Chapter 22.3 - Absolute value of a Number

In the previous section we saw the distance between two points. In this section we will see the midpoint between two points. Later in this section, we will learn about Absolute value.

Consider the fig.22.13 below. Two points A and B are marked in red. C is the midpoint between A and B. 

Fig.22.13

Our aim is to find the number corresponding to the midpoint C. The following steps can be used:
1. The numbers corresponding to A and B are x₁ and x₂. 
2. Point A is on the left side of point B. So x₁ is less than x₂. 
3. According to theorem 22.1 that we saw in the previous section, the distance between A and B is (x₂-x₁)
4. Half of this distance is ^(x₂-x₁)⁄₂
• This is the distance between A and C
• This is also the distance between B and C 
5. So, we have the following information:
• The number corresponding to the left side point A is x₁
• The distance between A and C is ^(x₂-x₁)⁄₂ 
• The number corresponding to the right side point B is x₂
• The distance between B and C is ^(x₂-x₁)⁄₂
6. To apply theorem 22.2, we can use the 'first two' or 'last two' of the above four information
7. Let us take the first two. Applying theorem 22.2, 
The number corresponding to C = number corresponding to A + distance between A and C
= x₁ + ^(x₂-x₁)⁄₂ = ^(x₁)⁄₂ + ^(x₂)⁄₂ = ^(x₁+x₂)⁄₂
8. Let us check the above result by using the last two information in (5)
Applying theorem 22.2, 
The number corresponding to C = number corresponding to B - distance between A and C
= x₂ - ^(x₂-x₁)⁄₂ =  ^(2x₂)⁄₂ - ^(x₂)⁄₂ + ^(x₁)⁄₂ = ^(x₂)⁄₂ + ^(x₁)⁄₂ =  ^(x₁+x₂)⁄₂

This is the same result in (7). So we have an easy method to find the midpoint between two points. We can write it in the form of a theorem.
Theorem 22.3:
• We are given two points
• The numbers corresponding to those points are also given
• Take the sum of those numbers
• Half of that sum will be the number corresponding to the midpoint between the given points

Let us see a sample calculation:
A and B are two points on the number line. The number corresponding to A is -2¹⁄₂. The number corresponding to B is 4³⁄₄. Find the number corresponding to the midpoint C between A and B
Solution: Sum of the numbers is (-2¹⁄₂ + 4³⁄₄ ) = 2¹⁄₄.
The number corresponding to the midpoint = half of the above sum = ¹⁄₂ × 2¹⁄₄ = 1¹⁄₈. This is shown in the fig.22.14 below:

Fig.22.14

Now we will see some solved examples
Solved example 22.1
Find the distance between two points on the number line denoted by each pair of numbers given below:
(i) 1, -5     (ii) ¹⁄₂ , ²⁄₃     (iii) -¹⁄₂ , -¹⁄₃     (iv) -¹⁄₂ , ³⁄₄     (v) -√2 , -√3
Solution:
We can find the distance by applying theorem 22.1
In each taking the smaller number as x₁ and the larger number as x₂ , the distance is (x₁-x₂)
(i) Distance = 1-(-5) = 1+5 = 6
(ii) Distance = ²⁄₃ - ¹⁄₂ = ⁴⁄₆ - ³⁄₆ = ¹⁄₆
(iii) Distance = -¹⁄₃ -(- ¹⁄₂) = -¹⁄₃ + ¹⁄₂ = -²⁄₆ + ³⁄₆ = ¹⁄₆
(iv) Distance = ³⁄₄ -(- ¹⁄₂) = ³⁄₄ + ¹⁄₂ = ³⁄₄ + ²⁄₄ = ⁵⁄₄ = 1¹⁄₄  
(v) Distance = -√2 -(-√3) = -√2 + √3  = (√3 - √2)

Solved example 22.2
Find the midpoint of each pair of points in the first problem
Solution:
We can use theorem 22.3 that we saw above in this section. We have to take half of the sum
(i) Midpoint = ^[1+(-5)]⁄₂  = -⁴⁄₂ = -2
(ii) Midpoint = ¹⁄₂ × (¹⁄₂ + ²⁄₃) = ¹⁄₂ × (³⁄₆ + ⁴⁄₆) = ¹⁄₂ × (⁷⁄₆) = ⁷⁄₁₂.
(iii) Midpoint = ¹⁄₂ × [(-¹⁄₂ )+(-¹⁄₃)] = ¹⁄₂ × [(-³⁄₆ )+(-²⁄₆)] = ¹⁄₂ × (-⁵⁄₆) = -⁵⁄₁₂. 
(iv) Midpoint = ¹⁄₂ × [(-¹⁄₂ )+(³⁄₄)] = ¹⁄₂ × [(-²⁄₄ )+(³⁄₄)] = ¹⁄₂ × (¹⁄₄) = ¹⁄₈.
(v) Midpoint = ^{[(-√2)+(-√3)]}⁄₂  = -^(√2+√3)⁄₂

Solved example 22.3
The part of the number line between the points denoted by numbers ¹⁄₃ and ¹⁄₂ is divided into four equal parts. Find the numbers denoting the ends of each such part
Solution:
1. It is better to draw a rough sketch as shown in fig.22.15 below.

Fig.22.15

¹⁄₃ is less than ¹⁄₂ . So ¹⁄₃ (marked as A) will be on the left of ¹⁄₂ (marked as B)
2. The distance between the two points A and B is (¹⁄₂ - ¹⁄₃) = (³⁄₆ - ²⁄₆) = ¹⁄₆.
3. This distance is divided into 4 equal parts. The division is done by three points C, D and E. Each of the 4 equal parts will be (¹⁄₄ × ¹⁄₆ ) = ¹⁄₂₄.
4. So, applying theorem 22.3, the number corresponding to C 
= Number corresponding to A + distance AC
= (¹⁄₃ + ¹⁄₂₄) = (⁸⁄₂₄ + ¹⁄₂₄) = ⁹⁄₂₄ = ³⁄₈.
5. Number corresponding to D = (⁹⁄₂₄ + ¹⁄₂₄) = ¹⁰⁄₂₄ = ⁵⁄₁₂
6. Number corresponding to E = (¹⁰⁄₂₄ + ¹⁄₂₄) = ¹¹⁄₂₄
7. Check: Number corresponding to B = (¹¹⁄₂₄ + ¹⁄₂₄) = ¹²⁄₂₄ = ¹⁄₂.
¹⁄₂ is the number corresponding to B. Thus we reach point B. The calculations are correct.

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Absolute value of numbers

• Consider the point A in the number line shown in fig.22.16 below. The number corresponding to A is -4. 
    ♦ The distance of A from zero is 4 units

Fig.22.16

• Consider point B. The number corresponding to B is 3. 
    ♦ The distance of B from zero is 3 units
In general:
• Consider any point on the right side of zero.
    ♦ It's distance from zero is same as it's number
   ♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
   ♦ The unit may be cm, m or any other appropriate unit
• Consider any point on the left side of zero
    ♦ It's distance from zero is same as it's 'number with out the negative sign'
   ♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
   ♦ The unit may be cm, m or any other appropriate unit

We know that, if we multiply a negative number by '-1', the number will become positive. So, for the points on the left we can say this:
• Consider any point on the left side of zero
    ♦ It's distance from zero is same as it's 'number multiplied by -1'
   ♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
   ♦ The unit may be cm, m or any other appropriate unit

For example, if the number corresponding to a point is -3, the distance of that point from zero is:
-1× -3 = 3 units

Note that if the point under consideration is zero, it's distance from zero is '0'. We can write the three situations together:

Consider any point on the number line. Let the number corresponding to the point be 'x'. Then:
• Distance of the point from zero = x (if x > 0)
• Distance of the point from zero = 0 (if x = 0)
• Distance of the point from zero = -1 x (if x < 0)

This distance of a point from zero is called Absolute value of that number. It is written using symbol as |x|. It is read as 'Absolute value of x'.

We can write the above points in the form of a theorem.
Theorem 22.4
• We have a point A on the number line.
• We know the number corresponding to the point A
• The distance of point A from zero, is the absolute value of the number 

Example:
1. A point is marked on the number line. The number corresponding to that point is 5. 
2. Then the distance of that point from zero = |5| = 5
Another example:
1. A point is marked on the number line. The number corresponding to that point is -2. 
2. Then the distance of that point from zero = |-2| = 2

A spread sheet showing the absolute values of some numbers can be seen here. The formula used is '=ABS()'

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Let us see some common situations where we use the absolute value:

• Consider the equation |x| = 2
• We want the value of x that will satisfy the above equation
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the equation will become equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also. 
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. Thus, there are two values for 'x' that will solve the given equation. They are x = 2 and -2
7. This is shown in the fig.22.17 below:

Fig.22.17

■ We started out to find 'the value' of x which will satisfy the equation |x| = 2. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation.

Another problem:
• Consider the inequality |x| ≤ 2
• We want the value of x that will satisfy the above inequality
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the inequality will become less than or equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also. 
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. But 2 and -2 are not the only points. Look at the inequality symbol '≤'. It is less than OR equal to. Points whose distances are less than 2 will also satisfy the inequality. 
7. For example consider 1.5 and -1.5. Both have a distance less than 2. There are numerous such points
8. All such points which satisfy the given inequality lies between -2 and 2. If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.18 below:

Fig.22.18

9. Note that, the inequality is 'less than OR equal to'. Because of the presence of 'equal to', the points 2 and -2 also qualify to be part of the graph
• We know that, on the red line, there are numerous points. Every point on the red line will satisfy the inequality |x| ≤ 2
    ♦For example weird integers like (-1¹⁸²⁄₂₅₇₄) fall on the red line. They will also satisfy the inequality |x| ≤ 2
10. If the inequality is |x| < 2, then only those points whose distances are less than 2 should be included in the graph. In such a situation, the points at a distance 'exact 2' will be shown in 'hollow circles'.
11. This is shown in the lower number line in fig.22.18 above. Such hollow circles indicate that, those points are not part of the graph.
■ We started out to find 'the value' of x which will satisfy the inequality |x| ≤ 2. When all steps are completed, we find that there are 'numerous values' of x that will satisfy the inequality

One more problem:
• Consider the inequality |x| ≥ 2
• We want the value of x that will satisfy the above inequality
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the inequality will become greater than or equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also. 
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. But 2 and -2 are not the only points. Look at the inequality symbol '≥'. It is greater than OR equal to. Points whose distances are greater than 2 will also satisfy the inequality. 
7. For example consider 2.5 and -2.5. Both have a distance greater than 2. There are numerous such points
8. • All points which lie to the left of -2 are at a distance greater than 2 from zero. They satisfy the inequality
• All points which lie to the right of 2 are at a distance greater than 2 from zero. They satisfy the inequality
If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.19 below:

Fig.22.19

• But the red line is broken. This is because, the points which lie in between -2 and 2 does not satisfy the given inequality
• Arrows are given at the left and right ends of the line. This is to show that the graph extends upto infinity on both sides.
    ♦ All numbers on the left of -2 upto infinity satisfies the inequality
    ♦ All numbers on the right of 2 upto infinity satisfies the inequality
9. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points 2 and -2 also qualify to be part of the graph
• We know that, on the red line, there are numerous points. Every point on the red line will satisfy the inequality |x| ≥ 2
    ♦For example weird integers like (-201⁷⁴²⁄₂₅₇₄) fall on the red line. They will also satisfy the inequality |x| ≥ 2
10. If the inequality is |x| > 2, then only those points whose distances are greater than 2 should be included in the graph. In such a situation, the points at a distance 'exact 2' will be shown in 'hollow circles'.
11. This is shown in the lower number line in fig.22.19 above. Such hollow circles indicate that, those points are not part of the graph.
■ We started out to find 'the value' of x which will satisfy the inequality |x| ≥ 2. When all steps are completed, we find that there are 'numerous values' of x that will satisfy the inequality

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So we have seen the distance between a point and zero. In the next section we will see the distance between two points in terms of absolute value.

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Chapter 22.2 - Real numbers - Distance between points on a Number line

In the previous sections we saw how rational numbers and irrational numbers are represented on a number line. We also saw the results when operations are performed on them. We can now write this:

■ All rational numbers can be represented on a number line

     ♦ All rational numbers represented on a number line can be expressed as a fraction

■ All irrational numbers can be represented on a number line

     ♦ No irrational number can be expressed as a fraction

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We can now take up the topic on real numbers.

■ All numbers which can be represented on a number line are called real numbers.

So both rational numbers and irrational numbers are real numbers

In physics we have learned that:

• Images which can be caught on a screen are called 'real images'. 

    ♦ Virtual images cannot be caught on a screen. 

• In a similar way, numbers which can be represented on a number line are called real numbers.

    ♦ Similar to virtual images, are there virtual or unreal numbers? We will learn about them in higher classes.

Operations on Real numbers

We have seen that real numbers can be represented on a number line. The number line is a geometrical representation of numbers. Positive numbers are marked on the right of zero and negative numbers are marked on the left of zero. This is shown in the fig.22.6 below:

Fig.22.6

Let us see some important facts related to a number line:

• In a number line, as we move towards the right from zero, the numbers get larger and larger
Example: 5 is larger than 4
• As we move towards the left from zero, the numbers get smaller and smaller
Example: -2 is smaller than -1

Instead of zero, we can start our journey from any number. We will get the same experience. 
• Towards the right, the numbers become larger and larger
• Towards the left, the numbers become smaller and smaller

So if we are given any two numbers, we are able to say this:
■ On the number line, the larger number will be on the right of the smaller number
For example, if we are given -34 and 12,
12 is on the right of -34 on the number line
Another example:
-114 is on the right of -150

Distance between numbers on a number line

Number line is a geometrical representation of numbers. So we will be able to measure the distances between points. But we do not have to make actual measurements. We will be able to find distances by simple addition or subtraction.

We know that the distance between any two adjacent integers will be the ‘standard 1 unit’. This standard is to be fixed up before drawing the number line.

An example for finding distances:
• The distance of 5¹⁄₄ from zero is 5¹⁄₄ units
• The distance of 2¹⁄₂ from zero is 2¹⁄₂ units
These are marked in fig.22.7 below:

Fig.22.7

From the fig., it is clear that, the distance between the points representing 5¹⁄₄ and 2¹⁄₂ is obtained as:
5¹⁄₄ - 2¹⁄₂ = 2³⁄₄ units. This is shown in blue colour.

Another case:
This time 5¹⁄₄ and 2¹⁄₂ are both marked on the left of zero. So the points are -5¹⁄₄ and -2¹⁄₂. This is shown in fig.22.8 below:

Fig.22.8

Here also the distance between the two points is 2³⁄₄ units.

Another case:
This time 2¹⁄₂ is marked on the left side of zero and 5¹⁄₄ is marked on the right side of zero. This is shown in the fig.22.9 below. 

Fig.22.9

From the fig. it is clear that, the distance between the points is 7³⁄₄ units. 

One more case:
This time 5¹⁄₄ is marked on the left side of zero and 2¹⁄₂ is marked on the right side of zero. This is shown in the fig.22.10 below:

Fig.22.10

From the fig., it is clear that, the distance between the points is: 7³⁄₄ units.

Let us write a summary of the four cases:

In the above table we have four pairs, corresponding to the four cases. 
■ Each pair has two numbers. 
■ Each of these numbers denotes a point. 
For example, the number '-5¹⁄₄' denotes a point which is 5¹⁄₄ units to the left of zero.

Based on this, let us analyse the table:
Case 1:
1. The number denoting the first point is 2¹⁄₂
2. The number denoting the second point is 5¹⁄₄
3. Distance between the two points is obtained as follows:
• Smaller of the two = 2¹⁄₂
• Subtract it from the larger. That is: 5¹⁄₄ – 2¹⁄₂
• The result is 2³⁄₄
Case 2:
1. The number denoting the first point is -2¹⁄₂
2. The number denoting the second point is -5¹⁄₄
3. Distance between the two points is obtained as follows:
• Smaller of the two = -5¹⁄₄
• Subtract it from the larger. That is: -2¹⁄₂ - (-5¹⁄₄) 
⇒ -2¹⁄₂ + 5¹⁄₄ ⇒ 5¹⁄₄ - 2¹⁄₂ 
• The result is 2³⁄₄ 
Case 3:
1. The number denoting the first point is -2¹⁄₂
2. The number denoting the second point is 5¹⁄₄
3. Distance between the two points is obtained as follows:
• Smaller of the two = -2¹⁄₂
• Subtract it from the larger. That is: 5¹⁄₄ - (-2¹⁄₂)
⇒ 5¹⁄₄ + 2¹⁄₂
• The result is 7³⁄₄
Case 4:
1. The number denoting the first point is 2¹⁄₂
2. The number denoting the second point is -5¹⁄₄
3. Distance between the two points is obtained as follows:
• Smaller of the two = -5¹⁄₄
• Subtract it from the larger. That is: 2¹⁄₂ - (-5¹⁄₄)
⇒ 5¹⁄₄ + 2¹⁄₂
• The result is 7³⁄₄

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So we can see that, in all four cases, the step 3 is the same. That means the procedure for finding the distance is the same. We can write it in the form of a theorem.
Theorem 22.1
1. We have two points on the number line
2. Each point is denoted by a number. Let the numbers be x₁ and x₂
3. Take out the smaller number. Let x₁ be this smaller number
4. Subtract it from the larger
5. The difference (x₂ - x₁) thus obtained is the distance between the points

This is shown in fig.22.11 below:

Fig.22.11

The above procedure is applicable for any positions of the points:
• The two points may both be on the left of zero
• The two points may both be on the right of zero
• One point may be on the left and the other on the right
■ That is why, zero is not shown in the above fig.22.11

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The theorem 22.1 above gives us the method to find the distance between any two points. 
So we obtained ‘distance’  from ‘numbers’. Can we do the reverse? That is., can we obtain number from distance?
The answer is yes. But we can obtain only one number. The distance, and the other number should be given. Let us analyse this:

In the fig.22.12 below, A and B are two points on the number line. Let the numbers denoting them be x₁ and x₂. That is., A is at a distance of x₁ from zero, and B is at a distance of x₂ from zero.

Fig.22.12

[As mentioned earlier, to find the distance, we need only the numbers. The position of zero does not have any role to play. So it is not shown in the above fig.22.12]
In the fig., A is on the left of B. So x₁ is less than x₂
Then, according to theorem 22.1, the distance between A and B is (x₂ - x₁)
■ Suppose that we are given the following two information:
• The number (x₁) corresponding to point A
• The distance between A and B
With the two information, we can find the number corresponding to point B. The steps are as follows: 
1. We know that the number corresponding to B will be greater than x₁. Because B is on the right side of A
2. So add the distance to x₁ 
We will get: x₁ + (x₂ - x₁) = x₁ + x₂ - x₁ = x₂
This is the number corresponding to point B
■ Suppose that we are given the following two information:
• The number (x₂) corresponding to point B
• The distance between A and B
With the two information, we can find the number corresponding to point A. The steps are as follows: 
1. We know that the number corresponding to A will be less than x₂. Because B is on the right side of A
2. So subtract the distance from x₂ 
We will get: x₂ - (x₂ - x₁) = x₂ - x₂ + x₁ = x₁
This is the number corresponding to point A

We can write this in the form of a theorem
Theorem 22.2:
• We are given two points
• The distance between them is available
• But the number corresponding to only one of them is available
■ If the left side number is available then:
Right side number = Left side number + distance
■ If the right side number is available then:
Left side number = Right side number - distance

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So it is clear that, if we are given one point and the distance, we can find the other point. This is useful for finding the midpoint between two points. We will see it in the next section.

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