier “antier” ─ “whole”). For example, = 5, [π ] = 3,

From the definition it follows that [x] ≤ x, since the integer part does not exceed x.

On the other hand, because [x] is the largest integer that satisfies the inequality, then [x] +1>x. Thus, [x] is an integer defined by the inequalities [x] ≤ x< [x] +1, а значит 0 ≤ х ─ [x] < 1.

The number α = υ ─ [x] is called the fractional part of the number x and is designated (x). Then we have: 0 ≤ (x)<1 и следовательно, х = [x] + {х}.

2. Some properties of antie.

1. If Z is an integer, then = [x] + Z.

2. For any real numbers x and y: ≥ [x] + [y].

Proof: since x = [x] + (x), 0 ≤ (x)<1 и у = [у] + {у}, 0 ≤ {у}<1, то х+у= [x] + {х} + [у] + {у}= [x] + [у] + α, где α = {х} + {у} и 0 ≤ α <2.

If 0 ≤ α<1. ς о = [x] + [у].

If 1≤ α<2, т.е. α = 1 + α` , где 0 ≤ α` < 1, то х+у = [x] + [у] +1+ α` и

= [x] + [y]+1>[x] + [y].

This property extends to any finite number of terms:

≥ + + + … + .

The ability to find the integer part of a quantity is very important in approximate calculations. In fact, if we know how to find the integer part of the value x, then, taking [x] or [x]+1 as an approximate value of the value x, we will make an error whose value is not greater than one, since

≤ x – [x]< [x] + 1 – [x]=1,
0< [x] + 1– x ≤[x] + 1 – [x] =1.

Moreover, the value of the integer part of the quantity allows you to find its value with an accuracy of 0.5. For this value you can take [x] + 0.5.

The ability to find the whole part of a number allows you to determine this number with any degree of accuracy. Indeed, since

≤ Nx ≤ +1, then

For larger N the error will be small.

IV. Problem solving.

(They are obtained by extracting roots with an accuracy of 0.1 with deficiency and excess). Adding these inequalities, we get

1+0,7+0,5+0,5+0,4 < х < 1+0,8+0,6+0,5+0,5.

Those. 3.1< x <3,4 и, следовательно, [x]=3.

Note that the number 3.25 differs from x by no more than 0.15.

Task 2. Find the smallest natural number m for which

Checking shows that for k = 1 and k = 2 the resulting inequality does not hold for any natural m, and for k = 3 it has a solution m = 1.

This means the required number is 11.

Answer: 11.

Antje in Eqs.

Solving equations with a variable under the “integer part” sign usually comes down to solving inequalities or systems of inequalities.

Task 3. Solve the equation:

Task 4. Solve the equation

By the definition of the integer part, the resulting equation is equivalent to the double inequality

Task 5. Solve the equation

Solution: if two numbers have the same integer part, then their difference in absolute value is less than 1, and therefore the inequality follows from this equation

And therefore, firstly, x≥ 0, and secondly, in the sum in the middle of the resulting double inequality, all terms, starting from the third, are equal to 0, so x < 7 .

Since x is an integer, all that remains is to check the values ​​from 0 to 6. The solutions to the equation are the numbers 0.4 and 5.

c) marking.

VI. Homework.

Additional task (optional).

Someone measured the length and width of a rectangle. He multiplied the whole part of the length by the whole part of the width and got 48; multiplied the whole part of the length by the fractional part of the width and got 3.2; multiplied the fractional part of the length by the whole part of the width and got 1.5. Determine the area of ​​the rectangle.

days (months, years) hours (minutes, seconds)

The type of separator between date elements is determined by the locale settings of the Windows operating system. In the Russian version, for date elements this is usually a dot (if you use the “–“ or “/” icons when entering, they will also be converted to dots after pressing the Enter key); for time elements it is a colon. Days are separated from hours by a space.

The basic unit of time in Excel is one day. Each day has a serial number, starting with 1, which corresponds to January 1, 1900 (the beginning of date counting in Excel). For example, January 1, 2001 stored as the number 36892, since that is how many days have passed since January 1, 1900. The described method of storing dates allows them to be processed in exactly the same way as ordinary numbers, for example, to find a date that is distant from any other date by the desired number of days in the future or past, to find the time interval between two dates, i.e. implement date arithmetic.

Date formats allow you to display them, for example, in one of the usual views: 1.01.98; 1.Jan.98; 1.Jan; January '98 and will be described later. It must be said that if you enter data directly in the form of a date, the appropriate format will be assigned automatically. So, the value entered into the cell 5.10.01 will be correctly perceived by the system as October 5, 2001. When entering dates, only the last two digits of the year are allowed. In this case, they are interpreted as follows depending on the range in which they lie:

00¸29– from 2000 to 2029; 30¸99– from 1930 to 1999

It is permissible not to indicate the year of the date. In this case, it is considered the current year (system year of the computer). So, input like 5.10 will set in the cage October 5 of the current year, for example 2004.

Time is the fractional part of the day. Since there are 24 hours in a day, one hour corresponds to 1/24, 12 hours corresponds to a value of 0.5, etc. Similar to entering a date, you can enter time directly in time format. For example, entering the form 10:15:28 will correspond to 10 hours 15 minutes 28 seconds on January 0, 1900, which in numerical format is equal to 0.420138888888889. Date arithmetic is naturally supported at the time level.

You can ignore seconds and minutes when specifying time. In the latter case, be sure to enter a colon after the hours. For example, if we enter the characters 6: , in the cell we will find 6:00 (i.e. 6 hours 0 minutes). It is possible to combine date and time, separated by a space. Yes, input 7.2.99 6:12:40 corresponds to February 7, 1999, 6 hours 12 minutes 40 seconds.

There is a quick way to enter the current date and time stored on the computer - these are keyboard shortcuts Ctrl+; And Ctrl+Shift+: respectively.

LOGICAL DATA have one of two meanings - TRUE or LIE. They are used as indicators of the presence/absence of any feature or event, and can also be arguments for some functions. In many cases, the numbers 1 or 0 can be used instead of these values, respectively.

ARRAYS are not actually a data type, but only form an organized set of cells or constants of any type. Excel treats an array (possibly containing many cells) as a single element to which mathematical and relational operations can generally be applied. An array can contain not only many cells, but also many constants, for example, the expression (7;-4;9) describes an array of constants of three numeric elements. We will return to the issue of array processing later.

Creating formulas

The power of spreadsheets lies in the ability to put not only data into them, but also formulas.

All formulas must begin with the “=” sign and can include constants, operation signs, functions, cell addresses (for example =5+4/35, =12%*D4, =12*A4-SIN(D3)^2).

The following operators are valid in Excel:

Arithmetic operators(listed in order of priority):

– inversion (multiplying by minus 1), ^ exponentiation,

% is the percentage operation, *, / multiplication, division, +, – addition, subtraction.

Operations are performed from left to right in order of priority, which can be modified by parentheses. Examples of formulas:

formulas in regular notation: cellular formulas:

=7+5^3/(6*8)

=A5/(C7-4)+(4+F4)/(8-D5)*2.4

2 + SinD32 =2+(SIN(D3))^2.

Notes on the % sign.

If you enter a number with a % sign in a cell, its actual value will be 100 times smaller. For example, if 5% is entered, the number 0.05 will be remembered. Thus, the percentage is entered and the coefficient is stored. This action is equivalent to setting the percentage cell format for the number 0.05.

Entering percentages in a formula (that is, in an expression that begins with an equal sign) can be useful for clarity. Let's say you need to get 5% of the number 200. You can write it like this =0.05*200, or you can =5%*200 or =200*5%. In both cases the result will be the same - 10. The percentage sign can also be applied to cells, for example =E4%. The result will be one hundredth of the contents of E4.

Text operator–&. The operator is used to concatenate two strings into one. So, for example, the result of applying the concatenation operator in the formula = “Peter”&” Kuznetsov” will be the phrase “Peter Kuznetsov”.

Relational operators:=, <, >, <=, >=, < >. Operators can be used with both numeric and text data. Their meaning is obvious, except, perhaps, for the signs < > . They mean a relationship of inequality.

Using relation signs, you can build formulas like ="F">"D" and =3>8.

Their result in the first case will be the word TRUE, since the letter F in the alphabet comes after the letter D (the code of the letter F is greater than the code of the letter D). In the second case, for obvious reasons, the word is FALSE.

The use of such formulas in practice seems to be of little use, but this is not so. Let, for example, you need to find out the fact that all the numbers contained in the table in cells A1, A2, A3 and A4 are greater than zero. This can be done using a simple expression of the form (parentheses are required) =(A1>0)*(A2>0)*(A3>0)*(A4>0).

If this is indeed the case, the result of the calculations will be

TRUE*TRUE*TRUE*TRUE=1*1*1*1=1.

Since in arithmetic operations the logical value TRUE is interpreted as 1 and FALSE as 0, here we will get the number 1. Otherwise - 0. Later (inside the IF() function), this circumstance can be correctly processed.

Another example. Find out the fact that only one of A1, A2, A3, A4 is greater than zero. The expression =(A1>0)+(A2>0)+(A3>0)+(A4>0) is useful here.

If, for example, only A2 is greater than zero then = FALSE + TRUE + FALSE + FALSE = 0 + 1 + 0 + 0 = 1.

If all numbers are negative, the result will be 0. If there is more than one positive number, then the result will be greater than 1 (from 2 to 4).

Comment. In Excel, it is possible to compare letters and numbers with each other and it is accepted that a letter is always “greater” than a number. So, for example, the value of a cell containing a space will be greater than any number. If you do not pay attention to this, a hard-to-recognize error may occur because a cell containing a space appears the same as an empty cell whose value is considered to be zero. In addition to operators, Excel has many functions that are the most important computing tool of spreadsheets. These will be discussed in Chapter 4.

Cell references can be entered directly from the keyboard, but can be more reliably and more quickly specified with a mouse, which is used as a pointer. Here, correct input is guaranteed, since the user directly sees (the selected objects are framed by a running dotted line) and selects exactly the data that he wants to include in the expression.

Suppose we need to enter a formula of the form =A2+D4·C1 into cell A1. Here (Fig. 2.4-1) you should perform the following chain of actions:

Similarly, you can include links to blocks in formulas. Let's assume that in A1 you need to enter the following (Fig. 2.4-2) summation function: =SUM(A2:D8;E3). The name of the function is entered in Russian letters, and the addresses of the cells, naturally, in Latin.

The Excel toolbar has special tools that make it easier to enter formulas. They are accessible via icons Function Wizard And Autosummation(for summation).

In view of its great importance, let us now consider the latter. Auto summation is available via button å on the toolbar. With its help, you can very easily implement the summation function, practically without touching the keyboard. Let (line 2 in Fig. 2.4-3) we need to calculate in cell G2 the sum of adjacent cells of area B2:F2. To do this, stand on cell G2 and click on the auto-sum button. Excel itself will enter the name of the function and its arguments into G2, and will also highlight the intended summation area with a running dotted line, so all you have to do is press the Enter button. Excel includes (circles with a running dotted line) in the summation area a continuous section of the table up to the first non-numeric value up or to the left.

Suppose that in G4 you need to summarize the data from the range of cells B4:F4, among which there are (for now) empty ones. Clicking a button å in cell G4 will create a summation function only for cells E4:F4. However, it is easy to correct the situation by immediately selecting the desired summation area B4:F4 with the mouse and pressing Enter. If the cell where the sum is being calculated is not adjacent to the top/left of any cell candidate for summation (line 6 in the figure), the autosum button will only enter the function name. Here you should proceed as before - point the summation object with the mouse (here B6:F6).

Processing arrays. Formulas that use the representation of data as arrays are usually entered into a block in all its cells at once. For example, let’s say in column C (Fig. 2.4-4) you want to get the product of the elements of columns A and B. A typical method is to enter a formula of the form =A1*B1 into C1 and then copy it down. However, you can do it differently. Select area C1:C3 of the future work, enter the formula =A1:A3*B1:B3 and press the keys Ctrl+Shift+Enter. You will find that in all cells of the area C1:C3 the corresponding pairwise products have been obtained, and in the formula bar you will see the same expression for all of them (=A1:A3*B1:B3).

Integer and fractional parts of a real number.
T.S. Karmakova, Associate Professor, Department of Algebra, Kharkiv State Pedagogical University
In various questions of number theory, mathematical analysis, the theory of recursive functions and other questions of mathematics, the concepts of integer and fractional parts of a real number are used.
In the program of schools and classes with in-depth study Mathematics includes questions related to these concepts, but only 34 lines are devoted to their presentation in the algebra textbook for grade 9. Let's take a closer look at this topic.
Definition 1
The integer part of a real number x is the largest integer not exceeding x.
The integer part of a number is denoted by the symbol [x] and is read as follows: “integer part of x” or: “integer part of x”. Sometimes the integer part of a number is denoted by E(x) and is read as follows: “antier x” or “antier from x”. The second name comes from French word entiere - whole.
Example.
Calculate [x] if x takes the values:
1,5; 3; -1.3; -4.
Solution
From the definition of [x] it follows:
= 1, because 1 Z, 1 1.5
[ 3 ] = 3, because 3 Z, 3 3
[-1.3]=-2, because -2 Z, -2 -1.3
[-4] =-4, because -4 Z, -4 -4.
Properties of the integer part of a real number.
1*. [ x ] = x if x Z
2*. [ x ] x * [ x ] + 1
3*. [ x + m ] = [ x ] + m , where m Z
Let's look at examples of using this concept in various tasks.
Example 1
Solve equations:
1.1[ x ] = 3
[ x + 1.3 ] = - 5
[ x + 1 ] + [ x - 2] - = 5
1.4 [x] - 7 [x] + 10 = 0
Solution
1.1 [ x ] = 3. By property 2*, this equation is equivalent to the inequality 3 x * 4
Answer: [ 3 ; 4)
[ x + 1.3 ] = - 5. By property 2*:
- 5 x + 1.3 * - 4 - 6.3 x * - 5.3
Answer: [ -6.3 ; -5.3)
[ x + 1 ] + [ x - 2 ] - [ x + 3 ] = 5. By property 3*:
[ x ] + 1 + [ x ] - 2 - [ x ] - 3 = 5
[ x ] = 9 9 x * 10 (2* each)
Answer: [ 9 ; 10)
1.4 [x] - 7 [x] + 10 = 0 Let [x] = t, then t - 7 t + 10 = 0, i.e.

Answer: [ 2 ; 3) [ 5 ; 6)
Example 2.
Solve inequalities:
2.1[x]2
[ x ] > 2
[ x ] 2
[ x ] [ x ] - 8 [ x ] + 15 0

Solution
2.1 According to the definition of [ x ] and 1*, this inequality is satisfied by x
Answer: [ 2 ;).
2.2 The solution to this inequality: x.
Answer: [ 3 ;).
2.3 x 2.4 x 2.5 Let [ x ] = t, then this inequality is equivalent to the system
3
Answer: [ 3; 6).
2.6 Let [x] = t, then we get.
Answer: (- .
Example 4.
Graph the function y = [ x ]
Solution
1). OOF: x R
2). MZF: y Z

3). Because at x * [ m ; m + 1), where m * Z, [ x ] = m, then y = m, i.e. the graph represents a collection of an infinite number of horizontal segments, from which their right ends are excluded. For example, x * [ -1 ; 0) * [ x ] = -1 * y = - 1 ; x * [ 0; 1) * [ x ] = 0 * y = 0.
Note.
1. We have an example of a function that is given by different analytical expressions in different areas.
2. Circles mark points that do not belong to the graph.
Definition 2.
The fractional part of a real number x is the difference x - [x]. The fractional part of a number x is represented by the symbol (x).
Example.
Calculate ( x ) if x takes the value: 2.37 ; -4 ; 3.14. . .; 5 .
Solution
(2.37) = 0.37, because ( 2.37 ) = 2.37 - [ 2.37 ] = 2.37 - 2 = 0.37.
, because
( 3.14...) = 0.14... , because ( 3.14...) = 3.14...-[ 3.14...] = 3.14...-3= 0.14...
(5) = 0, because ( 5 ) = 5 - [ 5 ] = 5 - 5 = 0.
Properties of the fractional part of a real number.
1*. ( x ) = x - [ x ]

2*. 0 ( x ) 3*. (x + m) = (x), where m * Z
4*. ( x ) = x if x * [ 0 ; 1)
5* If ( x ) = a, a * [ 0 ; 1), then x =a +m, where m * Z
6*. (x) = 0 if x * Z.
Let's look at examples of using the concept ( x ) in various exercises.

Example 1.
Solve equations:
1.1(x) = 0.1
1.2(x) = -0.7
(x) = 2.5
( x + 3 ) = 3.2
(x) - (x) +
Solution
For 5* the solution will be many
x = 0.1 + m, m * Z
1.2 By 2* the equation has no roots, x * *
1.3 By 2* the equation has no roots, x * *
By 3* the equation is equivalent to the equation
( x )+ 3 = 3.2 * ( x ) = 0.2 * x = 0.2 + m , m * Z
1.5 An equation is equivalent to a set of two equations
Answer: x =
x =
Example 2.
Solve inequalities:
2.1(x)0.4
2.2(x)0
(x+4)
( x ) -0.7 ( x ) + 0.2 > 0
Solution
2.1 By 5*: 0.4 + m x 2.2 By 1*: x * R
By 3*: (x) + 4 By 5*: m 2.4 Since (x) 0, then (x) - 1 > 0, therefore, we get 2 (x) + 1 2.5 Solve the corresponding quadratic equation:
( x ) - 0.7 ( x ) + 0.2 = 0 * This inequality is equivalent to the combination of two inequalities:
Answer: (0.5 + m; 1 + m) (k; 0.2 + k),
m*Z,k*Z
Example 3.
Graph the function y = ( x )
Construction.
1). OOF: x * R
2). MZF: y * [ 0 ; 1)
3). The function y = (x) is periodic and its period
T = m, m * Z, because if x * R, then (x+m) * R
and (x-m) * R, where m * Z and by 3* ( x + m ) =
(x - m) = (x).
The smallest positive period is 1, because if m > 0, then m = 1, 2, 3, . . . and the smallest positive value is m = 1.
4). Since y = ( x ) is a periodic function with period 1, it is enough to plot its graph on some interval of length 1, for example, on the interval [ 0 ; 1), then on the intervals obtained by shifting the selected one by m, m * Z, the graph will be the same.
A). Let x * [ 0 ; 1), then (x) = x and y = x. We obtain that on the interval [ 0 ; 1) the graph of this function represents the bisector segment of the first coordinate angle, from which the right end is excluded.

B). Using periodicity, we obtain an infinite number of segments forming an angle of 45* with the Ox axis, from which the right end is excluded.
Note.
Circles mark points that do not belong to the graph.
Example 4.
Solve Equation 17 [ x ] = 95 ( x )
Solution
Because ( x ) * [ 0 ; 1), then 95 ( x )* [ 0 ; 95), and, consequently, 17 [ x ]* [ 0 ; 95). From the relation
17 [ x ]* [ 0 ; 95) follows [ x ]* , i.e. [x] can be 0, 1, 2, 3, 4, and 5.
From this equation it follows that ( x ) = , i.e. taking into account the resulting set of values ​​for
[ x ] we conclude: ( x ), accordingly, can be equal to 0;
Since we need to find x, and x = [ x ] + ( x ), we find that x can be equal to
0 ;
Answer:
Note.
A similar equation was proposed in the 1st round of the regional mathematical Olympiad for tenth graders in 1996.
Example 5.
Graph the function y = [ ( x ) ].
Solution
OOF: x * R, because ( x )* [ 0 ; 1) , and the integer part of the numbers from the interval [ 0 ; 1) is equal to zero, then this function is equivalent to y = 0
y
0 x

Example 6.
Construct a set of points on the coordinate plane that satisfy the equation ( x ) =
Solution
Since this equation is equivalent to the equation x = , m * Z by 5*, then on the coordinate plane one should construct a set of vertical lines x = + m, m * Z
y

0 x
Bibliography
Algebra for 9th grade: Textbook. manual for students of schools and advanced classes. studying mathematics /N. Y. Vilenkin et al., ed. N. Ya. Vilenkina. - M. Education, 1995.
V. N. Berezin, I. L. Nikolskaya, L. Yu. Berezina Collection of problems for elective and extracurricular classes in mathematics - M. 1985
A. P. Karp I give mathematics lessons - M., 1982
Magazine “Kvant”, 1976, No. 5
Magazine “Mathematics at School”: 1973 No. 1, No. 3; 1981 No. 1; 1982 No. 2; 1983 No. 1; 1984 No. 1; 1985 No. 3.