The word "proportion" comes from the Latin root and means "proportion". People often use it in their daily life. They talk, for example, about the proportions of the human body or about proportions in cooking. Today we will find out what mathematicians mean by this word.
Consider two relationships. We remember that a ratio is a quotient of two numbers.
Note that in both the first and second cases, the quotient value is three. We have before us two equal relationships. Let's write equality.
Fifteen is as much to five as twenty-four to eight. This equality is called proportion. Sometimes this equality is written as an equality of ordinary fractions.
Let's formulate the definition: equality of two relations is called proportion.
With the help of letters, the proportion can be written:
Attitude a To b equal to ratio c To d. Sometimes the proportion is read differently: “ a so refers to b, how c refers to d». The numbers participating in the proportion are called the members of the proportion. All members are considered to be nonzero.
Numbers a and d are called the extreme terms of the proportion, and the numbers b and c- middle members. Indeed, in the first variant of writing the number b and c are in the middle, and the numbers a and d on the edge.
In the previously considered proportion 
find the product of its middle and extreme terms.
Note that the two products obtained are equal.
Let us formulate the main property of proportions in general terms.
In the correct proportion, the product of the extreme terms is equal to the product of the averages.
The converse is also true.
If the product of the extreme terms is equal to the product of the middle terms of the proportion, then the proportioncorrect.
Let's find the unknown term of the proportion, that is, we will solve the proportion.
The numbers 0.5 and 13 are extreme terms; the numbers a and 2 are the middle terms. Let's use the main property of proportion.
Let's solve the proportion.
Using the main property of proportion, we get:
To get rid of the decimal fraction in the denominator, multiply both the numerator and the denominator of the fraction by 10. Reduce the resulting fraction by 4, and then again by 4.
Check if the given proportions are correct:
In this task, you need to check whether the equality between the relations is actually true.
Let's find the product of the means and the product of the extreme terms for each proportion. If the products obtained are equal, then the proportion is correct. Otherwise, the proportion is incorrect.
correct proportion, because.
incorrect proportion, because .
If you swap the middle or extreme terms in the correct proportion, then the resulting new proportions are also correct.
This is because with such a permutation the product of the extreme and middle terms does not change.
Let's look at an example. From this proportion, get two new ones by rearranging the extreme and middle terms. First, we rearrange the middle terms (Fig. 1).
Rice. 1. Rearrangement of middle terms
Indeed, the product of the middle and extreme has not changed, which means that the resulting proportion is correct. Let us rearrange the extreme terms (Fig. 2).
Rice. 2. Rearrangement of extreme terms
And in this case, the product of the middle and extreme did not change. We got the right proportion.
Bibliography
- Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M .: Mnemosina, 2012.
- Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics grade 6. - Gymnasium. 2006.
- Depman I. Ya., Vilenkin N. Ya. Behind the pages of a mathematics textbook. - M .: Education, 1989.
- Rurukin A.N., Tchaikovsky I.V. Assignments for the course mathematics grade 5-6. - M .: ZSH MEPhI, 2011.
- Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for 6th grade students of the MEPhI correspondence school. - M .: ZSH MEPhI, 2011.
- Shevrin L.N., Gein A.G., Koryakov I.O., Volkov M.V. Mathematics: Textbook-companion for grades 5-6 of high school. - M .: Education, Library of the teacher of mathematics, 1989.
- Maths ().
- Internet portal Math-portal.ru ().
Homework
- Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M .: Mnemosina, 2012: No. 762 (a, d, e), No. 765, No. 777.
- Other assignments: No. 767, No. 775.
The equality of two relationships is called proportion.
a: b = c: d... This is proportion. Read: a so refers to b, how c refers to d... Numbers a and d are called extreme terms of proportion, and the numbers b and c – average members of the proportion.
Proportion example: 1 2 : 3 = 16 : 4 . This is the equality of the two ratios: 12: 3 = 4 and 16: 4 = 4 . They read: twelve refers to three as sixteen refers to four. Here 12 and 4 are the extreme terms of the proportion, and 3 and 16 are the middle terms of the proportion.
The main property of proportion.
The product of the extreme terms of the proportion is equal to the product of its middle terms.
For proportion a: b = c: d or a / b = c / d the main property is written like this: a d = b c.
For our proportion 12: 3 = 16: 4, the main property will be written as follows: 12 4 = 3 16 ... It turns out the correct equality: 48 = 48 .
To find the unknown extreme term of the proportion, you need to divide the product of the middle terms of the proportion by the known extreme term.
Examples.
1) x: 20 = 2: 5... We have NS and 5 Are the extreme terms of the proportion, and 20 and 2 - medium.
Solution.
x = (20 2): 5- you need to multiply the middle terms ( 20 and 2 ) and divide the result by the known extreme term (the number 5 );
x = 40: 5- the product of the middle terms ( 40 ) we divide by the known extreme term ( 5 );
x = 8. Got the desired extreme term of the proportion.
It is more convenient to write down the finding of the unknown member of the proportion using an ordinary fraction. This is how the example we considered would then be written:
The desired extreme term of the proportion ( NS) will be equal to the product of the middle terms ( 20 and 2 ) divided by the known extreme term ( 5 ).
Reduce the fraction by 5 (divide by 5 NS.
Still such examples for finding the unknown extreme term of the proportion.
To find the unknown average term of the proportion, you need to divide the product of the extreme terms of the proportion by the known average term.
Examples. Find the unknown mean term of the proportion.
5) 9: x = 3: 14. Number 3 Is the known middle term of a given proportion, number 9 and 14 - extreme members of the proportion.
Solution.
x = (914): 3 - we multiply the extreme terms of the proportion and divide the result by the known middle term of the proportion;
x = 136: 3;
x = 42.
The solution to this example can be written differently:
The sought-for middle term of the proportion ( NS) will be equal to the product of extreme terms ( 9
and 14
) divided by the known middle term ( 3
).
Reduce the fraction by 3 (divide by 3 and the numerator and denominator of the fraction). Find the value NS.
If you forgot how to reduce ordinary fractions, then repeat the topic: ""
Still such examples for finding the unknown middle term of the proportion.
The ratio of two numbers
Definition 1
By the ratio of two numbers is their private.
Example 1
the ratio of $ 18 $ to $ 3 $ can be written as:
$ 18 \ div 3 = \ frac (18) (3) = 6 $.
the ratio of $ 5 to $ 15 $ can be written as:
$ 5 \ div 15 = \ frac (5) (15) = \ frac (1) (3) $.
By using ratio of two numbers you can show:
- how many times one number exceeds another;
- which part represents one number from another.
When compiling the ratio of two numbers in the denominator of the fraction, write down the number with which the comparison is made.
Most often, such a number follows the words "in comparison with ..." or the preposition "to ...".
Let's recall the basic property of a fraction and apply it to a relation:
Remark 1
When multiplying or dividing both terms of the ratio by the same number, other than zero, we get the ratio, which is equal to the original one.
Consider an example that illustrates the use of the relationship between two numbers.
Example 2
The amount of precipitation in the previous month was $ 195 mm, and in the current month - $ 780 mm. How many times has the amount of precipitation increased in the current month compared to the previous month?
Solution.
Let's compose the ratio of the amount of precipitation in the current month to the amount of precipitation in the previous month:
$ \ frac (780) (195) = \ frac (780 \ div 5) (195 \ div 5) = \ frac (156 \ div 3) (39 \ div 3) = \ frac (52) (13) = 4 $.
Answer: The amount of precipitation in the current month is $ 4 $ times more than in the previous month.
Example 3
Find how many times $ 1 \ frac (1) (2) $ is contained in $ 13 \ frac (1) (2) $.
Solution.
$ 13 \ frac (1) (2) \ div 1 \ frac (1) (2) = \ frac (27) (2) \ div \ frac (3) (2) = \ frac (27) (2) \ cdot \ frac (2) (3) = \ frac (27) (3) = 9 $.
Answer: $ 9 $ times.
The concept of proportion
Definition 2
Proportion the equality of two relations is called:
$ a \ div b = c \ div d $
$ \ frac (a) (b) = \ frac (c) (d) $.
Example 4
$ 3 \ div 6 = 9 \ div 18 $, $ 5 \ div 15 = 9 \ div 27 $, $ 4 \ div 2 = 24 \ div 12 $,
$ \ frac (8) (2) = \ frac (36) (9) $, $ \ frac (10) (40) = \ frac (9) (36) $, $ \ frac (15) (75) = \ frac (1) (5) $.
In the proportion $ \ frac (a) (b) = \ frac (c) (d) $ (or $ a: b = c \ div d $) the numbers a and d are called extreme members proportions, and the numbers $ b $ and $ c $ - middle members proportions.
The correct proportion can be converted as follows:
Remark 2
The product of the extreme terms of the correct proportion is equal to the product of the middle terms:
$ a \ cdot d = b \ cdot c $.
This statement is the main property of proportion.
The converse is also true:
Remark 3
If the product of the extreme terms of the proportion is equal to the product of its middle terms, then the proportion is correct.
Remark 4
If you rearrange the middle members or extreme members in the correct proportion, then the proportions that will turn out will also be correct.
Example 5
$ 6 \ div 3 = 18 \ div 9 $, $ 15 \ div 5 = 27 \ div 9 $, $ 2 \ div 4 = 12 \ div 24 $,
$ \ frac (2) (8) = \ frac (9) (36) $, $ \ frac (40) (10) = \ frac (36) (9) $, $ \ frac (75) (15) = \ frac (5) (1) $.
Using this property, it is easy to find the unknown term from proportion if the other three are known:
$ a = \ frac (b \ cdot c) (d) $; $ b = \ frac (a \ cdot d) (c) $; $ c = \ frac (a \ cdot d) (b) $; $ d = \ frac (b \ cdot c) (a) $.
Example 6
$ \ frac (6) (a) = \ frac (16) (8) $;
$ 6 \ cdot 8 = 16 \ cdot a $;
$ 16 \ cdot a = 6 \ cdot 8 $;
$ 16 \ cdot a = 48 $;
$ a = \ frac (48) (16) $;
Example 7
$ \ frac (a) (21) = \ frac (8) (24) $;
$ a \ cdot 24 = 21 \ cdot 8 $;
$ a \ cdot 24 = 168 $;
$ a = \ frac (168) (24) $;
$ 3 for the gardener - $ 108 for trees;
$ x $ Gardeners - $ 252 Wood.
Let's make the proportion:
$ \ frac (3) (x) = \ frac (108) (252) $.
Let's use the rule for finding the unknown member of the proportion:
$ b = \ frac (a \ cdot d) (c) $;
$ x = \ frac (3 \ cdot 252) (108) $;
$ x = \ frac (252) (36) $;
Answer: It will take $ 7 for gardeners to prune $ 252 trees.
Most often, the proportion properties are used in practice in mathematical calculations in cases where it is necessary to calculate the value of an unknown proportional term, if the values of the other three terms are known.
To solve most problems in high school mathematics, knowledge of proportioning is required. This simple skill will help you not only perform complex exercises from the textbook, but also delve into the very essence of mathematics. How to make the proportion? Let's take a look at it now.
The simplest example is a problem where three parameters are known, and the fourth must be found. The proportions are, of course, different, but often you need to find some number by percentage. For example, the boy had ten apples in total. He gave the fourth part to his mother. How many apples does the boy have left? This is the simplest example that will allow you to compose the proportion. The main thing is to do it. There were originally ten apples. Let it be 100%. We marked all his apples. He gave away one fourth. 1/4 = 25/100. This means that he has left: 100% (it was originally) - 25% (he gave) = 75%. This figure shows the percentage of the number of fruits remaining to the number of the first available. Now we have three numbers, by which it is already possible to solve the proportion. 10 apples - 100%, NS apples - 75%, where x is the required amount of fruit. How to make the proportion? You need to understand what it is. Mathematically, it looks like this. The equal sign is put for your understanding.
10 apples = 100%;
x apples = 75%.
It turns out that 10 / x = 100% / 75. This is the main property of proportions. After all, the larger x, the more percent this number is from the original. We solve this proportion and we get that x = 7.5 apples. Why the boy decided to give a non-integer amount is unknown to us. Now you know how to proportion. The main thing is to find two relations, one of which contains the unknown unknown.
Solving proportions often comes down to simple multiplication, and then to division. In schools, children are not explained why this is exactly the case. While it is important to understand that proportional relationships are a mathematical classic, it is the very essence of science. To solve proportions, you need to be able to handle fractions. For example, you often have to convert percentages to fractions. That is, a 95% record will not work. And if you write 95/100 right away, then you can make solid reductions without starting the main count. It should be said right away that if your proportion turned out to be with two unknowns, then it cannot be solved. No professor can help you here. And your task, most likely, has a more complex algorithm of correct actions.
Consider another example where there is no interest. The motorist bought 5 liters of gasoline for 150 rubles. He wondered how much he would pay for 30 liters of fuel. To solve this problem, let x denote the required amount of money. You can solve this problem yourself and then check the answer. If you have not yet figured out how to make the proportion, then take a look. 5 liters of gasoline is 150 rubles. As in the first example, we will write down 5L - 150r. Now let's find the third number. Of course, this is 30 liters. Agree that a pair of 30 liters - x rubles is appropriate in this situation. Let's move on to mathematical language.
5 liters - 150 rubles;
30 liters - x rubles;
We solve this proportion:
x = 900 rubles.
So we decided. In your task, do not forget to check the adequacy of the answer. It happens that with the wrong decision, cars reach unrealistic speeds of 5000 kilometers per hour and so on. Now you know how to proportion. You can also solve it. As you can see, this is not difficult.
§ 125. The concept of proportion.
Proportion is the equality of two relationships. Here are examples of equalities called proportions:
Note. The names of the quantities in the proportions are not indicated.
The proportions are usually read as follows: 2 refers to 1 (one), as 10 refers to 5 (the first proportion). You can read it differently, for example: 2 is as many times more than 1, how many times is 10 more than 5. The third proportion can be read as follows: - 0.5 is as many times less than 2, how many times 0.75 is less than 3.
The numbers included in the proportion are called members of the proportion... Hence, the proportion consists of four members. The first and last terms, that is, the members at the edges, are called extreme, and the members of the proportion in the middle are called average members. This means that in the first proportion, the numbers 2 and 5 will be extreme terms, and the numbers 1 and 10 will be the middle terms of the proportion.
§ 126. The main property of proportion.
Consider the proportion:
Let's multiply separately its extreme and middle terms. The product of the extreme 6 4 = 24, the product of the average 3 8 = 24.
Consider another proportion: 10: 5 = 12: 6. Multiply here, too, the extreme and middle terms.
The product of the extreme 10 6 = 60, the product of the average 5 12 = 60.
The main property of proportion: the product of the extreme members of the proportion is equal to the product of its average members.
In general, the main property of proportion is written as follows: ad = bc .
Let's check it in several proportions:
1) 12: 4 = 30: 10.
This proportion is correct, since the relations from which it is composed are equal. At the same time, taking the product of the extreme terms of the proportion (12 10) and the product of its average terms (4 30), we will see that they are equal to each other, i.e.
12 10 = 4 30.
2) 1 / 2: 1 / 48 = 20: 5 / 6
The proportion is correct, as can be easily seen by simplifying the first and second relationships. The main property of proportion will take the form:
1 / 2 5 / 6 = 1 / 48 20
It is easy to make sure that if we write an equality in which on the left side there is the product of some two numbers, and on the right side the product of two other numbers, then from these four numbers you can make up a proportion.
Suppose we have an equality, which includes four numbers, multiplied in pairs:
these four numbers can be members of the proportion, which is easy to write if the first product is taken as the product of extreme terms, and the second as the product of averages. Published equality can be made, for example, the following proportion:
In general, from the equality ad = bc you can get the following proportions:
Do the following exercise yourself. Having the product of two pairs of numbers, write the proportion corresponding to each equality:
a) 1 6 = 2 3;
b) 2 15 = b 5.
§ 127. Calculation of unknown terms of proportion.
The main property of proportion allows you to calculate any of the terms of the proportion, if it is unknown. Let's take the proportion:
NS : 4 = 15: 3.
One extreme term is unknown in this proportion. We know that in any proportion the product of the extreme terms is equal to the product of the middle terms. On this basis, we can write:
x 3 = 4 15.
After multiplying 4 by 15, we can rewrite this equality like this:
NS 3 = 60.
Consider this equality. In it, the first factor is unknown, the second factor is known, and the product is known. We know that to find an unknown factor, it is enough to divide the product by another (known) factor. Then it will turn out:
NS = 60: 3, or NS = 20.
Let us check the found result by substituting the number 20 instead of NS in this proportion:
The proportion is correct.
Let's think about what actions we had to perform to calculate the unknown extreme term of the proportion. Of the four terms of the proportion, only one extreme was unknown to us; the two middle and the second extreme were known. To find the extreme term of the proportion, we first multiplied the middle terms (4 and 15), and then divided the found product by the known extreme term. Now we will show that the actions would not change if the desired extreme term of the proportion was not in the first place, but in the last one. Let's take the proportion:
70: 10 = 21: NS .
Let's write down the main property of the proportion: 70 NS = 10 21.
Multiplying the numbers 10 and 21, we rewrite the equality in this form:
70 NS = 210.
Here one factor is unknown; to calculate it, it is sufficient to divide the product (210) by another factor (70),
NS = 210: 70; NS = 3.
So we can say that each extreme term of the proportion is equal to the product of the averages divided by the other extreme.
We now turn to the calculation of the unknown mean term. Let's take the proportion:
30: NS = 27: 9.
Let's write the main property of proportion:
30 9 = NS 27.
Let's calculate the product of 30 by 9 and rearrange the parts of the last equality:
NS 27 = 270.
Let's find the unknown factor:
NS = 270: 27, or NS = 10.
Let's check by substitution:
30: 10 = 27: 9. The proportion is correct.
Let's take another proportion:
12: b = NS : 8. Let's write the main property of the proportion:
12 . 8 = 6 NS ... Multiplying 12 and 8 and rearranging the parts of the equality, we get:
6 NS = 96. Find the unknown factor:
NS = 96: 6, or NS = 16.
Thus, each middle term of the proportion is equal to the product of the extreme, divided by the other average.
Find the unknown terms of the following proportions:
1) a : 3= 10:5; 3) 2: 1 / 2 = x : 5;
2) 8: b = 16: 4; 4) 4: 1 / 3 = 24: NS .
The last two rules can be summarized as follows:
1) If the proportion looks like:
x: a = b: c , then
2) If the proportion looks like:
a: x = b: c , then
§ 128. Simplification of proportion and rearrangement of its members.
In this section, we will deduce the rules to simplify the proportion in the case when it includes large numbers or fractional terms. Among the transformations that do not violate the proportion are the following:
1. Simultaneous increase or decrease of both members of any relationship by the same number of times.
EXAMPLE 40: 10 = 60: 15.
Having increased by 3 times both terms of the first relation, we get:
120:30 = 60: 15.
The proportion was not violated.
Reducing both terms of the second relation by 5 times, we get:
We got the correct proportion again.
2. Simultaneous increase or decrease of both previous or both subsequent members by the same number of times.
Example. 16: 8 = 40:20.
Let's double the previous terms of both relations:
We got the right proportion.
Let us reduce by 4 times the subsequent terms of both relations:
The proportion was not violated.
The two conclusions obtained can be briefly stated as follows: The proportion will not be violated if we simultaneously increase or decrease by the same number of times any extreme term of the proportion and any average.
For example, having reduced by 4 times the 1st extreme and 2nd middle terms of the proportion 16: 8 = 40:20, we get:
3. Simultaneous increase or decrease of all members of the proportion by the same number of times. Example. 36:12 = 60:20. Let's double all four numbers:
The proportion was not violated. Let's reduce all four numbers by 4 times:
The proportion is correct.
The listed transformations make it possible, firstly, to simplify the proportions, and secondly, to free them from fractional terms. Here are some examples.
1) Let there be a proportion:
200: 25 = 56: x .
In it, the members of the first relation are relatively large numbers, and if we wanted to find the value NS , then we would have to perform calculations on these numbers; but we know that the proportion will not be violated if both terms of the relationship are divided by the same number. Let's divide each of them by 25. The proportion will take the form:
8:1 = 56: x .
We have thus obtained a more comfortable proportion, from which NS can be found in the mind:
2) Take the proportion:
2: 1 / 2 = 20: 5.
In this proportion there is a fractional term (1/2), from which you can get rid of. To do this, you will have to multiply this term, for example, by 2. But we have no right to increase the average term of the proportion; one of the extreme members must be increased with it; then the proportion will not be violated (based on the first two points). Let's increase the first of the extreme terms
(2 2): (2 1/2) = 20: 5, or 4: 1 = 20: 5.
Let's increase the second extreme term:
2: (2 1/2) = 20: (2 5), or 2: 1 = 20: 10.
Consider three more examples for freeing a proportion from fractional terms.
Example 1.1 / 4: 3/8 = 20:30.
Let's bring the fractions to a common denominator:
2 / 8: 3 / 8 = 20: 30.
Multiplying both terms of the first relation by 8, we get:
Example 2. 12: 15/14 = 16: 10/7. Let's bring the fractions to a common denominator:
12: 15 / 14 = 16: 20 / 14
We multiply both subsequent terms by 14, we get: 12:15 = 16:20.
Example 3.1 / 2: 1/48 = 20: 5/6.
Multiply all the terms of the proportion by 48:
24: 1 = 960: 40.
When solving problems in which there are some proportions, it is often necessary to rearrange the terms of the proportion for different purposes. Consider which permutations are legal, that is, do not violate the proportions. Let's take the proportion:
3: 5 = 12: 20. (1)
Rearranging the extreme terms in it, we get:
20: 5 = 12:3. (2)
Let us now rearrange the middle terms:
3:12 = 5: 20. (3)
Let us rearrange both extreme and middle terms at the same time:
20: 12 = 5: 3. (4)
All these proportions are correct. Now let's put the first relation in the place of the second, and the second in the place of the first. The proportion will turn out:
12: 20 = 3: 5. (5)
In this proportion, we will make the same permutations that we did earlier, that is, we will first rearrange the extreme terms, then the middle ones, and finally, simultaneously both the extreme and the middle ones. Three more proportions will turn out, which will also be fair:
5: 20 = 3: 12. (6)
12: 3 = 20: 5. (7)
5: 3 = 20: 12. (8)
So, from one given proportion by rearranging, you can get 7 more proportions, which together with this one is 8 proportions.
It is especially easy to discover the validity of all these proportions when writing in letters. The 8 proportions obtained above take the form:
a: b = c: d; c: d = a: b;
d: b = c: a; b: d = a: c;
a: c = b: d; c: a = d: b;
d: c = b: a; b: a = d: c.
It is easy to see that in each of these proportions the main property takes the form:
ad = bc.
Thus, these permutations do not violate the fairness of the proportion and they can be used if necessary.