# Fractions

Fractions are an alternative way to denote division between two numbers and are generally used to express proportions, for example, to express three-quarters of a quantity we write $\frac{3}{4}$ or to denote half of a cake we simply write $\frac{1}{2}$. It is possible to represent the fractions graphically to make them easier to understand.

Formally, if we consider two integers $a$ and $b neq 0$, then we will say that $a$ is the numerator of the fraction and $b$ is the denominator of the fraction, and so, the division $a div b$ will be represented by the following expression

$\frac{a}{b}$

The number above is called numerator and the number below is called denominator.

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## Fraction properties

When we work with fractions, we will find very particular expressions that we can identify when we want to simplify mathematical operations. Let’s consider $a$ an non-zero integer and see below what these fractions are.

One divided by one equals one. In general, if we consider any non-zero real, the division of this number by itself, is equal to one, then,

$\dfrac{1}{1} = 1$.

$\dfrac{a}{a} = 1$.

Any integer number can be expressed as the division of itself with one, this information will be useful when we are presented with operations between numbers expressed in fractions and integers.

$\dfrac{a}{1} = a$.

When dividing zero by any non-zero real number, the result will always be the same, zero.

$\dfrac{0}{a} = 0$

On the contrary, if we take any real number, it cannot be divided by zero because this operation is not defined, that is, division by zero is not defined.

$\dfrac{a}{0}$

not defined.

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## The Law of Signs for Fractions

Since fractions represent divisions, we can also establish the law of signs for division, if $a$ and $b$ are integers such that $b$ is non-zero, then

$\dfrac{a}{b} = \dfrac{a}{b}$.

$\dfrac{-a}{b} = -\dfrac{a}{b}$.

$\dfrac{a}{-b} = -\dfrac{a}{b}$.

$\dfrac{-a}{-b} = \dfrac{a}{b}$.

The advantage in the use of fractions is that they provide rigidity in the results and thus avoid approximation or rounding errors when making divisions, which is why it is necessary to master the operations of sum, subtraction, multiplication and division between the fractions.

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## Proper and improper fractions

One way to classify fractions is by considering the size of their numerator and denominator, because these will determine the portion they really represent. If $a$ and $b$ are two integers such that $b neq 0$, we have to

• If $a < b$, we will say that the fraction is $\frac{a}{b}$ is proper, that is, if the numerator is smaller than the denominator.
• If $a \geq b$, we will say that the fraction is $\frac{a}{b}$ is improper, that is, if the numerator is greater or equal than the denominator.

To clarify this idea, let’s see some examples.

### Examples

#### Example 1

The fraction $\frac{1}{2}$, is a proper fraction, because its numerator is less than its denominator.

#### Example 2

The fraction $\frac{7}{15}$, is an proper fraction, because its numerator is smaller than its denominator.

#### Example 3

The fraction $\frac{4}{9}$, is an proper fraction, because its numerator is smaller than its denominator.

#### Example 4

The fraction $\frac{6}{20}$, is an proper fraction, because its numerator is smaller than its denominator.

#### Example 5

The fraction $\frac{5}{3}$, is an improper fraction, because its numerator is greater than its denominator.

#### Example 6

The fraction $\frac{10}{4}$, is an improper fraction, because its numerator is bigger than its denominator.

#### Example 7

The fraction $\frac{20}{12}$, is an improper fraction, because its numerator is bigger than its denominator.

#### Example 8

The fraction $\frac{75}{44}$, is an improper fraction, because its numerator is bigger than its denominator.

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## Mixed fractions

When reading a recipe it is common to find measurements for ingredients such as one and a half cup of sugar or, that is why we can find containers with measurements of $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$ or $\frac{1}{8}$. This also occurs when buying foods that must be weighed, such as one kilograms and a quarter cheese or three and a half kilograms of meat.

Fractions are ideal to express this type of measurements, they are designed to measure portions, for example, to write a cup and a half you can write $1 + \frac{1}{2}$ which in turn is equal to $\frac{3}{2}$. However, the way they are written may not present comfort or clarity in practice, that is why the mixed fractions (or mixed numbers) are defined, then, that instead of writing $1 + \frac{1}{2}$, one writes

$1 \tfrac{1}{2}$

In this way, we define mixed fractions to separate the whole part from its non-integer part, the latter usually represented with a fraction of its own. Any mixed fraction can be rewritten as an improper fraction, because if $a$, $b$ and $c$ are positive integers, then the following mixed fraction

$a \tfrac{b}{c}$

is rewritten as an improper fraction adding $a$ with $\frac{b}{c}$, that is,

$a + \frac{b}{c} = \frac{a \cdot c + b}{c}$

Let’s see some examples of how to rewrite mixed fractions.

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### Examples

#### Example 9

Rewrite the mixed fraction $1 \tfrac{1}{2}$ as an improper fraction.

$1 \tfrac{1}{2} = 1 + \frac{1}{2} = \frac{1 \cdot 2 + 1}{2} = \frac{3}{2}$

#### Example 10

Rewrite the mixed fraction $1 \tfrac{1}{8}$ as an improper fraction.

$1 \tfrac{1}{8} = 1 + \frac{1}{8} = \frac{1 \cdot 8 + 1}{2} = \frac{9}{2}$

#### Example 11

Rewrite the mixed fraction $2 \tfrac{3}{4}$ as an improper fraction.

$2 \tfrac{3}{4} = 2 + \frac{3}{4} = \frac{2 \cdot 4 + 3}{4} = \frac{11}{4}$

#### Example 12

Rewrite the mixed fraction $2 \tfrac{3}{4}$ as an improper fraction.

$3 \tfrac{1}{2} = 3 + \frac{1}{2} = \frac{3 \cdot 2 + 1}{2} = \frac{7}{2}$

#### Example 13

Rewrite the mixed fraction $5 \tfrac{9}{16}$ as an improper fraction.

$5 \tfrac{9}{16} = 5 + \frac{9}{16} = \frac{5 \cdot 16 + 9}{16} = \frac{89}{16}$

# The Conjugate of a Sum

Next we will define an expression that is closely related to the difference of squares, because when we find the addition (or subtraction as the case may be) of two real numbers, we can define an expression that will allow us to write that subtraction as a difference of squares.

Formally, if $a$ and $b$ are two real numbers, the conjugate of the sum $(a+b)$ is defined as $(a-b)$. Similarly, the conjugate of the subtraction $(a-b)$ is defined as $(a+b)$. That is, the sign between the two is changed. The importance of the conjugate lies in the fact that the product of an addition by its conjugate is equal to a difference of squares, that is,

This equality can be deduced by performing the distributive property of the real numbers, let’s see then,

This type of expressions is often found in the development of algebraic operations and is used mainly to simplify operations, let’s see in the following examples how to identify the conjugation of some expressions:

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### Examples

#### Example 1

Identify the conjugate of $12 - 5$. It does not have much sense to identify the conjugate of this expression because we can simply make the subtraction and obtain 7 as a result.

#### Example 2

Identify the conjugate of $\sqrt{12} - 5$. Note that one of the involved sums is the square root of twelve, so it cannot be subtracted with five, so we conclude that its conjugate is $\sqrt{12} + 5$.

#### Example 3

Identify the conjugate of $3 + \sqrt{8}$. Note that one of the summands involved is the square root of eight, so it cannot be added with three, so we conclude that its conjugate is $3 - \sqrt{8}$.

#### Example 4

Identify the conjugate of $3x - 7$. Note that one of the sums involved is three multiplied by one unknown, so it cannot be subtracted with seven, then, we conclude that its conjugate is $3x + 7$.

#### Example 5

Identify the conjugate of $15 + 4x$. Let’s notice that one of the involved sums is four multiplied by one unknown, therefore it cannot be added with 15, then, we conclude that its conjugate is $15 - 4x$.

#### Example 6

Identify the conjugate of $6 + \sqrt{x+2}$. This subtraction cannot be done, so we conclude that your conjugate is $6 - \sqrt{x+2}$. Noting that the sign inside the root does not change.

# Difference of two squares

When carrying out mathematical operations it is common to find subtractions between two numbers, however, when finding the subtraction of the squares of two numbers we will say that this is a difference of squares and it is of our particular interest because through the distributive property, we can express it as the product of two factors.

Formally, if $a$ and $b$ are two real numbers, then the difference of their squares will be equal to the sum of the first plus the second, multiplied by the subtraction of the first by the second, that is,

This equality can be deduced by performing the distributive property of the real numbers, let’s see then,

This type of expression is often found in the development of algebraic operations and is used mainly for factoring operations, let’s see in the following examples how to apply this operation

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## Examples

### Example 1

Factorize the expression $5^2 - 3^2$. Note that in this case, we can simply apply the power of each of the summands and perform the subtraction directly.

$5^2 - 3^2 = 25 - 9$

$= 16$

### Example 2

Factorize the expression $x^2 - 9$. We notice that in this case, one of the summands is an x squared and the other one is a nine, so we cannot make the subtraction between them so we apply the difference of squares noting that nine is equal to three squared.

$x^2 - 9 = x^2 - 3^2$

$= (x-3)(x+3)$

### Example 3

Factorize the expression $x^2 - 2$. We notice that in this case, one of the summands is an x-squared and the other is two, so we cannot perform the subtraction between them so we apply the difference of squares noting that two can be rewritten as $2 = \left( \sqrt{2} \right)^2$.

$x^2 - 2 = x^2 -\left( \sqrt{2} \right)^2$

$= \left(x-\sqrt{2} \right) \left(x+\sqrt{2} \right)$

In this way, we can notice that if the square root of a number is not exact, it can be rewritten to use the difference of squares.

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### Example 4

Factorize the expression $8 - x^6$. We notice that in this case, one of the summands is 8 and the other one is x to six, so we cannot make the subtraction between them so we apply the difference of squares noting that eight can be rewritten as $8 = \left( \sqrt{8} \right)^2$ and x to six as $x^6 = \left( x^3 \right)^2$.

$8 - x^6 = \left( \sqrt{8} \right)^2 - \left(x^3 \right)^2$

$= \left(\sqrt{8}-x^3 \right) \left(\sqrt{8}+x^3 \right)$

### Example 5

Factorize the expression $36x^4 - 5x^8$. We notice that in this case, we cannot make the subtraction between them so we apply the difference of squares using the observations exposed in the previous examples.

$36x^4 - 5x^8 = \left( 6x^2 \right)^2 - \left( \sqrt{5}x^4 \right)^2$

$= \left(6x^2-\sqrt{5}x^4 \right) \left(6x^2+\sqrt{5}x^4 \right)$

# The Notable Product

The remarkable product is a particular case of the distributive property that gives us the perfect square trinomial as a result and establishes that, if $a$ and $b$ are two real numbers, the square of the sum of them is equal to the first squared plus twice the product of the first times the second plus the second squared, that is,

This equality can be deduced by performing the distributive property when we multiply the sum of two numbers by that same sum, let’s see then,

Similarly, if $a$ and $b$ are two real numbers, the square of the subtraction between the two is equal to the first squared minus twice the product of the first times the second plus the second squared, that is,

This equality can be deduced by performing the distributive property when we multiply the substraction of two numbers by that same substraction, let’s see then,

This type of expression is often found in the development of algebraic operations because we cannot always carry out the sum that is inside the parentheses, let’s see in the following examples how to apply this operation:

## Ejemplos

### Ejemplo 1

Apply the notable product to expand the expression $(3 + 2)^2$. We add the two elements within the parentheses and square as follows:

$(3 + 2)^2$
$= 5^2$
$= 25$

### Ejemplo 2

Apply the notable product to expand the expression $(3 + \sqrt{2})^2$. Note that one of the addends involved is the square root of two, therefore it cannot be added with three.

$(3 + sqrt{2})^2$
$= 3^2 + 2(3)(sqrt{2}) + (sqrt{2})^2$
$= 9 + 6sqrt{2} + 2$
$= 11+6sqrt{2}$

### Ejemplo 3

Apply the notable product to expand the expression $(\sqrt[3]{6} - 4)^2$. Note that one of the addends involved is the cube root of six, therefore it cannot be subtracted with four.

$(sqrt[3]{6} - 4)^2$
$= (sqrt[3]{6})^2 -2(sqrt[3]{6})(4) + 4^2$
$= (sqrt[3]{6})^2 -8sqrt[3]{6} +16$

### Ejemplo 4

Apply the notable product to expand the expression $(x + 7)^2$. Note that one of the addends involved is an unknown, therefore it cannot be added with seven.

$(x+7)^2$
$= x^2 + 2(x)(7) + 7^2$
$= x^2 +14x + 49$

### Ejemplo 5

Apply the notable product to expand the expression $(2x-8)^2$. Note that one of the addends involved is an unknown multiplied by two, therefore it cannot be subtracted with eight.

$(2x-8)^2$
$= (2x)^2 - 2(2x)(8) + 8^2$
$= 4x^2 - 32x + 64$

### Ejemplo 6

Apply the notable product to expand the expression $(x^2 + x^5)^2$. Note that one of the addends involved is x squared and the other is x raised to five, therefore they cannot be added.

$(x^2 + x^5)^2$
$= (x^2)^2 + 2(x^2)(x^5) + (x^5)^2$
$= x^4 + 2x^7 + x^{10}$

# The Distributive Property

When adding real numbers we have the freedom to associate the numbers involved smoothly, the same happens if we are multiplying real numbers, however, we must be cautious when we come across mixed operations, that is, sums and products at the same time. We will see a property that allows us to operate sums and products at the same time:

The distributive property states that if a number multiplies the sum of two numbers, then the factor involved is distributed among each of the addends. Formally, if $a$, $b$ and $c$ are real numbers, then

We can also apply this property if a subtraction is involved instead of an addition within the parentheses, as follows:

We notice that if we observe this equality from right to left, we are taking the common factor that exists in both addends and we are taking it out to multiply:

$a \cdot b \pm a \cdot c = a \cdot (b \pm c)$

This is one of the most used properties in the calculation of mixed operations and from them, some cases are deduced that facilitate the simplification of mathematical expressions. Let’s see some examples to understand this property well:

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## Examples

### Example 1

Use the distributive property to expand the expression $2 \cdot (1 + 6)$. In this case, it is not necessary to use the distributive property since we can add the numbers that are inside the parentheses and then multiply in the following way:

$2 \cdot (1 + 6) = 2 \cdot 7 = 14$

### Example 2

Use the distributive property to expand the expression $2 \cdot \left (1 + \sqrt {6} \right)$. Note that one of the addends involved is the square root of 6, therefore it cannot be added with 1, so we distribute the factor involved

$2 \cdot \left( 1 + \sqrt{6} \right) = 2 \cdot 1 + 2 \cdot \sqrt{6} = 2 + 2 \sqrt{6}$

### Example 3

Use the distributive property to expand the expression $5 \cdot \left (x - \sqrt {10} \right)$. Note that one of the addends involved is the square root of 10 and the other is an unknown, therefore they cannot be subtracted, so we distribute the factor involved

$5 \cdot \left( x - \sqrt{10} \right) = 5 \cdot x - 5 \cdot \sqrt{10} = 5x - 5\sqrt{10}$

### Example 4

Use the distributive property to expand the expression $x \cdot \left (x + x^2 \right)$. Note that one of the addends involved is an unknown and the other is an unknown squared, therefore they cannot be added, then we distribute the factor involved

$x \cdot \left( x + x^2 \right) = x \cdot x + x \cdot x^2 = x^2 + x^3$

### Example 5

Use the distributive property to take out the common factor of the expression $18 + 3 \sqrt {7}$. Note that $18 = 3 \cdot 6$, then,

$18 + 3\sqrt{7} = 3 \cdot 6 + 3 \sqrt{7} = 3 \cdot \left( 6 + \sqrt{7} \right)$

### Example 6

Use the distributive property to take out the common factor of the expression $x^4 - 8x$. Note that one of the addends involved is an unknown raised to four and the other is 8 times said unknown, therefore they cannot be subtracted, then

$x^4 - 8x = x \cdot x^3 - x \cdot 8 = x \cdot \left( x^3 - 8 \right)$

### Example 7

Use the distributive property to take the common factor of the expression $12x^7 + 15x^4$. These two elements cannot be added, so

$12x^7 + 15x^4 = 3 \cdot 4 \cdot x^4 \cdot x^3 + 3 \cdot 5 \cdot x^4 = 3 x^4 \cdot \left( 4x^3 + 5 \right)$