In this lesson we will look at the concept of modulus of a number in more detail.

## 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

|β3|= 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:

|x1 β x2|

Where x1 and x2 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| = |β3| = 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:

|x1 β x2| = |x2 β x1|

This equation can be read as follows: The distance from x1 to x2 equals the distance from x2 to x1.

## Expansion the module

When we say that |3|= 3 or |-3|= 3 we are performing an action called modulus expansion.
The rule of expansion of a module looks like this:

We haven't used this notation before. The point is that equality can be given by several formulas. The curly bracket indicates that two cases are possible depending on the condition. In this case, the conditions are the entries "if x β₯ 0" and "if x < 0".
Depending on what is substituted for 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.
The second case may seem contradictory, because the notation |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.
For example, find the modulus of the number -7 using the modulus expansion rule:

So x = -7

|β7|

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:

|β3| = β(β3) = 3

|β9| = β(β9) = 9

|β15| = β(β15) = 15

Example 3. Let x = β4 - 6. That is, we consider the modulus of the expression β4 - 6,

|β4 β 6|

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

|β4 β 6| = |2 β 6| = |β4|

x which was equal to β4-6 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:

|β4 β 6| = |2 β 6| = |β4| = β(β4) = 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 (β6) = 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 = β5 β (β5) β 3 = β3

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| = 2x+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+ 3 = 2 Γ (β3) + 3 = β6 + 3 = β3

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

Solution:
Solution:
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