Kvant 1970, No. 10 · Pages

We will never be able to reach absolute zero. The physical, school, definite definition of absolute zero says that this is the temperature at which the thermal motion of matter stops. But this definition is imprecise. From a modern point of view, based on quantum theory, the existence of movements at absolute nULE is allowed. The energy of this movement is completely defined and is the minimum molecular movement that can exist in a given substance. Let me give you a simple example. If a substance is heated strongly, then the electrons of the atoms, which move around the atomic nucleus in certain orbits, will, under the influence of temperature movements, be torn off, separated, and so-called dissociation will occur. When a substance is cooled, the movement of atoms slows down, the electrons begin to circulate again in their orbits and maintain their movement until absolute zero. But, in addition to the movement of electrons along the orbits of each atom separately, there is also a goal of a series of combined movements in a solid body, which from a modern point of view should be maintained to the lowest temperatures. Thanks to this so-called degeneracy of motion in this temperature range, completely new phenomena can appear that we cannot observe at ordinary temperatures.

One of these interesting phenomena, which has already become widely known, was discovered by Kamerlingh Onnes - the phenomenon of superconductivity. It lies in the fact that at very low temperatures, electric current is able to flow through some conductors without resistance, without forming a body. Experience shows that if a current is excited inductively in a closed superconductor, it flows without emitting a body or decreasing, only for as long as the experimenter was able to observe it. Another of these phenomena

which can only be detected at very low temperatures is the superfluidity we discovered 5 years ago in liquid helium.

Studies of this and other phenomena were carried out using liquid helium itself as a refrigerant. Liquid helium is the only known substance that, even at the lowest temperatures down to thousandths of a degree from absolute zero at normal pressure, remains liquid and does not turn into a solid state. It can only be turned into a solid under pressure starting at 25 amps.

Liquid helium itself is an extremely interesting object to study.

Helium is expected at a temperature of 4.8°K and forms a light, weighing 7-8 times less than water, and transparent liquid. Due to its low thermal capacity, during the experiment, liquid helium should be held in a vacuum Dewar flask surrounded by another similar flask with liquid air. Experimenting with liquid helium presents significant technical difficulties. This explains the fact that until now only a few cold laboratories in all countries produce liquid helium in sufficient quantities.

If the concept of the temperature of liquid helium is from the point of its transition (4.8°K), then when we reach a temperature of 2.19°K, it will undergo changes, and it is customary to say that helium I turns into helium II. This temperature is called the λ-point. Being in its original state, liquid helium usually boils continuously due to the small access of the body, which is difficult to avoid even in the presence of a heat ash. Below the λ-point 肢 the circle stops boiling, its surface becomes smooth; this is due to

5

viscosity *). All that remained was to test this by experiment.

Observing low viscosity, especially at low temperatures, turned out to be a difficult experimental task. It was necessary to find a special method for measuring it. When the necessary method was found and developed, the observation itself did not take much time and showed that the viscosity of the liquid gel was indeed vanishingly small. The liquid gel turned out to be over a billion times more fluid than water. Such a flowing medium is very difficult to imagine, and yet the given number indicates the limit not of viscosity, but only of the sensitivity of our measurements. We do not yet have a more sensitive method. Therefore, I assumed that there was every reason to believe that the liquid gel had no viscosity; I call it supercurrent. At first it was met with great opportunity. They looked for experimental errors in the methodology, measurements, etc. in my experiments. The discovery of superfluidity in liquid gels has thus been discussed on all sides, and now, I think, the existence of a supercurrent state in gels II can be considered recognized.

When this phenomenon was first formed, it seemed to us that the superfluidity of gel II was quite sufficient to explain the large thermal conductivity observed in liquid gel in accordance with the picture of the existence of convection currents that I just sketched for you. But the matter turned out to be much more interesting and complicated than we initially thought.

The story of how our views on this issue further developed presents some difficulties. I tried to tell you what contradictions we came into contact with, how our views changed and how we gradually developed ideas that looked like

*) That is, with his theory there will be no energy consumption during training. (Editor’s note).

They would be incongruous fiction if they were presented without connection with real experiments.

If we take the point of view of our ordinary mechanical concepts, which completely comprehensively describe the behavior of ordinary substances under ordinary conditions, then it turns out that supercurrent gels, as experience shows, cannot transfer heat as intensely as a change in convection currents requires. We are coping with the difficulty of finding a mechanism that could cause the necessary rapid flow of the gel by convection. In the general mechanism of body transfer by convection, we owe the movement of the medium to the fact that a more heated liquid or gas becomes somewhat less dense, so it tends upward, as if floating up in a denser medium, while colder, more dense ones tend downwards, “taunt”. Mixing occurs, and it is very important that the reason causing the movement is the force of gravity. But calculations show that this force in gels II is not enough to cause such a large thermal conductivity, which was observed in the experiment. This made the phenomenon again incomprehensible. It was necessary to look for some other, new mechanisms to explain it. A number of wholesalers finally managed to come across a completely new mechanism for the movement of liquid gel II.

It turned out that, under the influence of the temperature, very strong currents arose in liquid gel II, somewhat filling the convection currents. Under the influence of the disposition, the liquid begins to move, but this movement is of a special kind, specific to liquid gel II, unknown in any other liquid and under no other conditions.

Before trying to explain the existence of this movement, let’s get acquainted with its features. Let’s see how it looks in an experiment. I will not describe in detail the technical details of this experiment.

10

the possible thickness of this layer so that the flow rate of the gel in it does not take on an excessively large value. Further, as stated, I tried in my experiments to discover the thickness of this layer experimentally. To do this I applied the gel in a very thin layer. Gradually I reached a gel layer thickness of 0.00014 mm, but experience showed that the nature of all phenomena was preserved. Thus, the explanation of adaptation moved, and this led to the development of new views on the nature of hydrodynamic phenomena in Gelyan II. The first sketches of these ideas were expressed by Tissa, but their scientific development, the provision of theoretical justification for them and the creation of a hydrodynamic theory of the phenomenon belong to our scientist L. Landd.

I will try to give the most general picture of these views. According to this theory, the countercoccus, which I tried to explain the flow of the gel in one energy state along the wall of another inside the bigule, is replaced by the countercoccus of the gel, originating in itself.

The explanation of this phenomenon given by L. D. Landd is as follows.

Liquid gel is a mixture of two liquids. These two components of the liquid gel are in two different quantum states. Thanks to this, he showed that simultaneous counter flows of the same liquid can exist, which we observe in the mountain of the vessel at figure 1.

If this theoretical position were not so fully supported by experimental evidence, it would sound like an idea that is very difficult to recognize as reasonable.

Landdu’s theory well describes the physical existence of those two states in which a gel can simultaneously exist at temperatures below the l-point. As I already said, if the gel continues to cool after compression, then it will be in a state of general liquid up to 2.19 ° K, that is, the λ-point. Then, according to Landdu’s theory, a gel appears in this liquid as an impurity in a new state. This new state is characterized by the absence of adulthood. This gel represents liquid gel II in the state in which it was released at absolute zero. But at any other temperature, simultaneously with this state, there exists a gel, as if displaced with it, in a normal state. As the temperature decreases, the gel concentration begins to predominate. Only at absolute zero the helium scales, according to theory, should be allowed in the following way. Since the gel in a superfluid state does not experience friction either against the walls or against the gel in a normal state, the flow flowing through the capiller does not create a friction reaction and can, as it were, slightly fill the vessel *. On the contrary, the gel in its normal state flows from a vessel with friction, and its flow is a normal fluid flow, long studied by hydrodynamics. This normal flow is caught by a wing placed in front of the top of the tube at figure 1, while the idea of ​​a flow of helium in a superfluid state towards it cannot be detected by ordinary methods.

Based on the same picture, one can explain the high thermal conductivity

12

AXIOMS ​​AND FIRST EXAMPLES

Let us dado some set $X$. We say that a consideration is defined on it if each two elements $a$ and $b$ can be assigned some non-negative number $\rho(a, b)$ - “the distance from $a$ to $b$” - and the following three conditions are satisfied (see Fig. 1, $a, \delta, \theta$):

1°. $\rho(a, b) = 0$ if and only if $a = b$.
2°. $\rho(a, b) = \rho(b, a)$ for any two elements $a$ and $b$ of $X$.
3°. $\rho(a, c) \leqslant \rho(a, b) + \rho(b, c)$ for any three elements $a, b$ and $c$ of $X$.

A set defined on it by a distance (“metric”) is called a metric space $^*$. The elements $x$ themselves are generally called washers of the metric space.

Let us read again the formulations of the axioms that must be satisfied by the function $\rho$ of pairs of points given the distance.

1°. The distance from $a$ to $b$ is 0 if and only if $a$ is the same as $b$.
2°. The distance from $a$ to $b$ is equal to the distance from $b$ to $a$ (“symmetry axiom”).

In these pictures, for every two points, the “distance” between them is indicated. Does the axiom $3^\circ$ of metric space hold for this “distance”?

*) The root “meter” is found in many Russian words and comes from the word meter - measure, size.

13

3°. The distance from $a$ to $c$ is not greater than the sum of the distances from $a$ to $b$ and from $b$ to $c$ (“triangle axiom”).

Let’s start with the simplest examples of metric spaces.

Example 1. $X$ is the number line, that is, the set of all real numbers*). The distance $\rho$ is determined by the formula

$$\rho(x, y) = |x - y|.$$

(1)

Recall that $|a - b| = a - b$ if $a \geqslant b$ and $b - a$ if $b \geqslant a$, so in all cases $\rho(a, b)$ is equal to the length of the segment of the real line with ends $a$ and $b$ (Fig. 2, a).

Let’s check that $\rho$ satisfies all three requirements $1^\circ - 3^\circ$. It is obvious that $|a - b| = 0$ if and only if $a = b$, and that always $|a - b| = |b - a|$.

The inequality $3^\circ$ is also almost obvious (Fig. 2, b). In each of the possible cases in the relative position of points $a, b$ and $c$ on a line, including also when two of the points $a, b$ and $c$ coincide, the inequality

$$|a - c| \leqslant |a - b| + |b - c|$$

(2)

easy to prove formally without referring to figure. For example, if $b \leqslant a \leqslant c$ (lower Fig. 2, b), then

$$|a - c| = c - a,$$

$$|a - b| + |b - c| = a - b + c - b = a + c - 2b \geqslant a + c - 2a = c - a,$$

and therefore (2) is true.

Example 2. $X$ is any set,

$$\rho(a, b) = \begin{cases} 0, & \text{ если } a = b, \\ 1, & \text{ если } a \neq b. \end{cases}$$

(3)

All axions $1^\circ - 3^\circ$ are obviously fulfilled. This is, as they say, a “discrete” space, in which all the points seem to be separate, not too close to each other.

Example 3, which we already talked about a little above. Let’s assume that we have a map of Moscow, which lists all the stops of public transport and for each interval between stops it is indicated how many minutes it takes the bus to cover this interval (respectively, tram, metro, trolleybus). Using such a plan for any two

*) This set is generalized by the letter $R$.

14

stops $a$ and $b$ can we find the value $t(a, b)$ - the minimum time required to get from $a$ to $b$ (we won’t use it in our model)? take into account the time spent transferring from one type of transport to another and waiting, although this can be done). Let’s check whether the properties $1^\circ - 3^\circ$ are satisfied for the function $t$; in other words, whether the set of all stops $X$ with distance $t$ is a metric space. Property $1^\circ$ is, as always, obvious; $3^\circ$ is also beyond doubt: it is clear that it is possible to move from $a$ to $c$ via $b$, so the minimum time $t(a, c)$ depends on no more than $t(a, b) + t(b, c)$. But property $2^\circ$, especially now that there are many one-way streets in Moscow, can, of course, be violated. Check that for the function

$$t'(a, b) = \frac{t(a, b) + t(b, a)}{2}$$

all properties $1^\circ - 3^\circ$ are already fulfilled.

Here is another example where the distance does not change in kilometers.

Example $3'$. Let $X$ be the set of cities in the USSR where planes fly, and $\rho(a, b)$ be the cost of a ticket (in rubles) from city $a$ to city $b$ (on the cheapest route). It is easy to see that this is a metric space.

METRICS ON THE PLANE

Having taken up several exotic examples of metric spaces, we left aside the most natural one.

Example $4$. $X$ is the set of all points of the plane, $\rho$ is the general distance with which we deal in school geometry, that is, $\rho(A, B)$ is the length of the segment connecting two points $A$ and $B$. The properties of $1^\circ$ and $2^\circ$ here and in all the following examples are completely obvious, and we will not say more about them. And the property $3^\circ$ here is nothing more than the statement “in a triangle, each side is no greater than the sum of the other two”*). In the general construction of a geometry course, this statement is a simple theorem.

You probably know that on a plane with a rectangular coordinate system $Oxy$, the distance $\rho(M_1, M_2)$ between two points $M_1(x_1, y_1)$ and $M_2(x_2, y_2)$ is expressed by the following formula (see Fig. 3):

$$\rho((x_1, y_1), (x_2, y_2)) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.$$

Thus, the same metric space can be described without any reference to geometry as follows: $X$ is the set of all pairs $(x, y)$ of real numbers**), the distance between the pairs $(x_1, y_1)$ and $(x_2, y_2)$ is given by formula (4). But at the same time, in order to check the property $3^\circ$,

*) We have already discussed the case when three points lie on the same line (example 1).
**) This set has a special designation $R^2$ and is called the number plane.

16

NEIGHBORHOOD

Let us now return to two-dimensional space—to the $R^2$ plane—and discuss one visual way to compare different distances (4)–(6).

Let point $O=(0, 0)$ be the origin. Let us find the set of points located from the point $O$ at a distance less than a given number $r$. Everyone knows that this set is the interior of a circle with center $O$ of radius $r$ (Fig. 4, a); Of course, we are talking about distance (4), that is, about the set of points $(x, y)$ for which $\sqrt{x^2+y^2}<r$. And what will be the “circles of radius $r$” if the distance is determined by formula (5) or (6)? The set of points $(x, y)$ for which $|x|+|y|<r$ is the interior of a square with vertices $(0, r)$, $(-r, 0)$, $(0, -r)$ and $(r, 0)$ (Fig. 4, b); and the set of the point $\max\{|x|, |y|\}<r$, in other words, the set of points for which $|x|<r$ and $|y|<r$ are obviously a square with vertices $(-r, -r)$, $(-r, r)$, $(r, -r)$ and $(r, r)$ (Fig. 4, c).

Obviously, when talking about metric spaces, instead of the words “circles of radius $r$ with center $a$” they say “$r$ neighborhoods of the point $a$”.

Definition 1. Let $X$ be a metric space, $\rho$ the distance in $X$, $r$ a positive number. Then the $r$-neighborhood of the point $a$ is called the set of all points $m$ from $X$ for which $\rho(m, a)\leqslant r$. Briefly, this set can be written as follows: $\{m: \rho(m, a)\leqslant r\}$.

Thus, in Figures 4, a, b, c, the $r$-neighborhoods of the point $(0, 0)$ are depicted, if the distance on the plane is given by the corresponding formulas (4), (5) and (6); it is not difficult to report that the $r$-neighborhood of any other point $(x_0, y_0)$ in these metric spaces looks exactly the same as the neighborhood of the point $(0, 0)$ - simply the center of the circle or square is shifted to the point $(x_0, y_0)$.

Let’s see what the $r$-neighborhoods are in the metric spaces we talked about earlier. In Example 1, $r$ is the neighborhood of the point $a$ on the number axis

$$\{x; |x-a|\leqslant r\}$$

is a segment of length $2r$, the middle of which lies at the point $a$ (Fig. 5). In example 3′, the 20-neighborhood of Moscow is a set of cities where you can actually fly for no more than 20 rubles; in Example 2, the $r$-neighborhood consists of only one point if $r<1$, and contains the entire set $X$ if $r\geqslant 1$.

17

DISTANCES BETWEEN FUNCTIONS

Distance is conveniently determined not only for numbers or points of the plane and space, but also for many other mathematical objects. Here are two examples from geometry. For the set of all lines passing through a given point O, the distance between two lines can be taken to be the smaller of the angles it forms (in figure 6 we show what a neighborhood of one of the lines looks like). The distance between two convex polygons $M_1$ and $M_2$ on a plane can be determined as follows: for each vertex of polygons $M_1$ and $M_2$ we find the distance $^*$ to the vertex of another polygon closest to it, and from all these numbers we take the most 贴近; Thus, the $r$-neighborhood of a given polygon $M_0$ (Fig. 7) contains polygons $M$ such that all vertices lie in circles of radius $r$ with centers at vertices $M_0$, and each circle contains at least one vertex $M$; check that the set of convex polygons with such a distance is a metric space. The number of such examples could easily be increased.

But the most important applications, which the theory of metric spaces owes its emergence and development, are associated not with geometrics, but with the analysis and theory of functions.

Very often, in order to examine the data of functions or simply calculate their values, it is convenient to approximately replace them with other, simpler functions, say, polynomials. You’ve probably heard that when $x$ is close to zero, sin$x$ is approximately equal to $x$ (here sin$x$ means the sine of the number $x$, that is, the sine of an angle of $x$ radians). An even more accurate formula: $\sin x \approx x - \frac{x^3}{6}$. Let, for example, $x$ change on the segment $0 \leqslant x \leqslant \frac{\pi}{4}$.

How to evaluate how well function $f_1(x) = x - \frac{x^3}{6}$ approximates function $f_2(x) = \sin x$, how large is the “distance” between these functions?

One of the natural distances is also: for each $x$ we find the difference $f_1(x) - f_2(x)$, take that $x = x_0$, where this difference is largest (in absolute value), and put $\rho(f_1, f_2) = |f_1(x_0) - f_2(x_0)|$.

*) This refers to the “obvious” distance (4).

18

For our specific functions

$$\rho(f_1, f_2) = |\sin \frac{\pi}{4} - \left( \frac{\pi}{4} - \frac{\pi^3}{384} \right)| \approx 0.0025.$$

(It can be proven that the maximum is reached at the point $x = \frac{\pi}{4}$.)

In the same way, in the general case, we take as the distance between the functions $f_1, f_2$ defined on the segment $[a, b]$ the quantities

$$\rho(f_1, f_2) = \max_{a \leqslant x \leqslant b} |f_1(x) - f_2(x)|.$$

(7)

Check that the axioms $1^\circ - 3^\circ$ are satisfied here! Moreover, the $r$-neighborhood of this function (Fig. 8) consists of all such functions whose graphs lie in a strip of width $2r$ around the graph of the function $f$.

The following distances are often used:

$$\rho(f_1, f_2) = s(f_1, f_2)$$

where $s$ is the value of the area enclosed between graphs $f_1$ and $f_2$ (8) (Fig. 9) and especially this:

$$\rho(f_1, f_2) = \sqrt{S(f_1, f_2)},$$

(9)

where $S(f_1, f_2)$ is the area enclosed between the graph of the function $y = (f_1(x) - f_2(x))^2$ and the axis $Ox$.

Distance (7) is small when the values ​​of the function $f_1$ and $f_2$ are close for all values ​​of the argument, and distances (8) and (9) show how close the functions $f_1$ and $f_2$ are “on average” (on small segments they can differ significantly from each other). Let, for example, $f_1(t)$ and $f_2(t)$ be the potentials of two specific points of the electric circuit at time $t$; then the power released in the section of the circuit between these points during the period of time $a \leqslant t \leqslant b$ is proportional to $S(f_1, f_2)$ (resistance is constant; power is proportional to $(f_1(t) - f_2(t))^2$); thus, the greater the distance (9), the greater the power. And if it is important for us that the voltage $f_1(t) - f_2(t)$ does not always exceed a certain value $V$, then we must estimate the distance using formula (7): it is necessary that the value max $|f_1(t) - f_2(t)|$ does not exceed the length $V$.

20

Fig. 10.

in Russian ), is now presented in the first chapters of textbooks with approximately such titles as “Elements of functional analysis”, “Introduction to the theory of sets and functions”, “Fundamentals of modern analysis”, “Topological spaces”. Of course, most of this theory, associated with the concept of limit, is meaningful only for spaces with an infinite number of points. But at the same time, the general concept of “distance” is also useful for some problems about finite sets; in particular, the “distance between words”, which we talked about at the beginning of the article, turned out to be a very convenient tool in the now rapidly developing theory of error-correcting codes. For those who want to get acquainted with the theory of metric spaces in more detail, we recommend reading the book by Yu. A. Schrader*) “What is distance.”

Exercise

1. Prove that for any four points $A, B, C, D$ of the metric space $\rho(A, C) + \rho(B, D) \leq \rho(A, B) + \rho(B, C) + \rho(C, D) + \rho(D, A)$.
2. Prove that for any $n$ points $A_1, A_2, \dots, A_n$ ($n \geq 3$) $\rho(A_1, A_2) + \rho(A_2, A_3) + \dots + \rho(A_{n-1}, A_n) \geq \rho(A_1, A_n)$.
3. Dr. Charadec, who knew strategy well, was interested in the last war and in 1940 became acquainted with a map of the French theater of war. From here, it is believed, the next task arose. The distance (by air, like all distances in this problem) from Chalons to Vitry is 30 km, from Vitry to Chaumont 80 km, from Chaumont to Saint-Quentin 235 km, from Saint-Quentin to Rems 86 km, from Rems to Chalons 40 km. Calculate the distance from Rems to Chaumont in this closed polygon. Only Dr. Charadek *** can do this without a map.
4. a) Prove that if $\rho(M_1, M_2) = r$ and $r_1 + r_2 < r$, then the $r_1$-neighborhood of the point $M_1$ has no common points with the $r_2$-neighborhood of the point $M_2$ (figure 10 illustrates this fact for distances (4) and (6) on the plane).

6. Prove that if $\rho(M_1, M_2) = r$ and $r_1 - r > r_2$, then the $r_2$-neighborhood of the point $M_2$ is entirely contained in the $r_1$-neighborhood of the point $M_1$.

5. The set $E$ of points of the metric space $X$ is called a $\varepsilon$-net if the $\varepsilon$-neighborhoods of the points of the set $E$ (all together) completely cover the set $X$; in other words, if for each point $x$ from $X$ there is at least one point of the set $E$, distant from $x$ by no more than $\varepsilon$. (Here $\varepsilon$ is some positive number.)

For example, the set of black dots on figure 11, $a$ is a $\frac{1}{10}$-net for the real line segment $0 \leq x \leq 1$ with the usual distance (2). Of course, it is also a $\varepsilon$-network for

*) Hausdorff, Set Theory, ONTI, 1934.
**) Yu. A. Schrader, “What is distance.” M. Fizmatiz, 1963.
***) G. Steinhauz, One Hundred Problems, Fizmatiz, 1959, problem No. 100.

21

any $\varepsilon > \frac{1}{10}$. On figure 11, 6 the set of black dots is $\frac{1}{10}$—a network for the square $0 \leq x \leq 1$, $0 \leq y \leq 1$ on the plane $Oxy$ with distance (6). For which $\varepsilon$ is this set a $\varepsilon$-net in the sense of distances (4) and (5) on the same square?

6. A metric space $X$ is called bounded if there is a number $c$ such that the distance between any two points $X$ does not exceed $c$.

Prove that if for some $\varepsilon$ a space has a finite $\varepsilon$-net, then it is bounded.

7. Let $N(\varepsilon)$ be the smallest number of points in the $\varepsilon$-network of the space $X$, $M(\varepsilon)$ be the smallest number of points in $X$, the distance between any two of which is at least $\varepsilon$.

Prove that $M(2\varepsilon) \leq N(\varepsilon) \leq M(\varepsilon)$.

8. Let $C$ be a set of functions defined on the interval [0, 1] and taking values ​​on the same interval, whose graphs are broken lines. We will determine the distance in $C$ using formula (7).

Prove that it is possible to extract an infinite number of functions from $C$, all pairwise distances between which are equal to one. Deduce from this that for $\varepsilon < \frac{1}{2}$ in $C$ it is impossible to steal the finite $\varepsilon$-net.

9. In some metric space, can it be that the 3-neighborhood of a point $A$ is entirely contained in the 2-neighborhood of another point $B$ and does not fill it entirely? What will be the answer if we replace 3 and 2 with other numbers?

10. Prove that among $n$-digit numbers, it is impossible to extract more than $\frac{2^n}{n+1}$ numbers from two digits 1 and 2 so that any two of them differ from each other in at least three digits.

11. Using problem 4 a) prove that two different points $A$ and $B$ cannot be limits of the same sequence.

12. Construct an example of a sequence of functions defined on the interval [0, 1], which tends to the limit $f_0$, if we use distance (8), and does not have $f_0$ limit, if we use distance (7).

13. Let $\rho_1$ and $\rho_2$ — two distances on some set $X$ — have the property that $\rho_1(A, B) \leq k\rho_2(A, B)$ for any two points $A$ and $B$, where $k$ is some positive number (the same for all $A$ and $B$).

Prove that if $P$ is the limit of the sequence $M_1, M_2, M_3, \dots$ in the sense of the distance $\rho_2$, then $P$ will be the limit of this sequence in the sense of the distance $\rho_1$. Thus, prove that on the plane the statement “the limit of the sequence $M_1, M_2, M_3, \dots$ is the point $P$” has the same meaning, regardless of which of the distances (4), (5), (6) we obtain.

14 a). Think of a distance $\rho$ to the set of all lines in the plane for which the following conditions would be satisfied: if the sequences of points $A_1, A_2, A_3, \dots$ and $B_1, B_2, B_3, \dots$ have two distinct points $A$ and $B$ as limits, then the sequence of lines $A_1B_1, A_2B_2, A_3B_3, \dots$ has (in the sense of the distance $\rho$) the limit of the line $AB$.

6. Prove that such a distance $\rho$ cannot be specified so that the distance between any two intersecting lines depends only on the angle between them.

24

COLLISION OF MOLECULES WITH A WALL

When a molecule collides with a wall and bounces off it, the wall receives some momentum. Let us first assume that the molecule is reflected from the wall in an elastic manner. figure 1 shows a diagram of such a blow. Axis $Oz$ is directed perpendicular to the wall, axes $Ox$ and $Oy$ are somehow located in the plane. Let us decompose the speed of molecules into components along those axes. During an elastic impact, the angle of incidence $\alpha$ is equal to the angle of reflection $\beta$, therefore the velocity components $v_x$ and $v_y$ do not change. Component $v_z$ changes sign during an elastic impact. Thus, it is easy to calculate that the molecule will change its momentum (amount of motion) by the amounts
$$mv_z - mv'_z = mv_z - (-mv_z) = 2mv_z = 2mv \cos \alpha,$$
where $v$ is the speed of the molecule.

The problem, as you can see, is not difficult, and Waterston solved it correctly. The next person to take up this task (and publish his work in 1856) was Crang. He incorrectly calculated that the molecule upon impact transfers all its momentum $mv_z$ to the wall, and got a result that was half as large.

One might, however, think that our conclusion made one very significant simplification: we assumed that the impact was elastic. In reality, this assumption, surprisingly, is insignificant. a molecule can be reflected as carbon, but the result does not change. The assumption about the elastic nature of the impact does not change the result; this makes the conclusion very simple.

This remarkable property - the independence of the result from the law of reflection - is a consequence of the thermal equilibrium between the gas and the walls of the vessel (which are maintained at a constant temperature). Again (as with the collision of molecules), the final result - thermal equilibrium - does not change depending on how exactly the molecules are reflected from the wall.

the visual apparatus in unusual conditions allows us to deeply understand the complex processes occurring in the eyes and in our consciousness during the synthesis of images of the reality around us.

For those who want to get acquainted with experiments in physical optics in more detail, we can recommend the recently translated book by J. Gregg, “Olytics with Sight” (Mir Publishing House, Moscow, 1970). This book provides a description of a large number of experimental elements for studying the functioning of the eye. All of them are quite accessible to schoolchildren, and most of their weight is remunerative. The reader should only be warned that in those places where the author touches on physical optics, the language of the book becomes far from scientific.

In the library you will probably also find Minart’s book “Light and Color in Nature” (Nauka Publishing House, Moscow, 1969) and Artamonov’s book “The Illusion of Vision” (Nauka Publishing House, Moscow, 1969).

Fig. 3.

45

Fig. 9.

arrangement of circles and all possible quadrilateral shapes.

The idea of ​​the solution given below belongs to N. Vasiliev.

Let us denote the circles with centers A, B, C, D at the vertices of a given quadrilateral ABCD by Kₐ, Kₑ, K₃, K₄, and their radii by rₐ, rₑ, r₃, r₄. Let’s assume that the problem statement is false. Then each circle has a common point with the part of the triangle assigned to it that is not covered by triangles:

for example, Kₐ must contain points △BCD, none are covered

Fig. 10.

from circles Kₐ, Kₑ, K₃, K₄. Let O be the point of intersection of the diagonals. Let us prove, under our assumption, that rₐ ≥ OA, that is, that the circle Kₐ contains the point O. The rest is already clear: in the same way it will be possible to prove that the point O belongs to Kₐ and K₃, and from here it obviously follows that these circles cover △ABC. Look at figure 9,10: if OL and ON are perpendiculars dropped onto lines AB and BC, then Kₐ covers △OAL, K₃ - △OCN, K₄ - △OBL and △OBN, and these four right triangles cover △ABC, even if

Fig. 11, a.

Fig. 11, b.

46

one of the triangles AOB or OBC (as in figure 10) is obtuse.

We come to a contradiction with our assumption; therefore, the statement of the problem is correct.

So, it remains to prove that $r_A \geqslant OA$. We will need the following, almost obvious one.

Lemma. Let a convex quadrilateral PQRS and a circle $K_R$ (not containing $\triangle PQR$ entirely) be given. Then, of all the points lying inside $\triangle PQR$ and outside $K_R$, those close to the point $S$ will be:

a) the base $F$ perpendicular $SF$ dropped from the point $S$ to the line $PR$, if this point $F$ lies outside the circle $K_R$ (Fig. 11, a);

b) point $E$ of intersection of circle $K_R$ with segment $PR$, if point $F$ lies inside circle $K_R$ (Fig. 11, b). (It is important here that the QRS angle of the quadrilateral is less than $180^\circ$; we leave the proof of the lemma to the reader.)

We can assume that $\angle AOB \leqslant 90^\circ$ (otherwise we would take not $B$, but $D$). Let $BB_1 \perp AC$; $AA_1 \perp BD$; $B_1A_2 \perp BD$ (Fig. 12). Suppose that $r_A < OA$ and $E - \text{ближайшая } K$ are on the segment $OA$ and $AE = r_A$, $EM - \text{перpendикуляр}, \text{опущенный } E$ on $BD$. It is clear that then $AM > AE$. Let us prove that, nevertheless, $r_A > AM$, it will follow that our assumption ($r_A < OA$) is incorrect.

Let us first prove that $r_B > BM$. Let us apply the lemma to $\triangle ACD$, the circle $K_A$ and the point $B$. Distance two cases:

a) $E$ lies between $A$ and $B_1$. Then $r_B > BB_1 > BA_2 \geqslant BM$;

b) $E$ lies between $O$ and $B_1$. Then (by lemma)

$$r_B > BE > BM.$$

So we know that the point $M$ belongs to $K_B$. Applying the lemma to $\triangle BCD$, the circle $K_B$ and the point $A$ (note that $M$ lies between $O$ and $A_1$, so case 6 always holds)), we obtain $r_A > AM$. The circle is closed, we demand a contradiction.

F12. Two identical heavy steel balls rotate on light rods of length $l$ and $2l$ around the point $O_1$, $O$, the distance between which is 3$l$ (Fig. 13). At the initial moment, the balls are at points $A$ and $B$, having speeds $v$ and $2v$, respectively. How many times will the balls collide in time $t$? How long will it take for the balls to collide $k$ times? The coduar balls are considered to be absolutely elastic.

Solution. Since the balls are identical, during absolutely elastic collisions they will exchange velocities. This follows from the laws of conservation of energy and momentum (number of movements) of balls upon collision. The first time the balls collide is when each of them makes half a revolution around the point about which it is spinning.

This will take time $t_1 = \frac{\pi \cdot 2l}{2v} = \frac{\pi l}{v}$. After the collision, the left ball will have a speed of $2v$, and the right ball will have a speed of $v$.

The second collision of the balls will occur in time $t_2 = \frac{4\pi l}{v}$ after the first, when the right ball makes one revolution around the point $O_1$. During this time, the left ball will make four revolutions around the point $O$ - it has twice the speed, and it rotates around the circle in half the time

47

Fig. 13.

shero radius. Upon collision, the balls will again exchange velocities, and the third collision will occur in time $t_3 = 2t_1 = \frac{2\pi l}{v}$ after the second, when each of the balls has completed one revolution. Thus, after each odd collision, the left ball will have a speed of $2v$, and the right ball will have a speed of $v$, and therefore the next, even, collision will occur after a time of $t' = t_2 = t_4 = \cdots = \frac{4\pi l}{v}$.

After each odd collision, the left ball will have a speed of $v$, and the right one will have a speed of $2v$, and the next, now odd, collision of the balls will occur after a time $t'' = 2t_1 = t_3 = \cdots = \frac{2\pi l}{v}$ after the even one.

Now it is easy to calculate that if $k$ is an even number, then the $k$-th collision of the balls will occur in time
$$t_k = \left( \frac{k}{2} - \frac{1}{2} \right) t'' + \frac{k}{2} t' = (3k - 1) \frac{\pi l}{v},$$
since in total there will be $\frac{k}{2}$ even collisions between the balls and $\frac{k}{2}$ odd ones. This means that the right ball must make $\frac{k}{2} - \frac{1}{2}$ revolutions, having a speed of $2v$ (here we taught that before the first collision the ball makes only half revolutions) and $\frac{k}{2}$ revolutions around point $Q_1$, having a speed of $v$.

If $k$ is odd, then before the $k$-th collision the right ball must make a speed of $v$ and $\frac{k-1}{2} + 1 - \frac{1}{2} = \frac{k}{2}$ revolutions, having a speed of $2v$. Therefore, from the initial moment to the moment of the $k$-th collision in this case, time must pass
$$t_k = \frac{k}{2} t'' + \frac{k-1}{2} t' = \frac{(3k-2)\pi l}{v}.$$

Let us now calculate how many times the balls will collide in time $t$. We will assume that $t > t_1$. Between two consecutive odd collisions of balls, time $t_0 = t' + t''$ passes. There will be as many odd collisions after the first as how many times $t_0$ will collide in $t - t_1$:

$$n = \left[ \frac{t - t_1}{t_0} \right]$$

(the [ ] sign means “integer part of a number”).

A total of $n+1$ odd collisions will occur between the balls. The time elapsed before the last odd collision will be equal to $\left[ \frac{t - t_1}{t_0} \right] t_0 + t_1$.

During this time, $n$ even collisions should occur between the balls. Additionally, if the remaining time $t - \left( \frac{t - t_1}{t_0} \right] t_0 + t_1$ is greater than $t'$, then another odd collision will occur. Thus, in time $t$ the balls will collide

$$N = \left[ \frac{t - t_1}{t_0} \right] + 1 + \left[ \frac{t - t_1}{t_0} \right] + \left[ \frac{t - \left[ \frac{t - t_1}{t_0} \right] (t_1 + t_2)}{t_2} \right] = 2 \left[ \frac{vt - \pi l}{6\pi l} \right] + \left[ \frac{vt - \left[ \frac{vt - \pi l}{6\pi l} \right] 6\pi l}{4\pi l} \right] + 1$$

48

Fig. 14.

F13. Two identical cylinders with pistons are connected by a tube (Fig. 14). There is water in the cylinders. Identical cylindrical glasses with equal amounts of water are placed on top of the pistons. Then a body of mass $m$ is lowered into one of the glasses, and a body of mass $M$, which do not sink, is lowered into the other.

At what distances from each other will the ends of the pistons and the water levels in the glasses be when the system reaches equilibrium? The day area of ​​the glasses is $S_1$, the area of ​​the pistons is $S_2$.

Solution. Since the cylinders are connected by a tube, the water pressure in the cylinders at the same levels should be the same. Identical and pressure in the cylinders is at the level at which the lower edge of the left piston is located. But the pressure in the left cylinder (Fig. 14) at this level is $\frac{(M_0 + M)g}{S_2}$, and the pressure at this level in the right cylinder is $\frac{(M_0 + m)g}{S_2} + \rho gy$, where $M_0$ is the total mass of the piston, glass and water in the glass, and $\rho gy$ is the pressure of a column of water of height $y$. Then from the equilibrium condition we obtain:

$$\frac{(M_0 + M)g}{S_2} = \frac{(M_0 + m)g}{S_2} + \rho gy.$$

Solving this equation, we find that

$$y = \frac{M - m}{\rho S_2}.$$

Now let’s determine the values ​​of $x$. When a body is immersed in a glass, the pressure at the bottom of the glass increases by an amount equal to the weight of this body. On the other hand, since the body does not sink and does not itself press on the bottom, a change in the force of pressure on the bottom is a change in the force of water pressure on the bottom due to the fact that the water level increases. If the water level in a glass rose by $\Delta x$, then the water pressure at the bottom increased by $\rho g \Delta x$, and the pressure force by $\rho g \Delta x S_1$. Thus, for the left glass $\rho g \Delta x S_1 = Mg$, where $\Delta x_1$ is the change in the water level in the left glass: $\Delta x_1 = \frac{M}{\rho S_1}$. Similarly, we find that in the right glass the water level when the body was immersed in it rose by $\Delta x_2 = \frac{m}{\rho S_1}$. Since the distance between the bottom of the left and the bottom of the right glasses is $y$, then

$$x = y + \Delta x_2 - \Delta x_1 = \frac{(M - m)(S_1 - S_2)}{\rho S_1 S_2}.$$

F14. Two pipes with sections $S_1$ and $S_2$ are connected to each other and plugged with pistons whose masses are $m_1$ and $m_2$ (Fig. 15). After an explosion in the space between the pistons, the pistons fly out of the pipes. One of them flew out at a speed of $v$. At what speed did the second one fly out if: a) the pipes are fixed and cannot mix, b) the pipes are not fixed and their total mass is $M$? Neglect the friction of the pistons against the pipe walls.

Solution. Let us first consider the case when the pipe is fixed. The gas pressure on the pistons is the same, and the pressure force is related to the area of ​​the pistons. Since these forces act on the pistons for the same amount of time (we believe that after one of the pistons leaves the pipes, the pressure on the second piston drops abruptly to the pressure outside), then the ratio of the average pressure forces on the pistons is equal to the ratio of the areas of the pistons. This gives us the opportunity to find the ratio of the impulses of the pistons after they leave the pipes. The initial impulses of the pistons are zero, and the change in their impulses is equal to the average forces multiplied by

49

Fig. 15.

at the time of the explosion. Therefore, the ratio of the impulses of the pistons after they leave the pipes is equal to the ratio of the average pressure forces acting on the pistons, and, therefore, equal to the ratio of the areas of the pistons.

To determine this, we will assume that the left piston (Fig. 15) comes out at a speed of $v$. The speed of the right piston will be denoted by $u$. Then we can write that $\frac{m_1v}{m_2u} = \frac{S_1}{S_2}$; hence $u = v\frac{m_1S_2}{m_2S_1}$.

Now consider the case when the pipes are not fixed. Then after the explosion they will acquire a speed, which we will denote $u_1$. Comparing in the same way as we did for pistons, the average pressure forces acting on the left piston and on the joint of the pipes, we find that $u_1 = v\frac{m_1(S_2 - S_1)}{MS_1}$.

As for the pistons, our alienation is fair for them in this case as well.

F15. The amount of heat $Q=190$ kilocalories penetrates through the walls of the refrigerator in an hour. The temperature inside the refrigerator is $T_1=+5^\circ\text{C}$, and in the room $T_2=+20^\circ\text{C}$. What is the minimum power consumption of this refrigerator from the network?

Solution. A refrigerator is a heat engine operating on a processing cycle. If a direct heat engine absorbs an amount of $T_2$ and separates a lesser amount of heat $Q_1$ at a low temperature $T_1$, performing work $A = Q_2-Q_1$, then a refrigeration engine absorbs an amount of heat $Q'_1$ from cold heat at a temperature of $T_1$ and separates an amount of heat $Q'_2$ to a higher heat at a temperature $T_2$. In this case, you need, of course, to supply energy $A'=Q'_2-Q'_1$ to the refrigerator.

The efficiency coefficient of a direct heat engine $\eta = \frac{A}{Q_2} = \frac{Q_2-Q_1}{Q_2}$ cannot exceed the value $\frac{T_2-T_1}{T_2}$. Moreover, in order for c.p. The machine was at its maximum, it would have to work in a reverse cycle. If the refrigeration machine operates on the same cycle as the direct one, then $Q'_1 = Q_1$, $Q_2=Q_2$ and $A'=A$. K.p. d. of the refrigeration machine would in this case be equal to $\eta_x = \frac{Q'_1}{A} = \frac{Q_1}{Q_2-Q_1}$. But in this case, k.p. d. of a straight machine is $\frac{Q_2-Q_1}{Q_2} = \frac{T_2-T_1}{T_2}$, and therefore $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$. Using this correspondence, we find that k.p. d. of the refrigeration machine cannot exceed the value $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.

The k.p. calculated using this formula. refrigerator (sometimes called the “coefficient of performance”) can be more than one. This is due to the fact that, by separating the energy $Q_2$, we consume not only the energy $A$, but also the energy $Q_1$ received from cooling the body. When calculating the k.p. e machine, we consider only the energy $A$ consumed from the network, since this is the only energy we have to pay for.

51

Fig. 16.
at high speeds of automatic movement, there was no loss of power in the mechanisms and rolling friction of the wheels on the road. Consider how taking these factors into account would change equations (1) and (2).

F17. A plate having a charge $q$ was placed between the plates of a short-circuited flat capacitor. The plate is moved parallel to itself at a distance $x$ (Fig. 17). What charge passes through the external circuit of the capacitor if the distance between its plates is $d$?

To solve the problem, we need to know what field is created by a large uniformly charged plate (large compared to the distance from the plate to the point at which the field interests us). We will proceed from the fact that, as we know, the electric field in a flat capacitor is equal to $\frac{u}{d} = \frac{Q}{C_d} = \frac{Q}{\varepsilon_0 S}$, where $C$ is the capacitance of the capacitor, $u$ is the potential difference between its plates, $Q$ is the charge of the capacitor, $S$ is the area of the plates, $d$ is the distance between the plates and $\varepsilon_0$ is the electrical constant. But the field in the capacitor is equal to the sum of the fields created by its plates. This means that each of the capacitor plates creates a field equal to $\frac{1}{2} \frac{Q}{\varepsilon_0 S}$. Such a field should be created simply by one large uniformly charged plate.

Thus, we found that the middle movable plate creates a field $\frac{1}{2} \frac{q}{\varepsilon_0 S}$. But the field of a system of charges is equal to the sum of the fields created by each of the charges separately. Therefore, the field between the plates is equal to the sum of the fields created by the plates and the moving plate. We will assume that the total charge of the closed plates is zero *). Then, if the charge of one of the plates is equal to $Q$, then the charge of the second plate is equal to minus $Q$, and the field created by these plates is exactly the same as the field in a capacitor with charge $Q$ $E = \frac{Q}{\varepsilon_0 S}$. On one side of the plates, the fields of the plates and the movable plate are directed in one direction, on the other - in the opposite direction. Therefore, to the left of the plates (Fig. 18) (for definition we assume that $q > 0$) is the field $E_1 = \frac{1}{2} \frac{q}{\varepsilon_0 S} + \frac{Q}{\varepsilon_0 S} = \frac{q + 2Q}{\varepsilon_0 S}$, and to the right of it is the field

$$E_2 = \frac{Q}{\varepsilon_0 S} - \frac{1}{2} \frac{q}{\varepsilon_0 S} = \frac{2Q - q}{\varepsilon_0 S}.$$

*) It is not difficult to generalize the problem to the case when the total charge of the capacitor plates is not zero. The answer in this case remains the same.

52

The work required to transfer a single positive charge from one plate of our capacitor to the other is $A = E_1 d_1 + E_2 d_2$. But the capacitor is short-circuited, and the potential difference between the plates is zero. Therefore, $E_1 d_1 + E_2 d_2 = 0$.

Or $\frac{2Q + q}{2\varepsilon_0 S} d_1 + \frac{2Q - q}{2\varepsilon_0 S} (d - d_1) = 0$. Simplifying this equation, we get

$$Q (d - d_1) + q d_1 = 0. \tag{1}$$

When the middle dam moves over a distance $x$, a charge $\Delta Q$ will pass through the wire connecting the plates, and the charges of the plates will be equal to $Q + \Delta Q$ and $-Q - \Delta Q$. Since in this case the potential difference between the plates is zero, then, reasoning in the same way as in the first case, we obtain the equation

$$\frac{q + 2(Q - \Delta Q)}{2\varepsilon_0 S} (d_1 + x) + \frac{2(Q - \Delta Q) - q}{2\varepsilon_0 S} (d - d_1 - x) = 0$$

or

$$(Q - \Delta Q) (d - d_1 - x) + q (d_1 + x) = 0. \tag{2}$$

Subtracting equation (1) from equation (2), we find that $\Delta Q = q \frac{x}{d}$.

53

1. RESEARCH ALL CASES CAREFULLY

Suppose we need to solve the following linear inequality for $x$:

$$\frac{3ax + 4}{3a + 9} < \frac{x}{a + 3} + \frac{3a - 5}{3a - 9}.$$

It would seem, what could be simpler? Following the well-known school rule, we collect all terms containing $x$ on the left side of the inequality, and free terms on the right:

$$\frac{3ax}{3a + 9} - \frac{x}{a + 3} < \frac{3a - 5}{3a - 9} + \frac{4}{3a + 9},$$

and forgive the resulting inequality:

$$\frac{a - 1}{a + 3} x < \frac{a^2 - 1}{a^2 - 9}.$$

(1)

Now all that remains is to divide both sides of the inequality by the coefficient of the unknown.

But it is precisely at this step - when dividing both sides of inequality (1) by the coefficient of the unknown - that you need to be extremely careful, or the inequality contains a parameter.

An inequality with a parameter is essentially a set of inequalities, each of which is obtained from a parameter given at a specific value. Imagine that a friend asked you to solve an inequality with a parameter, and he himself thought of a specific number for the value of the parameter. You do not know the intended meaning, and your task is not to let yourself be taken by surprise, to carefully examine all the cases that may present themselves. Once you’ve solved the inequality, and buddy

announced to you the value of the parameter he intended, you must find in your answer a case that relates specifically to this value.

Let’s return to our example. There may be 4 cases:

1. one or both parts of the inequality do not make sense;
2. the coefficient for the unknown is equal to;
3. the coefficient for the unknown is positive;
4. the coefficient for the unknown is negative.

The first case will occur at $a = \pm 3$. For these values ​​of the parameter, inequality (1) has no solutions.

The second case will occur at $a = 1$. In this case, the value of the parameter inequality (1) takes the form $0 \cdot x < 0$ and obviously has no solutions.

The third case will occur at $a < -3$ or at $a > 1$, $a \neq 3$. Dividing both sides of inequality (1) by the positive number $\frac{a - 1}{a + 3}$, we obtain:

$$x < \frac{a + 1}{a - 3}.$$

The fourth case will occur at $-3 < a < 1$. Dividing both sides of inequality (1) by the negative number $\frac{a - 1}{a + 3}$, we obtain: $x > \frac{a + 1}{a - 3}$.

Combining the results obtained, we write down

Answer: if $a = 1$, $a = \pm 3$, then there are no solutions;

if $a < -3$, $1 < a < 3$, $a > 3$, then

$$x < \frac{a + 1}{a - 3};$$

54

if $-3 < a < 1$, then $x > \frac{a+1}{a-3}$.

The given answer contains a solution to inequality (1) for any value of the parameter. This is precisely the main goal when solving an inequality with a parameter.

Using figure 1, solve the inequality $\cos x \geq a$ and prove that the answer is:
if $a \leq -1$, then $-\infty < x < \infty$;
if $-1 < a < 1$ then $m_1 + 2k\pi \leq x \leq m_2 + 2k\pi$,
where $m_1 = -\arccos a$, $m_2 = \arccos a$;
if $a = 1$, then $x = 2k\pi$;
if $a > 1$, then there are no solutions.
($k = 0, \pm 1, \dots$)

2. DON’T RUSH TO CONCLUSIONS

Let’s look at another simple example. Let’s solve the inequality $\sqrt{x} > a$. It is tempting to write $x > a^2$ and claim that this entry contains all the solutions to the inequality. But one should not rush to conclusions when solving an inequality with a parameter: after all, $a$ can be less than zero; then the solution to the inequality $\sqrt{x} > a$ will be any non-negative value $x$; if $a \geq 0$, then squaring both sides of the inequality $\sqrt{x} > a$ is an equivalent transformation; having polished it, we get $x > a^2$.

Answer: if $a < 0$, then $x \geq 0$;
if $a \geq 0$, then $x > a^2$.

Now let’s move on to more complex examples.

Example 1. Solve inequality $\sqrt{x^2 + x} < a - x$. (2)

Solved and e. The range of admissible values ​​of the unknown definition of inequalities $x^2 + x \geq 0$. In addition, it is clear that if $a - x \leq 0$, then inequality (2) has no solution, so we restrict ourselves to the ray $a - x > 0$. But if the condition is met
$$\begin{cases} x^2 + x \geq 0, \\ a - x > 0 \end{cases}$$ (3)

both sides of inequality (2) can be

55

enter into a square (this will not lead to the appearance of construction solutions). After squaring and bringing similar terms, we get:
$$(1+2a)x < a^2.$$
(4)

Inequality (4) has the following solutions:

$$\text{ если } a = -\frac{1}{2},$$
$$\text{ то } -\infty < x < \infty;$$
$$\text{ если } a > -\frac{1}{2}, \text{ то } x < \frac{a^2}{1+2a};$$
$$\text{ если } a < -\frac{1}{2}, \text{ то } x > \frac{a^2}{1+2a}.$$

But one should not rush to conclusions and assume that this has resulted in solutions to inequality (2). From the found solutions to inequality (4), it is necessary to display those that satisfy conditions (3). To facilitate further work, let us solve for $x$ the system of inequalities (3) for any value $a$.

The first inequality of system (3) has solutions $x \leq -1$ or $x \geq 0$. The second inequality has solutions $x < a$. Further reasoning depends on how the number $a$ is positioned relative to the numbers $-1$ and $0$.

Having analyzed the cases $a \leq -1$, $-1 < a \leq 0$, $a > 0$, we obtain the following solutions to system (3):

a) if $a \leq -1$, then $x < a$;
b) if $-1 < a \leq 0$, then $x \leq -1$;
b) if $a > 0$, then $x \leq -1$ or $0 \leq x < a$.

Let us now select from the solutions to inequality (4) found above those that satisfy the just obtained conditions $a$), $6$), $b$). These will be the solutions to the original inequality (2).

For $a = -\frac{1}{2}$, inequality (4) is satisfied by all real numbers. Condition (6) forces us to limit ourselves to the values ​​of $x \leq -1$. So, if $a = -\frac{1}{2}$, then inequality (2) has solutions: $x \leq -1$.

For $a > -\frac{1}{2}$, inequality (4) has solutions $x < \frac{a^2}{1+2a}$. Let us consider, in accordance with conditions (6) and (b), two cases:

$$-\frac{1}{2} < a \leq 0 \text{ и } a > 0.$$

In the first case, we need to find all $x$ satisfying the system of inequalities $x \leq -1$ and $x < \frac{a}{1+2a}$. Thus the values ​​of $x \leq -1$.

So, if $-\frac{1}{2} < a \leq 0$, then inequality (2) has solutions $x \leq -1$.

In the second case, that is, for $a > 0$, we need to take such $x$ that satisfy one of the systems of inequalities:

$$\begin{cases} x < \frac{a^2}{1+2a}, \\ x \leq -1, \end{cases}$$
$$\begin{cases} x < \frac{a^2}{1+2a}, \\ 0 \leq x < a. \end{cases}$$

The first of these systems has solutions $x \leq -1$. To solve the second system, you need to compare the numbers $a$ and $\frac{a^2}{1+2a}$. Let’s make the difference $a - \frac{a^2}{1+2a} = \frac{a(a+1)}{1+2a}$.

It is clear that for $a > 0$ this difference is positive, hence $\frac{a^2}{1+2a} < a$. This means that the second system has the following solutions: $0 \leq x < \frac{a^2}{1+2a}$.

So, if $a > 0$, then inequality (2) has solutions $x \leq -1$;
$$0 \leq x < \frac{a^2}{1+2a}.$$

For $a < -\frac{1}{2}$, inequality (4) has solutions $x > \frac{a^2}{1+2a}$. Let us consider, in accordance with conditions (6) and (b), two cases:

$$a \leq -1 \text{ и } -1 < a < -\frac{1}{2}.$$