## What is a module?

Modulus is the distance from the origin to some number on the coordinate line. Since the distance is never negative, the modulus is always non-negative. Thus, the modulus of the number 3 is 3, just as the modulus of the number -3 is 3| 3 |= 3

|β3|= 3

Imagine that on the coordinate line the distance between the integers is one step. Now if you mark the numbers -3 and 3, the distance to them from the origin of coordinates will be equal to three steps: Modulus is not only the distance from the origin to some number. Modulus is also the distance**between any two numbers**on a coordinate line. This distance is expressed as the difference between these numbers, enclosed under the modulo sign:

|*x*_{1} β *x*_{2}|

_{1}and x

_{2}are numbers on the coordinate line. For example, mark the numbers 2 and 5 on the coordinate line. The distance between the numbers 2 and 5 can be written using a modulus. To do this, write down the difference of numbers 2 and 5 and put this difference under the modulo sign:

|2 β 5| = |β3| = 3

We see that the distance from number 2 to number 5 is three steps: If the distance from 2 to 5 is 3, then the distance from 5 to 2 is also 3 That is, if you swap the numbers in the expression |5 - 2|, the result will not change:|5 β 2| = | 3 | = 3

Then we can write that |2 - 5| = |5 - 2|. In general, the following equality is true:|*x*_{1 }β *x*_{2}| = |*x*_{2 }β *x*_{1}|

_{1}to x

_{2}equals the distance from x

_{2}to x

_{1}.

## Expansion the module

When we say that |3|= 3 or |-3|= 3 we are performing an action called**modulus expansion**.

*x*, the expression |x| will be equal to

*x*if the number substituted is greater than or equal to zero. And if a number less than zero is substituted for

*x*, the expression |x| equals -

*x*.

*|x| = -x*looks as if the modulus became equal to a negative number. Keep in mind that when

*x < 0*, there is a negative number under the modulus sign. After the equal sign, put this negative number instead of

*x*and expand the brackets.

|β7|

In this case the second condition x < 0 is satisfied, because -7 < 0 Therefore we use the second formula. Namely |x| = -x. Instead of x we substitute the number -7 Hence: Therefore |-7| = 7.**Example 2.** Let x = 5. That is, we consider the modulus of the number 5

| 5 |

In this case, the first condition x β₯ 0 is satisfied, because 5 β₯ 0

So we use the first formula. Namely | x | = x. We get | 5 | = 5.

Zero is a kind of transition point at which the module changes its order of expansion and then retains its sign. Visually, this can be represented as follows:

In the figure, the red minus and plus signs indicate how the module |x| will unfold at intervals x < 0 and x β₯ 0.

For example, if you take the numbers 1, 9, and 13, and they belong to the interval x β₯ 0, then according to the figure the module |x| will unfold with a plus sign:

| 1 | = 1

| 9 | = 9

| 13 | = 13

And if we take numbers smaller than zero, for example -3, -9, -15, then according to the figure the module will expand with a minus sign:

|β3| = β(β3) = 3

|β9| = β(β9) = 9

|β15| = β(β15) = 15

**Example 3.** Let x = β4 - 6. That is, we consider the modulus of the expression β4 - 6,

|β4 β 6|

The root of number 4 is 2. Then the modulus will be

|β4 β 6| = |2 β 6| = |β4|

x which was equal to β4-6 is now equal to -4. In this case the second condition x < 0 is satisfied, because -4 < 0

Hence, we use the second formula |x| = -x. We continue the solution in the original example:

|β4 β 6| = |2 β 6| = |β4| = β(β4) = 4

In practice, they usually reason like this:

*"A module expands with a plus sign if the submodular expression is greater than or equal to zero; a module expands with a minus sign if the submodular expression is less than zero."*

**Examples:**

|2| = 2 - the modulus opened with a **plus** sign because 2 β₯ 0

|-4| = -(-4) = 4 - the module opened with a **minus** sign, because -4 < 0

In some textbooks you can find the following notation of the modulus disclosure rule:

In this entry, the first condition, which previously looked like x β₯ 0, is described in more detail, namely that if x > 0, then the expression |x| will equal x, and if x=0, then the expression |x| will equal zero.

**Example 4.** Let x = 0. That is, we consider a modulus of zero:

| 0 |

In this case the condition x=0 is fulfilled, because 0 = 0

Hence: |0| = 0

**Example 5.** Expand the modulus in the expression |x|+ 3

If x β₯ 0, then the modulus will expand with a plus sign, and then the original expression will look like x + 3.

If x < 0, then the modulus is expanded with a minus sign, and then the original expression is -x + 3. To make this expression more readable, swap the terms, we obtain 3 - x

Now let's write down the solution this way:

Let us check this solution for arbitrary values of x.

Suppose we need to find the value of the expression |x|+ 3 when x = 5. Since 5 β₯ 0, the modulus contained in the expression |x|+ 3 will unfold with a plus sign and then the solution will take the form:

|*x*|+ 3 = *x *+ 3 = 5 + 3 = 8

Find the value of the expression |x|+ 3 when x = -6. Since -6 < 0, the modulus contained in the expression |x|+ 3 will decompose with a minus sign and then the solution will take the form:

|*x*| + 3 = 3 β *x* = 3 β (β6) = 9

**Example 6.** Expand the modulus in the expression x +|x + 3|

If x + 3 β₯ 0, then the modulus |x + 3| expands with a plus sign and then the original expression takes the form x + x + 3, hence 2x + 3.

If x + 3 < 0, then the module |x + 3| expands with a minus sign and then the original expression takes the form x - (x + 3), where x - x - 3 = -3.

Let's write the solution this way:

Note that the conditions x + 3 β₯ 0 and x + 3 < 0 are inequalities. They can be reduced to a simpler form by solving them:

Then the conditions from the solution can be replaced by the equivalent conditions x β₯ -3 and x < -3

In the second case, when x is strictly less than -3, the expression x +|x + 3| will always be a constant number -3.

For example, find the value of the expression x +|x + 3| when x = -5. Since -5 < -3, according to our solution the value of expression x +|x + 3| will be -3

When x = -5,

*x* +|*x* + 3| = *x β x β* 3 = β5* β *(β5)* β* 3 = β3

Find the value of the expression x +|x + 3| when x = 4. Since 4 β₯ -3, according to our solution the modulus of expression x +|x + 3| is expanded with a plus sign, and then the original expression takes the form 2x+3, where by substituting 4 we obtain 11

When x = 4,

*x* +|*x* + 3| = 2*x*+3 = 2 Γ 4 + 3 = 8 + 3 = 11

Find the value of the expression x +|x + 3| at x=-3.

Since -3 β₯ -3, according to our solution, the modulus of the expression x +|x + 3| is expanded with a plus sign, and then the original expression takes the form 2x+3, where by substituting -3 we obtain -3

*x* +|*x* + 3| = 2*x *+ 3 = 2 Γ (β3) + 3 = β6 + 3 = β3

**Example 3.** Expand the module in expression

As before, we use the module disclosure rule:

But this solution will not be correct because in the first case the condition x β₯ 0 is written, which admits that when x = 0 the denominator of expression turns to zero, and you cannot divide by zero.

In this example, it is more convenient to use a detailed notation of the modulus disclosure rule, where the case in which x = 0 is considered separately

Let's rewrite the solution this way:

In the first case, the condition x > 0. Then expression becomes 1. For example, if x = 3, then the numerator and denominator equals 3, which results in 1

And so it will be for any x greater than zero.

In the second case the condition x = 0 is written. Then there will be no solutions, because you cannot divide by zero.

In the third case, the condition x < 0 is written. Then expression becomes -1. For example, if x = -4, then the numerator becomes 4 and the denominator becomes -4, from which we get one -1

**Example 4.** Expand the module in expression

If x β₯ 0, then the modulus contained in the numerator will expand with a plus sign, and then the original expression will take the form , which for any x greater than zero will be equal to one:

If x < 0, then the modulus expands with a minus sign, and then the original expression takes the form

But we must keep in mind that at x = - 1 the denominator of expression turns to zero. Therefore the second condition x < 0 should be supplemented with a note about what values x can take

## Conversion of expressions with modules

A module included in an expression can be treated as a full-fledged multiplier. It can be reduced and taken out of brackets. If a modulus is part of a polynomial, it can be added to a similar modulus.

Like an ordinary letter multiplier, a modulus has its own coefficient. For example, the coefficient of a module |x| is 1, and the coefficient of a module -|x| is -1. The coefficient of the module 3|x+1| is 3, and the coefficient of the module -3|x+1| is -3.

**Example 1.** Simplify the expression |x| + 2|x| - 2x + 5y and reveal the modulus in the resulting expression.

**Solution**

The expressions|x| and 2|x| are similar terms. Let us add them. Leave the rest unchanged:

Expand the modulus in the resulting expression. If x β₯ 0, then we get 3x - 2x + 5y, where x + 5y.

If x < 0, then we get -3x - 2x + 5y, whence -5x + 5y. We remove the factor -5 from the parentheses and obtain -5(x - y)

As a result, we have the following solution:

**Example 2.** Expand the modulus in the expression: -|x|

Solution

In this case there is a minus in front of the modulo sign. It can be understood as the minus one in front of the modulus sign. If x β₯ 0, then the modulus will expand with the plus sign, and then the original expression will look like -x

If x < 0, then the modulus expands with a minus sign, and then the original expression takes the form -(-x) where we get just x

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