So, we already know that the numerator and denominator of a fraction can be multiplied and divided by the same number, the fraction will not change. Let's consider three approaches:
Approach one.
To reduce, divide the numerator and denominator by a common divisor. Let's look at examples:
Let's shorten:
In the examples given, we immediately see which divisors to take for reduction. The process is simple - we go through 2,3,4,5 and so on. In most school course examples, this is quite enough. But if it’s a fraction:
Here the process of selecting divisors can take a long time;). Of course, such examples are outside the school curriculum, but you need to be able to cope with them. Below we will look at how this is done. For now, let's get back to the downsizing process.
As discussed above, in order to reduce a fraction, we divided by the common divisor(s) we determined. Everything is correct! One has only to add signs of divisibility of numbers:
- if the number is even, then it is divisible by 2.
- if a number from the last two digits is divisible by 4, then the number itself is divisible by 4.
— if the sum of the digits that make up the number is divisible by 3, then the number itself is divisible by 3. For example, 125031, 1+2+5+0+3+1=12. Twelve is divisible by 3, so 123031 is divisible by 3.
- if the end of a number is 5 or 0, then the number is divisible by 5.
— if the sum of the digits that make up the number is divisible by 9, then the number itself is divisible by 9. For example, 625032 =.> 6+2+5+0+3+2=18. Eighteen is divisible by 9, which means 623032 is divisible by 9.
Second approach.
To put it briefly, in fact, the whole action comes down to factoring the numerator and denominator and then reducing equal factors in the numerator and denominator (this approach is a consequence of the first approach):
Visually, in order to avoid confusion and mistakes, equal factors are simply crossed out. Question - how to factor a number? It is necessary to determine all divisors by searching. This is a separate topic, it is not complicated, look up the information in a textbook or on the Internet. You will not encounter any great problems with factoring numbers that are present in school fractions.
Formally, the reduction principle can be written as follows:
Approach three.
Here is the most interesting thing for the advanced and those who want to become one. Let's reduce the fraction 143/273. Try it yourself! Well, how did it happen quickly? Now look!
We turn it over (we change places of the numerator and denominator). We divide the resulting fraction with a corner and convert it into a mixed number, that is, we select the whole part:
It's already easier. We see that the numerator and denominator can be reduced by 13:
Now don’t forget to turn the fraction back over again, let’s write down the whole chain:
Checked - it takes less time than searching through and checking divisors. Let's return to our two examples:
First. Divide with a corner (not on a calculator), we get:
This fraction is simpler, of course, but the reduction is again a problem. Now we separately analyze the fraction 1273/1463 and turn it over:
It's easier here. We can consider a divisor such as 19. The rest are not suitable, this is clear: 190:19 = 10, 1273:19 = 67. Hurray! Let's write down:
Next example. Let's shorten it to 88179/2717.
Divide, we get:
Separately, we analyze the fraction 1235/2717 and turn it over:
We can consider a divisor such as 13 (up to 13 is not suitable):
Numerator 247:13=19 Denominator 1235:13=95
*During the process we saw another divisor equal to 19. It turns out that:
Now we write down the original number:
And it doesn’t matter what is larger in the fraction - the numerator or the denominator, if it is the denominator, then we turn it over and act as described. This way we can reduce any fraction; the third approach can be called universal.
Of course, the two examples discussed above are not simple examples. Let's try this technology on the “simple” fractions we have already discussed:
Two quarters.
Seventy-two sixties. The numerator is greater than the denominator; there is no need to reverse it:
Of course, the third approach was applied to such simple examples simply as an alternative. The method, as already said, is universal, but not convenient and correct for all fractions, especially for simple ones.
The variety of fractions is great. It is important that you understand the principles. There is simply no strict rule for working with fractions. We looked, figured out how it would be more convenient to act, and moved forward. With practice, skill will come and you will crack them like seeds.
Conclusion:
If you see a common divisor(s) for the numerator and denominator, use them to reduce.
If you know how to quickly factor a number, then factor the numerator and denominator, then reduce.
If you can’t determine the common divisor, then use the third approach.
*To reduce fractions, it is important to master the principles of reduction, understand the basic property of a fraction, know approaches to solving, and be extremely careful when making calculations.
And remember! It is customary to reduce a fraction until it stops, that is, reduce it as long as there is a common divisor.
Sincerely, Alexander Krutitskikh.
Division and the numerator and denominator of the fraction on their common divisor, different from one, is called reducing a fraction.
To reduce a common fraction, you need to divide its numerator and denominator by the same natural number.
This number is the greatest common divisor of the numerator and denominator of the given fraction.
The following are possible decision recording forms Examples for reducing common fractions.
The student has the right to choose any form of recording.
Examples. Simplify fractions.
Reduce the fraction by 3 (divide the numerator by 3;
divide the denominator by 3).
Reduce the fraction by 7.
We perform the indicated actions in the numerator and denominator of the fraction.
The resulting fraction is reduced by 5.
Let's reduce this fraction 4)
on 5·7³- the greatest common divisor (GCD) of the numerator and denominator, which consists of the common factors of the numerator and denominator, taken to the power with the smallest exponent.
Let's factor the numerator and denominator of this fraction into prime factors.
We get: 756=2²·3³·7 And 1176=2³·3·7².
Determine the GCD (greatest common divisor) of the numerator and denominator of the fraction 5) .
This is the product of common factors taken with the lowest exponents.
gcd(756, 1176)= 2²·3·7.
We divide the numerator and denominator of this fraction by their gcd, i.e. by 2²·3·7 we get an irreducible fraction 9/14
.
Or it was possible to write the decomposition of the numerator and denominator in the form of a product of prime factors, without using the concept of power, and then reduce the fraction by crossing out the same factors in the numerator and denominator. When there are no identical factors left, we multiply the remaining factors separately in the numerator and separately in the denominator and write out the resulting fraction 9/14 .
And finally, it was possible to reduce this fraction 5) gradually, applying signs of dividing numbers to both the numerator and denominator of the fraction. Let's think like this: numbers 756 And 1176 end in an even number, which means both are divisible by 2 . We reduce the fraction by 2 . The numerator and denominator of the new fraction are numbers 378 And 588 also divided into 2 . We reduce the fraction by 2 . We notice that the number 294 - even, and 189 is odd, and reduction by 2 is no longer possible. Let's check the divisibility of numbers 189 And 294 on 3 .
(1+8+9)=18 is divisible by 3 and (2+9+4)=15 is divisible by 3, hence the numbers themselves 189 And 294 are divided into 3 . We reduce the fraction by 3 . Further, 63 is divisible by 3 and 98 - No. Let's look at other prime factors. Both numbers are divisible by 7 . We reduce the fraction by 7 and we get the irreducible fraction 9/14 .
This article continues the topic of converting algebraic fractions: consider such an action as reducing algebraic fractions. Let's define the term itself, formulate a reduction rule and analyze practical examples.
Yandex.RTB R-A-339285-1
The meaning of reducing an algebraic fraction
In materials about common fractions, we looked at its reduction. We defined reducing a fraction as dividing its numerator and denominator by a common factor.
Reducing an algebraic fraction is a similar operation.
Definition 1
Reducing an algebraic fraction is the division of its numerator and denominator by a common factor. In this case, in contrast to the reduction of an ordinary fraction (the common denominator can only be a number), the common factor of the numerator and denominator of an algebraic fraction can be a polynomial, in particular, a monomial or a number.
For example, the algebraic fraction 3 x 2 + 6 x y 6 x 3 y + 12 x 2 y 2 can be reduced by the number 3, resulting in: x 2 + 2 x y 6 x 3 · y + 12 · x 2 · y 2 . We can reduce the same fraction by the variable x, and this will give us the expression 3 x + 6 y 6 x 2 y + 12 x y 2. It is also possible to reduce a given fraction by a monomial 3 x or any of the polynomials x + 2 y, 3 x + 6 y , x 2 + 2 x y or 3 x 2 + 6 x y.
The ultimate goal of reducing an algebraic fraction is a fraction of a simpler form, at best an irreducible fraction.
Are all algebraic fractions subject to reduction?
Again, from materials on ordinary fractions, we know that there are reducible and irreducible fractions. Irreducible fractions are fractions that do not have common numerator and denominator factors other than 1.
It’s the same with algebraic fractions: they may have common factors in the numerator and denominator, or they may not. The presence of common factors allows you to simplify the original fraction through reduction. When there are no common factors, it is impossible to optimize a given fraction using the reduction method.
In general cases, given the type of fraction it is quite difficult to understand whether it can be reduced. Of course, in some cases the presence of a common factor between the numerator and denominator is obvious. For example, in the algebraic fraction 3 x 2 3 y it is quite clear that the common factor is the number 3.
In the fraction - x · y 5 · x · y · z 3 we also immediately understand that it can be reduced by x, or y, or x · y. And yet, much more often there are examples of algebraic fractions, when the common factor of the numerator and denominator is not so easy to see, and even more often, it is simply absent.
For example, we can reduce the fraction x 3 - 1 x 2 - 1 by x - 1, while the specified common factor is not present in the entry. But the fraction x 3 - x 2 + x - 1 x 3 + x 2 + 4 · x + 4 cannot be reduced, since the numerator and denominator do not have a common factor.
Thus, the question of determining the reducibility of an algebraic fraction is not so simple, and it is often easier to work with a fraction of a given form than to try to find out whether it is reducible. In this case, such transformations take place that in particular cases make it possible to determine the common factor of the numerator and denominator or to draw a conclusion about the irreducibility of a fraction. We will examine this issue in detail in the next paragraph of the article.
Rule for reducing algebraic fractions
Rule for reducing algebraic fractions consists of two sequential actions:
- finding common factors of the numerator and denominator;
- if any are found, the action of reducing the fraction is carried out directly.
The most convenient method of finding common denominators is to factor the polynomials present in the numerator and denominator of a given algebraic fraction. This allows you to immediately clearly see the presence or absence of common factors.
The very action of reducing an algebraic fraction is based on the main property of an algebraic fraction, expressed by the equality undefined, where a, b, c are some polynomials, and b and c are non-zero. The first step is to reduce the fraction to the form a · c b · c, in which we immediately notice the common factor c. The second step is to perform a reduction, i.e. transition to a fraction of the form a b .
Typical examples
Despite some obviousness, let us clarify the special case when the numerator and denominator of an algebraic fraction are equal. Similar fractions are identically equal to 1 on the entire ODZ of the variables of this fraction:
5 5 = 1 ; - 2 3 - 2 3 = 1 ; x x = 1 ; - 3, 2 x 3 - 3, 2 x 3 = 1; 1 2 · x - x 2 · y 1 2 · x - x 2 · y ;
Since ordinary fractions are a special case of algebraic fractions, let us recall how they are reduced. The natural numbers written in the numerator and denominator are factored into prime factors, then the common factors are canceled (if any).
For example, 24 1260 = 2 2 2 3 2 2 3 3 5 7 = 2 3 5 7 = 2 105
The product of simple identical factors can be written as powers, and in the process of reducing a fraction, use the property of dividing powers with identical bases. Then the above solution would be:
24 1260 = 2 3 3 2 2 3 2 5 7 = 2 3 - 2 3 2 - 1 5 7 = 2 105
(numerator and denominator divided by a common factor 2 2 3). Or for clarity, based on the properties of multiplication and division, we give the solution the following form:
24 1260 = 2 3 3 2 2 3 2 5 7 = 2 3 2 2 3 3 2 1 5 7 = 2 1 1 3 1 35 = 2 105
By analogy, the reduction of algebraic fractions is carried out, in which the numerator and denominator have monomials with integer coefficients.
Example 1
The algebraic fraction is given - 27 · a 5 · b 2 · c · z 6 · a 2 · b 2 · c 7 · z. It needs to be reduced.
Solution
It is possible to write the numerator and denominator of a given fraction as a product of simple factors and variables, and then carry out the reduction:
27 · a 5 · b 2 · c · z 6 · a 2 · b 2 · c 7 · z = - 3 · 3 · 3 · a · a · a · a · a · b · b · c · z 2 · 3 · a · a · b · b · c · c · c · c · c · c · c · z = = - 3 · 3 · a · a · a 2 · c · c · c · c · c · c = - 9 a 3 2 c 6
However, a more rational way would be to write the solution as an expression with powers:
27 · a 5 · b 2 · c · z 6 · a 2 · b 2 · c 7 · z = - 3 3 · a 5 · b 2 · c · z 2 · 3 · a 2 · b 2 · c 7 · z = - 3 3 2 · 3 · a 5 a 2 · b 2 b 2 · c c 7 · z z = = - 3 3 - 1 2 · a 5 - 2 1 · 1 · 1 c 7 - 1 · 1 = · - 3 2 · a 3 2 · c 6 = · - 9 · a 3 2 · c 6 .
Answer:- 27 a 5 b 2 c z 6 a 2 b 2 c 7 z = - 9 a 3 2 c 6
When the numerator and denominator of an algebraic fraction contain fractional numerical coefficients, there are two possible ways of further action: either divide these fractional coefficients separately, or first get rid of the fractional coefficients by multiplying the numerator and denominator by some natural number. The last transformation is carried out due to the basic property of an algebraic fraction (you can read about it in the article “Reducing an algebraic fraction to a new denominator”).
Example 2
The given fraction is 2 5 x 0, 3 x 3. It needs to be reduced.
Solution
It is possible to reduce the fraction this way:
2 5 x 0, 3 x 3 = 2 5 3 10 x x 3 = 4 3 1 x 2 = 4 3 x 2
Let's try to solve the problem differently, having first gotten rid of fractional coefficients - multiply the numerator and denominator by the least common multiple of the denominators of these coefficients, i.e. on LCM (5, 10) = 10. Then we get:
2 5 x 0, 3 x 3 = 10 2 5 x 10 0, 3 x 3 = 4 x 3 x 3 = 4 3 x 2.
Answer: 2 5 x 0, 3 x 3 = 4 3 x 2
When we reduce general algebraic fractions, in which the numerators and denominators can be either monomials or polynomials, there can be a problem where the common factor is not always immediately visible. Or moreover, it simply does not exist. Then, to determine the common factor or record the fact of its absence, the numerator and denominator of the algebraic fraction are factored.
Example 3
The rational fraction 2 · a 2 · b 2 + 28 · a · b 2 + 98 · b 2 a 2 · b 3 - 49 · b 3 is given. It needs to be reduced.
Solution
Let's factor the polynomials in the numerator and denominator. Let's put it out of brackets:
2 a 2 b 2 + 28 a b 2 + 98 b 2 a 2 b 3 - 49 b 3 = 2 b 2 (a 2 + 14 a + 49) b 3 (a 2 - 49)
We see that the expression in parentheses can be converted using abbreviated multiplication formulas:
2 b 2 (a 2 + 14 a + 49) b 3 (a 2 - 49) = 2 b 2 (a + 7) 2 b 3 (a - 7) (a + 7)
It is clearly seen that it is possible to reduce a fraction by a common factor b 2 (a + 7). Let's make a reduction:
2 b 2 (a + 7) 2 b 3 (a - 7) (a + 7) = 2 (a + 7) b (a - 7) = 2 a + 14 a b - 7 b
Let us write a short solution without explanation as a chain of equalities:
2 a 2 b 2 + 28 a b 2 + 98 b 2 a 2 b 3 - 49 b 3 = 2 b 2 (a 2 + 14 a + 49) b 3 (a 2 - 49) = = 2 b 2 (a + 7) 2 b 3 (a - 7) (a + 7) = 2 (a + 7) b (a - 7) = 2 a + 14 a b - 7 b
Answer: 2 a 2 b 2 + 28 a b 2 + 98 b 2 a 2 b 3 - 49 b 3 = 2 a + 14 a b - 7 b.
It happens that common factors are hidden by numerical coefficients. Then, when reducing fractions, it is optimal to put the numerical factors at higher powers of the numerator and denominator out of brackets.
Example 4
Given the algebraic fraction 1 5 · x - 2 7 · x 3 · y 5 · x 2 · y - 3 1 2 . It is necessary to reduce it if possible.
Solution
At first glance, the numerator and denominator do not have a common denominator. However, let's try to convert the given fraction. Let's take out the factor x in the numerator:
1 5 x - 2 7 x 3 y 5 x 2 y - 3 1 2 = x 1 5 - 2 7 x 2 y 5 x 2 y - 3 1 2
Now you can see some similarity between the expression in brackets and the expression in the denominator due to x 2 y . Let us take out the numerical coefficients of the higher powers of these polynomials:
x 1 5 - 2 7 x 2 y 5 x 2 y - 3 1 2 = x - 2 7 - 7 2 1 5 + x 2 y 5 x 2 y - 1 5 3 1 2 = = - 2 7 x - 7 10 + x 2 y 5 x 2 y - 7 10
Now the common factor becomes visible, we carry out the reduction:
2 7 x - 7 10 + x 2 y 5 x 2 y - 7 10 = - 2 7 x 5 = - 2 35 x
Answer: 1 5 x - 2 7 x 3 y 5 x 2 y - 3 1 2 = - 2 35 x .
Let us emphasize that the skill of reducing rational fractions depends on the ability to factor polynomials.
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Last time we made a plan, following which you can learn how to quickly reduce fractions. Now let's look at specific examples of reducing fractions.
Examples.
Let's check whether the larger number is divisible by the smaller number (numerator by denominator or denominator by numerator)? Yes, in all three of these examples the larger number is divided by the smaller number. Thus, we reduce each fraction by the smaller of the numbers (by the numerator or by the denominator). We have:
Let's check if the larger number is divisible by the smaller number? No, it doesn't share.
Then we move on to checking the next point: does the entry of both the numerator and denominator end with one, two or more zeros? In the first example, the numerator and denominator end in zero, in the second example, two zeros, and in the third, three zeros. This means that we reduce the first fraction by 10, the second by 100, and the third by 1000:
We got irreducible fractions.
A larger number cannot be divided by a smaller number, and numbers do not end with zeros.
Now let’s check whether the numerator and denominator are in the same column in the multiplication table? 36 and 81 are both divisible by 9, 28 and 63 are divisible by 7, and 32 and 40 are divisible by 8 (they are also divisible by 4, but if there is a choice, we will always reduce by a larger one). Thus, we come to the answers:
All numbers obtained are irreducible fractions.
A larger number cannot be divided by a smaller number. But the record of both the numerator and the denominator ends in zero. So, we reduce the fraction by 10:
This fraction can still be reduced. We check the multiplication table: both 48 and 72 are divisible by 8. We reduce the fraction by 8:
We can also reduce the resulting fraction by 3:
This fraction is irreducible.
The larger number is not divisible by the smaller number. The numerator and denominator end in zero. This means we reduce the fraction by 10.
We check the numbers obtained in the numerator and denominator for and. Since the sum of the digits of both 27 and 531 is divisible by 3 and 9, this fraction can be reduced by either 3 or 9. We choose the larger one and reduce by 9. The resulting result is an irreducible fraction.
Reducing fractions is necessary in order to reduce the fraction to a simpler form, for example, in the answer obtained as a result of solving an expression.
Reducing fractions, definition and formula.
What is reducing fractions? What does it mean to reduce a fraction?
Definition:
Reducing Fractions- this is the division of a fraction's numerator and denominator by the same positive number not equal to zero and one. As a result of the reduction, a fraction with a smaller numerator and denominator is obtained, equal to the previous fraction according to.
Formula for reducing fractions basic properties of rational numbers.
\(\frac(p \times n)(q \times n)=\frac(p)(q)\)
Let's look at an example:
Reduce the fraction \(\frac(9)(15)\)
Solution:
We can factor a fraction into prime factors and cancel common factors.
\(\frac(9)(15)=\frac(3 \times 3)(5 \times 3)=\frac(3)(5) \times \color(red) (\frac(3)(3) )=\frac(3)(5) \times 1=\frac(3)(5)\)
Answer: after reduction we got the fraction \(\frac(3)(5)\). According to the basic property of rational numbers, the original and resulting fractions are equal.
\(\frac(9)(15)=\frac(3)(5)\)
How to reduce fractions? Reducing a fraction to its irreducible form.
To get an irreducible fraction as a result, we need find the greatest common divisor (GCD) for the numerator and denominator of the fraction.
There are several ways to find GCD; in the example we will use the decomposition of numbers into prime factors.
Get the irreducible fraction \(\frac(48)(136)\).
Solution:
Let's find GCD(48, 136). Let's write the numbers 48 and 136 into prime factors.
48=2⋅2⋅2⋅2⋅3
136=2⋅2⋅2⋅17
GCD(48, 136)= 2⋅2⋅2=6
\(\frac(48)(136)=\frac(\color(red) (2 \times 2 \times 2) \times 2 \times 3)(\color(red) (2 \times 2 \times 2) \times 17)=\frac(\color(red) (6) \times 2 \times 3)(\color(red) (6) \times 17)=\frac(2 \times 3)(17)=\ frac(6)(17)\)
The rule for reducing a fraction to an irreducible form.
- You need to find the greatest common divisor for the numerator and denominator.
- You need to divide the numerator and denominator by the greatest common divisor to obtain an irreducible fraction.
Example:
Reduce the fraction \(\frac(152)(168)\).
Solution:
Let's find GCD(152, 168). Let's write the numbers 152 and 168 into prime factors.
152=2⋅2⋅2⋅19
168=2⋅2⋅2⋅3⋅7
GCD(152, 168)= 2⋅2⋅2=6
\(\frac(152)(168)=\frac(\color(red) (6) \times 19)(\color(red) (6) \times 21)=\frac(19)(21)\)
Answer: \(\frac(19)(21)\) is an irreducible fraction.
Reducing improper fractions.
How to reduce an improper fraction?
The rules for reducing fractions are the same for proper and improper fractions.
Let's look at an example:
Reduce the improper fraction \(\frac(44)(32)\).
Solution:
Let's write the numerator and denominator into simple factors. And then we’ll reduce the common factors.
\(\frac(44)(32)=\frac(\color(red) (2 \times 2 ) \times 11)(\color(red) (2 \times 2 ) \times 2 \times 2 \times 2 )=\frac(11)(2 \times 2 \times 2)=\frac(11)(8)\)
Reducing mixed fractions.
Mixed fractions follow the same rules as ordinary fractions. The only difference is that we can do not touch the whole part, but reduce the fractional part or Convert a mixed fraction to an improper fraction, reduce it and convert it back to a proper fraction.
Let's look at an example:
Cancel the mixed fraction \(2\frac(30)(45)\).
Solution:
Let's solve it in two ways:
First way:
Let's write the fractional part into simple factors, but we won't touch the whole part.
\(2\frac(30)(45)=2\frac(2 \times \color(red) (5 \times 3))(3 \times \color(red) (5 \times 3))=2\ frac(2)(3)\)
Second way:
Let's first convert it to an improper fraction, and then write it into prime factors and reduce it. Let's convert the resulting improper fraction into a proper fraction.
\(2\frac(30)(45)=\frac(45 \times 2 + 30)(45)=\frac(120)(45)=\frac(2 \times \color(red) (5 \times 3) \times 2 \times 2)(3 \times \color(red) (3 \times 5))=\frac(2 \times 2 \times 2)(3)=\frac(8)(3)= 2\frac(2)(3)\)
Questions on the topic:
Can you reduce fractions when adding or subtracting?
Answer: no, you must first add or subtract fractions according to the rules, and only then reduce them. Let's look at an example:
Evaluate the expression \(\frac(50+20-10)(20)\) .
Solution:
They often make the mistake of reducing the same numbers in the numerator and denominator, in our case the number 20, but they cannot be reduced until you have completed the addition and subtraction.
\(\frac(50+\color(red) (20)-10)(\color(red) (20))=\frac(60)(20)=\frac(3 \times 20)(20)= \frac(3)(1)=3\)
What numbers can you reduce a fraction by?
Answer: You can reduce a fraction by the greatest common factor or the common divisor of the numerator and denominator. For example, the fraction \(\frac(100)(150)\).
Let's write the numbers 100 and 150 into prime factors.
100=2⋅2⋅5⋅5
150=2⋅5⋅5⋅3
The greatest common divisor will be the number gcd(100, 150)= 2⋅5⋅5=50
\(\frac(100)(150)=\frac(2 \times 50)(3 \times 50)=\frac(2)(3)\)
We got the irreducible fraction \(\frac(2)(3)\).
But it is not necessary to always divide by gcd; an irreducible fraction is not always needed; you can reduce the fraction by a simple divisor of the numerator and denominator. For example, the number 100 and 150 have a common divisor of 2. Let's reduce the fraction \(\frac(100)(150)\) by 2.
\(\frac(100)(150)=\frac(2 \times 50)(2 \times 75)=\frac(50)(75)\)
We got the reducible fraction \(\frac(50)(75)\).
What fractions can be reduced?
Answer: You can reduce fractions in which the numerator and denominator have a common divisor. For example, the fraction \(\frac(4)(8)\). The number 4 and 8 have a number by which they are both divisible - the number 2. Therefore, such a fraction can be reduced by the number 2.
Example:
Compare the two fractions \(\frac(2)(3)\) and \(\frac(8)(12)\).
These two fractions are equal. Let's take a closer look at the fraction \(\frac(8)(12)\):
\(\frac(8)(12)=\frac(2 \times 4)(3 \times 4)=\frac(2)(3) \times \frac(4)(4)=\frac(2) (3) \times 1=\frac(2)(3)\)
From here we get, \(\frac(8)(12)=\frac(2)(3)\)
Two fractions are equal if and only if one of them is obtained by reducing the other fraction by the common factor of the numerator and denominator.
Example:
If possible, reduce the following fractions: a) \(\frac(90)(65)\) b) \(\frac(27)(63)\) c) \(\frac(17)(100)\) d) \(\frac(100)(250)\)
Solution:
a) \(\frac(90)(65)=\frac(2 \times \color(red) (5) \times 3 \times 3)(\color(red) (5) \times 13)=\frac (2 \times 3 \times 3)(13)=\frac(18)(13)\)
b) \(\frac(27)(63)=\frac(\color(red) (3 \times 3) \times 3)(\color(red) (3 \times 3) \times 7)=\frac (3)(7)\)
c) \(\frac(17)(100)\) irreducible fraction
d) \(\frac(100)(250)=\frac(\color(red) (2 \times 5 \times 5) \times 2)(\color(red) (2 \times 5 \times 5) \ times 5)=\frac(2)(5)\)