The main function of brackets is to change the order of actions when calculating values. for example, in the numerical expression \(5 3+7\) the multiplication will be calculated first, and then the addition: \(5 3+7 =15+7=22\). But in the expression \(5·(3+7)\), addition in brackets will be calculated first, and only then multiplication: \(5·(3+7)=5·10=50\).

Example. Expand the bracket: \(-(4m+3)\).
Decision : \(-(4m+3)=-4m-3\).

Example. Expand the bracket and give like terms \(5-(3x+2)+(2+3x)\).
Decision : \(5-(3x+2)+(2+3x)=5-3x-2+2+3x=5\).

Example. Expand the brackets \(5(3-x)\).
Decision : We have \(3\) and \(-x\) in the bracket, and five in front of the bracket. This means that each member of the bracket is multiplied by \ (5 \) - I remind you that the multiplication sign between a number and a bracket in mathematics is not written to reduce the size of records.

Example. Expand the brackets \(-2(-3x+5)\).
Decision : As in the previous example, the bracketed \(-3x\) and \(5\) are multiplied by \(-2\).

Example. Simplify the expression: \(5(x+y)-2(x-y)\).
Decision : \(5(x+y)-2(x-y)=5x+5y-2x+2y=3x+7y\).

It remains to consider the last situation.

When multiplying parenthesis by parenthesis, each term of the first parenthesis is multiplied with every term of the second:

\((c+d)(a-b)=c (a-b)+d (a-b)=ca-cb+da-db\)

Example. Expand the brackets \((2-x)(3x-1)\).
Decision : We have a product of brackets and it can be opened immediately using the formula above. But in order not to get confused, let's do everything step by step.
Step 1. Remove the first bracket - each of its members is multiplied by the second bracket:

Step 2. Expand the products of the bracket by the factor as described above:
- the first one first...

Then the second.

Step 3. Now we multiply and bring like terms:

It is not necessary to paint all the transformations in detail, you can immediately multiply. But if you are just learning to open brackets - write in detail, there will be less chance of making a mistake.

Note to the entire section. In fact, you don't need to remember all four rules, you only need to remember one, this one: \(c(a-b)=ca-cb\) . Why? Because if we substitute one instead of c, we get the rule \((a-b)=a-b\) . And if we substitute minus one, we get the rule \(-(a-b)=-a+b\) . Well, if you substitute another bracket instead of c, you can get the last rule.

parenthesis within parenthesis

Sometimes in practice there are problems with brackets nested inside other brackets. Here is an example of such a task: to simplify the expression \(7x+2(5-(3x+y))\).

To be successful in these tasks, you need to:
- carefully understand the nesting of brackets - which one is in which;
- open the brackets sequentially, starting, for example, with the innermost one.

It is important when opening one of the brackets don't touch the rest of the expression, just rewriting it as is.
Let's take the task above as an example.

Example. Open the brackets and give like terms \(7x+2(5-(3x+y))\).
Decision:

Example. Expand the brackets and give like terms \(-(x+3(2x-1+(x-5)))\).
Decision :

\(-(x+3(2x-1\)\(+(x-5)\) \())\)		This is a triple nesting of parentheses. We start with the innermost one (highlighted in green). There is a plus in front of the parenthesis, so it is simply removed.
\(-(x+3(2x-1\)\(+x-5\) \())\)		Now you need to open the second bracket, intermediate. But before that, we will simplify the expression by ghosting similar terms in this second bracket.
\(=-(x\)\(+3(3x-6)\) \()=\)		Now we open the second bracket (highlighted in blue). There is a multiplier in front of the parenthesis - so each term in the parenthesis is multiplied by it.
\(=-(x\)\(+9x-18\) \()=\)
		And open the last parenthesis. Before the bracket minus - so all the signs are reversed.

Bracket opening is a basic skill in mathematics. Without this skill, it is impossible to have a grade above three in grades 8 and 9. Therefore, I recommend a good understanding of this topic.

In this video, we will analyze a whole set of linear equations that are solved using the same algorithm - that's why they are called the simplest.

To begin with, let's define: what is a linear equation and which of them should be called the simplest?

A linear equation is one in which there is only one variable, and only in the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest ones using the algorithm:

1. Open brackets, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Bring like terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$ .

Of course, this algorithm does not always help. The fact is that sometimes, after all these machinations, the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when you get something like $0\cdot x=8$, i.e. on the left is zero, and on the right is a non-zero number. In the video below, we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

And now let's see how it all works on the example of real problems.

Examples of solving equations

Today we deal with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to open the parentheses, if any (as in our last example);
2. Then bring similar
3. Finally, isolate the variable, i.e. everything that is connected with the variable - the terms in which it is contained - is transferred to one side, and everything that remains without it is transferred to the other side.

Then, as a rule, you need to bring similar on each side of the resulting equality, and after that it remains only to divide by the coefficient at "x", and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Usually, mistakes are made either when opening brackets, or when counting "pluses" and "minuses".

In addition, it happens that a linear equation has no solutions at all, or so that the solution is the entire number line, i.e. any number. We will analyze these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.

Scheme for solving simple linear equations

To begin with, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the parentheses, if any.
2. Seclude variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
3. We present similar terms.
4. We divide everything by the coefficient at "x".

Of course, this scheme does not always work, it has certain subtleties and tricks, and now we will get to know them.

Solving real examples of simple linear equations

Task #1

In the first step, we are required to open the brackets. But they are not in this example, so we skip this step. In the second step, we need to isolate the variables. Please note: we are talking only about individual terms. Let's write:

We give similar terms on the left and on the right, but this has already been done here. Therefore, we proceed to the fourth step: divide by a factor:

\[\frac(6x)(6)=-\frac(72)(6)\]

Here we got the answer.

Task #2

In this task, we can observe the brackets, so let's expand them:

Both on the left and on the right, we see approximately the same construction, but let's act according to the algorithm, i.e. sequester variables:

Here are some like:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task #3

The third linear equation is already more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they just have different signs in front of them. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's calculate:

We perform the last step - we divide everything by the coefficient at "x":

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, then I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, zero can get in among them - there is nothing wrong with that.

Zero is the same number as the rest, you should not somehow discriminate it or assume that if you get zero, then you did something wrong.

Another feature is related to the expansion of parentheses. Please note: when there is a “minus” in front of them, we remove it, but in brackets we change the signs to opposite. And then we can open it according to standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such actions is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complicated and a quadratic function will appear when performing various transformations. However, you should not be afraid of this, because if, according to the author's intention, we solve a linear equation, then in the process of transformation all monomials containing a quadratic function will necessarily be reduced.

Example #1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some like:

Obviously, this equation has no solutions, so in the answer we write as follows:

\[\variety \]

or no roots.

Example #2

We perform the same steps. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some like:

Obviously, this linear equation has no solution, so we write it like this:

\[\varnothing\],

or no roots.

Nuances of the solution

Both equations are completely solved. On the example of these two expressions, we once again made sure that even in the simplest linear equations, everything can be not so simple: there can be either one, or none, or infinitely many. In our case, we considered two equations, in both there are simply no roots.

But I would like to draw your attention to another fact: how to work with brackets and how to expand them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by "x". Please note: multiply each individual term. Inside there are two terms - respectively, two terms and is multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can the bracket be opened from the point of view that there is a minus sign after it. Yes, yes: only now, when the transformations are done, we remember that there is a minus sign in front of the brackets, which means that everything below just changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is no coincidence that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and learn to solve such simple equations again.

Of course, the day will come when you will hone these skills to automatism. You no longer have to perform so many transformations each time, you will write everything in one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task #1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do a retreat:

Here are some like:

Let's do the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, however, they mutually canceled out, which makes the equation exactly linear, not square.

Task #2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's do the first step carefully: multiply every element in the first bracket by every element in the second. In total, four new terms should be obtained after transformations:

And now carefully perform the multiplication in each term:

Let's move the terms with "x" to the left, and without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

We have received a definitive answer.

Nuances of the solution

The most important remark about these two equations is this: as soon as we start multiplying brackets in which there is more than a term, then this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we get four terms.

On the algebraic sum

With the last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: we subtract seven from one. In algebra, we mean by this the following: to the number "one" we add another number, namely "minus seven." This algebraic sum differs from the usual arithmetic sum.

As soon as when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

In conclusion, let's look at a couple more examples that will be even more complex than the ones we just looked at, and in order to solve them, we will have to slightly expand our standard algorithm.

Solving equations with a fraction

To solve such tasks, one more step will have to be added to our algorithm. But first, I will remind our algorithm:

1. Open brackets.
2. Separate variables.
3. Bring similar.
4. Divide by a factor.

Alas, this wonderful algorithm, for all its efficiency, is not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on the left and on the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be performed both before the first action and after it, namely, to get rid of fractions. Thus, the algorithm will be as follows:

1. Get rid of fractions.
2. Open brackets.
3. Separate variables.
4. Bring similar.
5. Divide by a factor.

What does it mean to "get rid of fractions"? And why is it possible to do this both after and before the first standard step? In fact, in our case, all fractions are numeric in terms of the denominator, i.e. everywhere the denominator is just a number. Therefore, if we multiply both parts of the equation by this number, then we will get rid of fractions.

Example #1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two brackets doesn't mean you have to multiply each of them by "four". Let's write:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's open it:

We perform seclusion of a variable:

We carry out the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, we pass to the second equation.

Example #2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

Problem solved.

That, in fact, is all that I wanted to tell today.

Key points

The key findings are as follows:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Do not worry if you have quadratic functions somewhere, most likely, in the process of further transformations, they will be reduced.
• The roots in linear equations, even the simplest ones, are of three types: one single root, the entire number line is a root, there are no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site, solve the examples presented there. Stay tuned, there are many more interesting things waiting for you!

"Opening brackets" - Mathematics textbook Grade 6 (Vilenkin)

Short description:

In this section, you will learn how to open parentheses in examples. What is it for? All for the same as before - to make it easier and easier for you to count, to make fewer mistakes, and ideally (your math teacher's dream) in order to solve everything without mistakes at all.
You already know that brackets in a mathematical notation are placed if two mathematical signs go in a row, if we want to show the union of numbers, their rearrangement. To expand brackets means to get rid of extra characters. For example: (-15)+3=-15+3=-12, 18+(-16)=18-16=2. Do you remember the distributive property of multiplication with respect to addition? After all, in that example, we also got rid of the brackets to simplify calculations. The named property of multiplication can also be applied to four, three, five or more terms. For example: 15*(3+8+9+6)=15*3+15*8+15*9+15*6=390. Have you noticed that when opening brackets, the numbers in them do not change sign if the number in front of the brackets is positive? After all, fifteen is a positive number. And if you solve this example: -15*(3+8+9+6)=-15*3+(-15)*8+(-15)*9+(-15)*6=-45+(- 120)+(-135)+(-90)=-45-120-135-90=-390. We had a negative number minus fifteen in front of the brackets, when we opened the brackets all the numbers began to change their sign to another - the opposite - from plus to minus.
Based on the above examples, two basic rules for opening brackets can be voiced:
1. If you have a positive number in front of the brackets, then after opening the brackets, all signs of the numbers in the brackets do not change, but remain exactly the same as they were.
2. If you have a negative number in front of the brackets, then after opening the brackets, the minus sign is no longer written, and the signs of all absolute numbers in the brackets are sharply reversed.
For example: (13+8)+(9-8)=13+8+9-8=22; (13+8)-(9-8)=13+8-9+8=20. Let's complicate our examples a bit: (13+8)+2(9-8)=13+8+2*9-2*8=21+18-16=23. You noticed that opening the second brackets, we multiplied by 2, but the signs remained the same as they were. And here is an example: (3+8)-2*(9-8)=3+8-2*9+2*8=11-18+16=9, in this example the number two is negative, it is before the brackets stands with a minus sign, therefore, opening them, we changed the signs of the numbers to the opposite ones (nine was with a plus, it became with a minus, eight was with a minus, it became with a plus).

In the fifth century BC, the ancient Greek philosopher Zeno of Elea formulated his famous aporias, the most famous of which is the aporia "Achilles and the tortoise". Here's how it sounds:

Let's say Achilles runs ten times faster than the tortoise and is a thousand paces behind it. During the time during which Achilles runs this distance, the tortoise crawls a hundred steps in the same direction. When Achilles has run a hundred steps, the tortoise will crawl another ten steps, and so on. The process will continue indefinitely, Achilles will never catch up with the tortoise.

This reasoning became a logical shock for all subsequent generations. Aristotle, Diogenes, Kant, Hegel, Gilbert... All of them, in one way or another, considered Zeno's aporias. The shock was so strong that " ... discussions continue at the present time, the scientific community has not yet managed to come to a common opinion about the essence of paradoxes ... mathematical analysis, set theory, new physical and philosophical approaches were involved in the study of the issue; none of them became a universally accepted solution to the problem ..."[Wikipedia," Zeno's Aporias "]. Everyone understands that they are being fooled, but no one understands what the deception is.

From the point of view of mathematics, Zeno in his aporia clearly demonstrated the transition from the value to. This transition implies applying instead of constants. As far as I understand, the mathematical apparatus for applying variable units of measurement has either not yet been developed, or it has not been applied to Zeno's aporia. The application of our usual logic leads us into a trap. We, by the inertia of thinking, apply constant units of time to the reciprocal. From a physical point of view, it looks like time slowing down to a complete stop at the moment when Achilles catches up with the tortoise. If time stops, Achilles can no longer overtake the tortoise.

If we turn the logic we are used to, everything falls into place. Achilles runs at a constant speed. Each subsequent segment of its path is ten times shorter than the previous one. Accordingly, the time spent on overcoming it is ten times less than the previous one. If we apply the concept of "infinity" in this situation, then it would be correct to say "Achilles will infinitely quickly overtake the tortoise."

How to avoid this logical trap? Remain in constant units of time and do not switch to reciprocal values. In Zeno's language, it looks like this:

In the time it takes Achilles to run a thousand steps, the tortoise crawls a hundred steps in the same direction. During the next time interval, equal to the first, Achilles will run another thousand steps, and the tortoise will crawl one hundred steps. Now Achilles is eight hundred paces ahead of the tortoise.

This approach adequately describes reality without any logical paradoxes. But this is not a complete solution to the problem. Einstein's statement about the insurmountability of the speed of light is very similar to Zeno's aporia "Achilles and the tortoise". We have yet to study, rethink and solve this problem. And the solution must be sought not in infinitely large numbers, but in units of measurement.

Another interesting aporia of Zeno tells of a flying arrow:

A flying arrow is motionless, since at each moment of time it is at rest, and since it is at rest at every moment of time, it is always at rest.

In this aporia, the logical paradox is overcome very simply - it is enough to clarify that at each moment of time the flying arrow is at rest at different points in space, which, in fact, is movement. There is another point to be noted here. From one photograph of a car on the road, it is impossible to determine either the fact of its movement or the distance to it. To determine the fact of the movement of the car, two photographs taken from the same point at different points in time are needed, but they cannot be used to determine the distance. To determine the distance to the car, you need two photographs taken from different points in space at the same time, but you cannot determine the fact of movement from them (naturally, you still need additional data for calculations, trigonometry will help you). What I want to point out in particular is that two points in time and two points in space are two different things that should not be confused as they provide different opportunities for exploration.

Wednesday, July 4, 2018

Very well the differences between set and multiset are described in Wikipedia. We look.

As you can see, "the set cannot have two identical elements", but if there are identical elements in the set, such a set is called a "multiset". Reasonable beings will never understand such logic of absurdity. This is the level of talking parrots and trained monkeys, in which the mind is absent from the word "completely." Mathematicians act as ordinary trainers, preaching their absurd ideas to us.

Once upon a time, the engineers who built the bridge were in a boat under the bridge during the tests of the bridge. If the bridge collapsed, the mediocre engineer died under the rubble of his creation. If the bridge could withstand the load, the talented engineer built other bridges.

No matter how mathematicians hide behind the phrase "mind me, I'm in the house", or rather "mathematics studies abstract concepts", there is one umbilical cord that inextricably connects them with reality. This umbilical cord is money. Let us apply mathematical set theory to mathematicians themselves.

We studied mathematics very well and now we are sitting at the cash desk, paying salaries. Here a mathematician comes to us for his money. We count the whole amount to him and lay it out on our table into different piles, in which we put bills of the same denomination. Then we take one bill from each pile and give the mathematician his "mathematical salary set". We explain the mathematics that he will receive the rest of the bills only when he proves that the set without identical elements is not equal to the set with identical elements. This is where the fun begins.

First of all, the deputies' logic will work: "you can apply it to others, but not to me!" Further, assurances will begin that there are different banknote numbers on banknotes of the same denomination, which means that they cannot be considered identical elements. Well, we count the salary in coins - there are no numbers on the coins. Here the mathematician will frantically recall physics: different coins have different amounts of dirt, the crystal structure and arrangement of atoms for each coin is unique ...

And now I have the most interesting question: where is the boundary beyond which elements of a multiset turn into elements of a set and vice versa? Such a line does not exist - everything is decided by shamans, science here is not even close.

Look here. We select football stadiums with the same field area. The area of ​​the fields is the same, which means we have a multiset. But if we consider the names of the same stadiums, we get a lot, because the names are different. As you can see, the same set of elements is both a set and a multiset at the same time. How right? And here the mathematician-shaman-shuller takes out a trump ace from his sleeve and begins to tell us about either a set or a multiset. In any case, he will convince us that he is right.

To understand how modern shamans operate with set theory, tying it to reality, it is enough to answer one question: how do the elements of one set differ from the elements of another set? I will show you, without any "conceivable as not a single whole" or "not conceivable as a single whole."

Sunday, March 18, 2018

The sum of the digits of a number is a dance of shamans with a tambourine, which has nothing to do with mathematics. Yes, in mathematics lessons we are taught to find the sum of the digits of a number and use it, but they are shamans for that, to teach their descendants their skills and wisdom, otherwise shamans will simply die out.

Do you need proof? Open Wikipedia and try to find the "Sum of Digits of a Number" page. She doesn't exist. There is no formula in mathematics by which you can find the sum of the digits of any number. After all, numbers are graphic symbols with which we write numbers, and in the language of mathematics, the task sounds like this: "Find the sum of graphic symbols representing any number." Mathematicians cannot solve this problem, but shamans can do it elementarily.

Let's figure out what and how we do in order to find the sum of the digits of a given number. And so, let's say we have the number 12345. What needs to be done in order to find the sum of the digits of this number? Let's consider all the steps in order.

1. Write down the number on a piece of paper. What have we done? We have converted the number to a number graphic symbol. This is not a mathematical operation.

2. We cut one received picture into several pictures containing separate numbers. Cutting a picture is not a mathematical operation.

3. Convert individual graphic characters to numbers. This is not a mathematical operation.

4. Add up the resulting numbers. Now that's mathematics.

The sum of the digits of the number 12345 is 15. These are the "cutting and sewing courses" from shamans used by mathematicians. But that is not all.

From the point of view of mathematics, it does not matter in which number system we write the number. So, in different number systems, the sum of the digits of the same number will be different. In mathematics, the number system is indicated as a subscript to the right of the number. With a large number 12345, I don’t want to fool my head, consider the number 26 from the article about. Let's write this number in binary, octal, decimal and hexadecimal number systems. We will not consider each step under a microscope, we have already done that. Let's look at the result.

As you can see, in different number systems, the sum of the digits of the same number is different. This result has nothing to do with mathematics. It's the same as if you would get completely different results when determining the area of ​​a rectangle in meters and centimeters.

Zero in all number systems looks the same and has no sum of digits. This is another argument in favor of the fact that . A question for mathematicians: how is it denoted in mathematics that which is not a number? What, for mathematicians, nothing but numbers exists? For shamans, I can allow this, but for scientists, no. Reality is not just about numbers.

The result obtained should be considered as proof that number systems are units of measurement of numbers. After all, we cannot compare numbers with different units of measurement. If the same actions with different units of measurement of the same quantity lead to different results after comparing them, then this has nothing to do with mathematics.

What is real mathematics? This is when the result of a mathematical action does not depend on the value of the number, the unit of measurement used, and on who performs this action.

Sign on the door Opens the door and says:

Ouch! Isn't this the women's restroom?
- Young woman! This is a laboratory for studying the indefinite holiness of souls upon ascension to heaven! Nimbus on top and arrow up. What other toilet?

Female... A halo on top and an arrow down is male.

If you have such a work of design art flashing before your eyes several times a day,

Then it is not surprising that you suddenly find a strange icon in your car:

Personally, I make an effort on myself to see minus four degrees in a pooping person (one picture) (composition of several pictures: minus sign, number four, degrees designation). And I do not consider this girl a fool who does not know physics. She just has an arc stereotype of perception of graphic images. And mathematicians teach us this all the time. Here is an example.

1A is not "minus four degrees" or "one a". This is "pooping man" or the number "twenty-six" in the hexadecimal number system. Those people who constantly work in this number system automatically perceive the number and letter as one graphic symbol.

Parentheses are used to indicate the order in which actions are performed in numeric and alphabetic expressions, as well as in expressions with variables. It is convenient to pass from an expression with brackets to an identically equal expression without brackets. This technique is called parenthesis opening.

To expand brackets means to rid the expression of these brackets.

Another point deserves special attention, which concerns the peculiarities of writing solutions when opening brackets. We can write the initial expression with brackets and the result obtained after opening the brackets as equality. For example, after opening the parentheses, instead of the expression
3−(5−7) we get the expression 3−5+7. We can write both of these expressions as the equality 3−(5−7)=3−5+7.

And one more important point. In mathematics, to reduce entries, it is customary not to write a plus sign if it is the first in an expression or in brackets. For example, if we add two positive numbers, for example, seven and three, then we write not +7 + 3, but simply 7 + 3, despite the fact that seven is also a positive number. Similarly, if you see, for example, the expression (5 + x) - know that there is a plus in front of the bracket, which is not written, and there is a plus + (+5 + x) in front of the five.

Bracket expansion rule for addition

When opening brackets, if there is a plus before the brackets, then this plus is omitted along with the brackets.

Example. Open the brackets in the expression 2 + (7 + 3) Before the brackets plus, then the characters in front of the numbers in the brackets do not change.

2 + (7 + 3) = 2 + 7 + 3

The rule for expanding brackets when subtracting

If there is a minus before the brackets, then this minus is omitted along with the brackets, but the terms that were in the brackets change their sign to the opposite. The absence of a sign before the first term in parentheses implies a + sign.

Example. Open brackets in expression 2 − (7 + 3)

There is a minus before the brackets, so you need to change the signs before the numbers from the brackets. There is no sign in brackets before the number 7, which means that the seven is positive, it is considered that the + sign is in front of it.

2 − (7 + 3) = 2 − (+ 7 + 3)

When opening the brackets, we remove the minus from the example, which was before the brackets, and the brackets themselves 2 − (+ 7 + 3), and change the signs that were in the brackets to the opposite ones.

2 − (+ 7 + 3) = 2 − 7 − 3

Expanding parentheses when multiplying

If there is a multiplication sign in front of the brackets, then each number inside the brackets is multiplied by the factor in front of the brackets. At the same time, multiplying a minus by a minus gives a plus, and multiplying a minus by a plus, like multiplying a plus by a minus, gives a minus.

Thus, parentheses in products are expanded in accordance with the distributive property of multiplication.

Example. 2 (9 - 7) = 2 9 - 2 7

When multiplying parenthesis by parenthesis, each term of the first parenthesis is multiplied with every term of the second parenthesis.

(2 + 3) (4 + 5) = 2 4 + 2 5 + 3 4 + 3 5

In fact, there is no need to remember all the rules, it is enough to remember only one, this one: c(a−b)=ca−cb. Why? Because if we substitute one instead of c, we get the rule (a−b)=a−b. And if we substitute minus one, we get the rule −(a−b)=−a+b. Well, if you substitute another bracket instead of c, you can get the last rule.

Expand parentheses when dividing

If there is a division sign after the brackets, then each number inside the brackets is divisible by the divisor after the brackets, and vice versa.

Example. (9 + 6) : 3=9: 3 + 6: 3

How to expand nested parentheses

If the expression contains nested brackets, then they are expanded in order, starting with external or internal.

At the same time, when opening one of the brackets, it is important not to touch the other brackets, just rewriting them as they are.

Example. 12 - (a + (6 - b) - 3) = 12 - a - (6 - b) + 3 = 12 - a - 6 + b + 3 = 9 - a + b