Chapter 27.12 - Chords inside a circle - Solved examples

In the previous section we saw how to draw a square whose area is same as that of a given rectangle. We also saw a method to represent irrational numbers. In this section we will see a few more solved examples.

Solved example 27.26

Draw a triangle of 4, 5, 6 cm and draw a square of equal area

Solution:

1. Draw the triangle ABC as shown in the fig.27.64(a) shown below

We have AB = 6 cm, BC = 5 cm and AC = 4 cm.  We will not write the construction steps for this triangle. However, it may be noted that, intersection of two arcs shown in green colour gives the vertex C

2. We want a square whose area is same as that of △ ABC. So first we must know the area of this triangle

3. Draw a line CD perpendicular to the side AB of the triangle. This is shown in the above fig.27.64(a)

4. Now we have AB as the base and CD as the altitude. So area of △ ABC =

¹⁄₂ × AB × CD ⇒ Area of △ABC = AB × (¹⁄₂ × CD)

5. So, if we get half of the altitude CD, we can multiply it with AB, to get the required area.

6. To get half of CD, draw a perpendicular bisector of CD. This is shown in red colour in fig.(b). Note that, the procedure for drawing the perpendicular bisector is not shown here.

7. Let the perpendicular bisector intersect CD at E. So we get:

 CE = ED = ¹⁄₂×CD

So we have to multiply ED with AB

8. Extend AB towards the right. It is shown by the red dotted line.

9. With ED as radius, and B as center, draw an arc, cutting the extension at F. 

So we have BF = ED

10. Now we want a semi circle with AF as the diameter. For that, draw the perpendicular bisector of AF. This is shown in magenta colour in fig.(b)

11. Let the perpendicular bisector intersect AF at O. Then O is the center of the required semi circle

12. With O as center and OA as radius, draw a semi circle. This is shown in green colour in fig. (c)

13. Erect a perpendicular at B. let it meet the semi circle at G. BG is shown in blue colour in fig(c).

14. By the properties of chords, we have:

AB × BF = BG × BG

15. But AB × BF is the area of △ABC.

• BG × BG is the area of a square whose side is BG

16. So, if we draw a square with side BG, the area of that square will be equal to the area of △ABC

That means, a square with side BG is our required square.   

17. With B as center and BG as radius, draw an arc cutting the red dotted line at H. 

Then BH = BG

18. With G as center and BG as radius, draw an arc

With H as center and BG as radius, draw an arc

These two arcs, which are shown in red colour in fig. (d), will intersect at I

19. So BHIG is the required square

Solved example 27.27
Draw an equilateral triangle of height 3 cm
Solution:
• If we are given the three sides of a triangle, we can easily construct it. 
    ♦ If it is an equilateral triangle, we will need only one side. 
• But in this problem, we are given the height. Consider fig27.65(a) below. An equilateral triangle of side 's' is drawn. Let it's height be 'h'. 

• This height will be perpendicular to the base. So we get two right triangles in side the given equilateral triangle. 
• Consider any one of those right triangles. Applying Pythagoras theorem, we get:  
h² =  s² - (^s⁄₂)² ⇒ h² =  s² - ^s2⁄₄ ⇒ h² =  ^3s2⁄₄ ⇒ h = ^√3s⁄₂.
• In the present problem we have h = 3 cm. So we can write:
3 = ^√3s⁄₂ ⇒ √3 × √3 = ^√3s⁄₂ 
⇒ √3 = ^s⁄₂ ⇒ s = 2 × √3 ⇒ s = √4 × √3 ⇒ s = √12  
• Thus, the side of an equilateral triangle whose height is 3 cm is √12 cm. Now we can do the construction:
1. Draw AB with length 4 cm and BC with length 3 cm. This is shown in fig.27.65(b)
2. Draw the perpendicular bisector of AC. It is shown in magenta colour in fig(b)
3. Let the perpendicular bisector intersect AC at O
4. Draw a semi circle with O as center and  AO as radius
5. Erect a perpendicular BD at B. Length of BD will be equal to √12 cm
6. With B as center and BD as radius, draw an arc. This is shown in fig(c)
With D as center and BD as radius, draw another arc
7. The two arcs will intersect at a point. Name this point as E
8. Join DE and BE. Then △BED is the required triangle.

Solved example 27.28
Draw an isosceles right triangle whose hypotenuse is 4 cm
Solution:
We have to draw an isosceles triangle which is right angled. We know that, in an isosceles triangle, two sides will be equal
• If we are given the length of equal sides and the base angle, we can easily draw it. 
• But in this problem, we are given the hypotenuse. Consider fig27.66(a) below. An isosceles right triangle is drawn

• Let the equal sides be 's'. Then we get:
4² =  s² + s² ⇒ 4² =  2s² ⇒ 16 = 2s² ⇒ s² = 8 ⇒ s = √8
• So the equal sides of an isosceles right triangle, whose hypotenuse is 4 cm is √8 cm. Now we can do the construction
1. Draw AB with length 4 cm and BC with length 2 cm. This is shown in fig.27.66(b)
2. Draw the perpendicular bisector of AC. It is shown in magenta colour in fig(b)
3. Let the perpendicular bisector intersect AC at O
4. Draw a semi circle with O as center and  AO as radius
5. Erect a perpendicular BD at B. Length of BD will be equal to √8 cm
6. Extend BC towards the right. This is shown as red dashed line in fig(c)
7. With B as centre and BD as radius, draw an arc, cutting the extension at E
8. Join DE. Then BED is the required triangle  

Solved example 27.28
Draw a line of length √12 cm. Construct a square with this length as the side. Can you construct a line of length √48 cm in the same figure
Solution:
1. Draw AB with length 4 cm and BC with length 3 cm. This is shown in fig.27.67

2. Draw the perpendicular bisector of AC. It is shown in magenta colour
3. Let the perpendicular bisector intersect AC at O₁
4. Draw a semi circle with O₁ as center and  AO₁ as radius
5. Erect a perpendicular BD at B. Length of BD will be equal to √12 cm
6. Once BD is obtained, the required square BEFD can be easily constructed
7. Now we want to draw √48. We will get 48 as 4 × 12
8.  We already have AB = 4 cm. So extend AC towards the left in such a way that, AG = 12 cm  
9. Draw the perpendicular bisector of BG. It is shown in magenta colour
10. Let the perpendicular bisector intersect AC at O₂
12. Draw a semi circle with O₂ as center and  GO₂ as radius
13. Erect a perpendicular AH at H. Length of AH will be equal to √48 cm

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In the next section, we will see chord which intersect at a point outside the circle.

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Chapter 27.11 - Rectangle into Square of Equal area

In the previous section we saw an application of theorem 27.10 . We also saw some solved examples. In this section we will see a special case of that application.

■ In the previous section we saw this:
• A rectangle was given to us
    ♦ We made a new rectangle of the same area but different dimensions
■ In this section we will see this:
• A rectangle will be given to us
    ♦ We will make a square of the same area.
■ We will learn the method by analysing an actual example:

Solved example 27.22

Draw a rectangle of Length 4 cm and width 2 cm. Draw a square of the same area.

Solution:

Consider the rectangle in fig.27.59(a) below:

Fig.27.59

• It has a length of 4 cm and width of 2 cm. We want to change it into a square. 
• But there is one condition: The area of the rectangle and the new square must be the same. 
• Let the side of the new square be 'x'. It is shown in fig.27.59(b). We want to find this 'x' graphically. Let us try:
• Consider fig27.59(c). Two chords AB and CD intersect at P. 
• Out of the two chords, AB is a diameter. Because it is passing through the centre 'O'
• This diameter intersects the other chord CD in a perpendicular direction. So CD is bisected. 
    ♦ That means, PC = PD = x. See theorem 17.1.
• The lengths of the four pieces are:
PA = 2, PB = 4, PC = x, PD = x
• Imagine a rectangle with length 4 and width 2
    ♦ It's area will be equal to 4×2 = PA×PB
• Imagine another rectangle with length x and width x  
    ♦ It's area will be equal to x×x = PD×PC
■ Based on theorem 27.10 , the two areas will be equal.
• A rectangle with length x and width also x is a square.
• So our next aim is to construct a circle and the chords shown in fig.27.59(c). 
• What we have, is the given rectangle with length 4 cm and width 2 cm. We have to begin our work from that rectangle.

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1. Consider fig.27.60(a) below. The base of the given rectangle is named as PB. So PB = 4. For the ease of construction, the given rectangle should be placed in such a way that the side PB is exactly horizontal.

Fig.27.60

2. With P as centre and PA as radius, draw an arc (shown in yellow colour) which will cut the horizontal through P at A. So PA = 2 cm

3. With AB as diameter, draw a circle. This is shown in fig(b).
• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'. 
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a vertical through P. It will intersect the circle at C and D.
• AB is a diameter of the circle. CD is a chord
• Since PB is horizontal and CD is vertical, ∠PBD = 90^o
• So diameter AB is perpendicular to CD. Then by theorem 17.1, AB is the perpendicular bisector of CD. Thus we get PC = PD
5. Once we get any one side of a square, we can easily construct it. Here, Both PC and PD are equal to the side of the required square. We can use any one of them. 
• We will use the lower PD So that the square will be distinct from the given rectangle
• Thus in the fig(c), the square PP'ED is constructed

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Now, there is an easier method to obtain the required square:

Consider fig.27.60(c) above. Our real aim is to obtain the side PD. If we can obtain this side directly, a lot of work can be saved. Let us see the method for doing it:

Consider fig.27.61(a) below:

Fig.27.61

1. Draw a line AB, 6 cm in length

2. Mark a point P such that PA = 2 cm and PB = 4 cm

3. Draw a semi-circle with AB as diameter

• For drawing the circle, first draw the perpendicular bisector of AB. It will cut AB at centre 'O'. 
• With centre 'O' and OA as radius, draw the circle. This step is not shown in the fig.
4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D
5. PD is the required side. The square PP'ED can then be easily drawn.
■ In this method, we do not even have to draw the original rectangle
■ Also note that, the semi-circle can be drawn on the upper side of AB 

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Solved example 27.23

Let a rectangle be of length 6 cm and width 4 cm. Draw a square of the same area.

Solution:

1. Draw a line AB, (6+4) = 10 cm in length. See fig.27.61(b) above.

2. Mark a point P such that PA = 4 cm and PB = 6 cm

3. Draw a semi-circle with AB as diameter

4. Draw a perpendicular to AB through P. This perpendicular will meet the semi-circle at D
5. PD is the required side. The square PP'ED can then be easily drawn.

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Irrational numbers

Let us analyse the above two solved examples.
(i) In the solved example 27.22, we got a square whose area is same as a rectangle of length 4 cm and width 2 cm
• So the area of the newly formed square is 8 cm².
• If the area of a square is 8 cm², then obviously, it's side would be √8 cm
• Thus the length of PD = √8 cm 
(ii) In the solved example 27.23, we got a square whose area is same as a rectangle of length 6 cm and width 4 cm
• So the area of the newly formed square is 24 cm².
• If the area of a square is 24 cm², then obviously, it's side would be √24 cm 
• Thus the length of PD = √24 cm 
■ So this is an excellent method to find the values of irrational numbers graphically. 
• Note that in an earlier chapter, we had learned another method for doing this. Details here.

Another example:
Consider fig.27.62(a) below:

Fig.27.62

1. Out of the two chords, chord AB is a diameter. Chord CD is drawn perpendicular to AB. 
• So AB will be the perpendicular bisector of CD. Thus PC = PD
2. Now let us apply theorem 27.10:
Multiplying opposite pieces of the same chord, we get: 
PA×PB = PC×PD ⇒ PA×PB = PC² (∵ PC = PD)
3. This is a very useful result. 
• We no longer need the piece PD. So we need not consider the lower part of the circle. 
• That means, all our further calculations will be related to the portion above the diameter AB. 
• That is., we will be dealing with a semi-circle only.
4. In fig.27.62(b) above, the diameter AB (= 8 cm) is split into two parts at 'P'. 
PA is 6 cm and PB is 2 cm. 
5. A perpendicular PC is erected at P in such a way that C lies on the semi-circle with AB as diameter. 
6. Using the equation in (2) above, we get:
 PA×PB = PC². That is., 6×2 = PC² ⇒ 12 = PC² ⇒ PC = √12
■ Let us write a summary of what we have done above:
• We drew a semi-circle with diameter AB = 8 cm
• We split the diameter at P such that PA = 6 cm and PB = 2 cm
• Finally we erected a perpendicular at P in such a way that it meets the circle at C
• We find that PC = √12 cm

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• We know that √12 is an irrational number. We cannot obtain √12 on a scale. 
• But using the above method, we are able to draw a line of length √12 cm with out using a scale.    
• Once we obtain a line of length √12 cm, if required, we can construct a square of area 12 cm². 
• Because area of a square of side √12 cm = √12 ×√12 = 12 cm²
■ The procedure for constructing the square is shown in fig(c)
1. First draw a horizontal line through C
2. Then draw an arc with C as centre and CP as radius. This arc will cut the horizontal line at a point. Name this point as D
3. Through D, drop a perpendicular to AB. Name the foot of this perpendicular as F
4. Then FPCD is a square of area 12 cm²

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Now let us see some solved examples: 

Solved example 27.24

Find the value of 'x' in the figs.27.63(a), (b) and(c) below:

Fig.27.63

Solution:
Case (a): x² = 8×18 ⇒ x² = 144 ⇒ x = √144 ⇒ x = 12
Case (b): x² = 9×4 ⇒ x² = 36 ⇒ x = √36 ⇒ x = 6
Case (c): x² = 2×5 ⇒ x² = 10 ⇒ x = √10

Solved example 27.25

In the fig.27.63(d) above, AD = 10 cm, BD = 6 cm, CD = 2 cm. Find the value of CP, CQ and PQ.

Solution:
• CQ² = AC×CD = (AD-CD)×CD = (10-2)×2 = 8×2 = 16 cm
⇒ CQ = √16 = 4 cm 
• CP² = BC×CD = (BD-CD)×CD = (6-2)×2 = 4×2 = 8 cm
⇒ CP = √8 = 2√2 cm 
• PQ = (CQ - CP) = (4 - 2√2) = 2(2-√2) cm

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A video presentation of a problem can be seen here:

Trapezium problem

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In the next section, we will see a few more solved examples.

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Chapter 22.1 - Rational and Irrational numbers

In the previous section we saw how rational numbers are represented on a number line. In this section we will see irrational numbers.

We have seen examples of irrational numbers in an earlier chapter. We were able to represent √2 on the number line. We saw it in fig.16.6. For convenience it is shown again in fig.22.4(a) below.

Fig.22.4

In the same manner we can represent √3, √5 √7 etc., √3 is shown in fig.b

Now we come to the question of accuracy. We know that √2 = 1.41421356... 
Can we mark a non-terminating non-recurring decimal like 1.41421356... accurately? 
The answer is yes. The method is the same 'zooming in' that we saw in the fig.22.2 in the previous section. Let us see the details:

We know that 1.41421356... lies in between 1 and 2. In the fig.22.5 below, ‘zoom level 1’ shows the portion between 1 and 2, at an enlarged scale.

In this zoom level, we have enough space to clearly mark ten subdivisions between 1 and 2. So the fourth subdivision will be 1.4 and the fifth subdivision will be 1.5.

Then we zoom in on the region between this 1.4 and 1.5. This is shown as 'zoom level 2'. In this zoom level, we have enough space to clearly mark ten subdivisions between 1.4 and 1.5. So the first subdivision will be 1.41 and the second subdivision will be 1.42.

Then we zoom in on the region between this 1.41 and 1.42. This is shown as 'zoom level 3'. In this zoom level, we have enough space to clearly mark ten subdivisions between 1.41 and 1.42. So the fourth subdivision will be 1.414 and the fifth subdivision will be 1.415.

So, if we take the left red subdivision in zoom level 3, it will represent 1.414. It is fairly accurate. We know that 0.004 is a small quantity. But we are able to mark it by 'zooming in'. We can mark a quantity, even if it is non-recurring and non-terminating, however small it may be. All we need to do, is to zoom in to the required level.

Like √3, √7, π etc., also have decimal forms. Even though they are non-recurring and non-terminating decimals, they can be represented on a number line.

So we can conclude that:
Any irrational number can be represented on a number line.

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Now we try to do the reverse: We are given an irrational number on the number line. Can we represent it as a fraction?

The answer is:
No irrational numbers cannot be expressed in the form ^p⁄_q. We have seen the proof for this, in the case of √2. (Details here)

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We will now see some important differences between rational and irrational numbers:
I. Sum of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers ^p⁄_q and ^r⁄_s. Their sum is:
^p⁄_q + ^r⁄_s = ^ps⁄_qs + ^rq⁄_sq = ^(ps+rq)⁄_qs.
(b) Let us analyse the above result. 
• ^p⁄_q and ^r⁄_s are rational numbers. So p, q, r and s are integers (... -4, -3, -2, -1, 0, 1, 2, 3, 4...)
• Also q and s not equal to zero
(c) So ps, rq and qs will be integers. (ps+rq) will also be an integer
(d) Thus, the result in (a) is an [ ^integer⁄_integer]. That means, the result in (a) is a rational number

II The sum of two irrational numbers may be rational or irrational
Example 1: Sum of √3 and √2 is irrational
Proof:

5. If 'a' is a rational number, [a + ¹⁄_a] will also be a rational number.
6. Then ¹⁄₂[a + ¹⁄_a] will also be a rational number
7. But from (4), we have '¹⁄₂[a + ¹⁄_a]' is equal to √3 which is an irrational number.
8. So 'a' is irrational. That means (√3 + √2) is irrational

Example 2: Sum of (1+√2) and (1-√2) is rational
Proof:
• This can be proved just by adding: 1 + √2 + 1 - √2 = 2
• '2' is a rational number

III. Difference of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers ^p⁄_q and ^r⁄_s. Their difference is:
^p⁄_q - ^r⁄_s = ^ps⁄_qs - ^rq⁄_sq = ^(ps-rq)⁄_qs.
(b) From (I) above, we can write:
The result in (a) is an [ ^integer⁄_integer]. That means, the result in (a) is a rational number

IV. The difference of two irrational numbers may be rational or irrational
This can be proved in the same way as in (II)

V. Product of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers ^p⁄_q and ^r⁄_s. Their product is:
(^p⁄_q) × (^r⁄_s) = ^pr⁄_qs. This is a rational number

VI. The product of two irrational numbers may be rational or irrational
Example 1: Consider 6√5. 
1. We have √5 = 2.236...
2. So 6√5 = 6 × 2.236... = 13.416...
3. 13.416... is an irrational number. So 6√5 is an irrational number
4. Similarly we can prove 2√5 is also an irrational number
5. Now take the products. 6√5 × 2√5 = 12 × 5 = 60
6. 60 is a rational number. So we got a rational number when two irrationals were multiplied

Example 2: 
1. √3 × √5 = √15
2. √15 is an irrational number
3. So we got an irrational number when we multiplied two irrational numbers

VII. Quotient of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers ^p⁄_q and ^r⁄_s. Their quotient is:
(^p⁄_q) ÷ (^r⁄_s) = (^p⁄_q) × (^s⁄_r) = ^ps⁄_qr. This is a rational number

VIII. The quotient of two irrational numbers may be rational or irrational
Example 1: ^(8√3)⁄_(√3) = 8 
'8' is a rational number
Example 2: ^(8√15)⁄_(2√3) = ^(8√15)⁄_(2√3) = ^{(8×√5×√3)}⁄_(2√3) = 4√5
'4√5' is an irrational number

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From the above discussion, we can make the following conclusions:

Conclusion 1:
From I and II, we can conclude this:
• Sum of two rational numbers will always be a rational number. 
• But sum of two irrational numbers may be rational or irrational
Conclusion 2:
From III and IV, we can conclude this:
• Difference of two rational numbers will always be a rational number. 
• But difference of two irrational numbers may be rational or irrational
Conclusion 3:
From V and VI, we can conclude this:
• Product of two rational numbers will always be a rational number. 
• But product of two irrational numbers may be rational or irrational
Conclusion 4:
From VII and VIII, we can conclude this:
• Quotient of two rational numbers will always be a rational number. 
• But quotient of two irrational numbers may be rational or irrational

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The above 4 conclusions are applicable when the rational numbers and irrational numbers are treated separately. We will now see what will be the result when they come together in calculations.

Conclusion 5

• Sum of a rational number and an irrational number will always be an irrational number. 

Conclusion 6

• Difference of a rational number and an irrational number will always be an irrational number.

Conclusion 7

• Product of a rational number and an irrational number will always be an irrational number.

    ♦ In this case it is important to mention that, the rational number should not be zero. Other wise, the product will be zero. 

Conclusion 8

• Quotient of a rational number and an irrational number will always be an irrational number.

    ♦ In this case it is important to mention that, the rational number should not be zero. Other wise, the quotient will be zero.

The reader may write his/her own examples related to the above conclusions 5, 6, 7 and 8

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This completes a main portion of our discussion. In the next section we will take up the topic of real numbers.

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Chapter 22 - Real Numbers

In the previous section we completed the discussion on Circles, arcs and sectors. In a previous chapter, we have seen different types of numbers. (see fig.16.1). In this chapter, we will see how and where 'Real numbers' fit into the list of numbers.

First let us recall how numbers are represented on a number line. In fig.22.1(a), a horizontal line segment is drawn between points A and B. We will fix this distance AB as 1 unit.

Fig.22.1

• So, if we take two line segments, each equal to AB, and put them end to end, we will get a distance of 2 units. This distance is the distance from zero to 2 on the number line shown in fig.d

• If we take 3 such segments and put them end to end, we will get a distance of 3 units. This distance is the distance from zero to 3 on the number line shown in fig.d. In this way, we can find the position of any number on the number line.

Now what about fractions? Consider fig.b. The segment AB is divided into 2 equal parts at C. So AC = BC = ¹⁄₂ unit. 
• If we take two line segments, one equal to AB, and the other equal to AC, and put them end to end, we will get a distance 1¹⁄₂ = ³⁄₂ units. If we measure out this end to end distance (from zero) on the number line, we can mark ³⁄₂. It is shown in fig.d

Another example: In fig.c, the segment AB is divided into 3 equal parts at D and E. So AD = DE = BE = ¹⁄₃ unit. 
• If we take three line segments, two of them equal to AB, and the third equal to AD, and put them end to end, we will get a distance 2¹⁄₃ = ⁷⁄₃ units.  If we measure out this end to end distance (from zero) on the number line, we can mark ⁷⁄₃. It is shown in fig.d

In the above examples, we marked points using a geometrical method. That is., we measured the distance from the 'standard 1 unit', or it's fractions, and then marked it on the number line. Can we do the marking with out actual measuring? Let us analyse:

• ³⁄₂ does not give us any problems because in decimal form, it is 1.5. It is a terminating decimal. 
• But ⁷⁄₃ is a different case. We discussed about such numbers here. It is a recurring decimal. It is written as ⁷⁄₃ = 2.333... Using this decimal form, we can represent ⁷⁄₃ on the number line by the following procedure: 

2.333... has infinite number of decimal places. The digit '3' repeats forever. If more number of decimal places are taken, we will get more accuracy. If we want such a greater accuracy, we will have to 'zoom in' on the region between 2 and 3. Because 2.333... lies in between 2 and 3. In the fig.22.2 below, ‘zoom level 1’ shows the portion between 2 and 3, at an enlarged scale.

Fig.22.2

In this zoom level, we have enough space to clearly mark ten subdivisions between 2 and 3. So the third subdivision will be 2.3 and the fourth subdivision will be 2.4. They are specially marked in red because 2.333... lies between them

Next, we zoom in on the region between this 2.3 and 2.4. This is shown as 'zoom level 2'. In this zoom level, we have enough space to clearly mark ten subdivisions between 2.3 and 2.4. So the third subdivision will be 2.33 and the fourth subdivision will be 2.34. They are specially marked in red because 2.333... lies between them

Next, we zoom in on the region between this 2.33 and 2.34. This is shown as 'zoom level 3'. In this zoom level, we have enough space to clearly mark ten subdivisions between 2.33 and 2.34. So the third subdivision will be 2.333 and the fourth subdivision will be 2.334. They are specially marked in red because 2.333... lies between them

So, if we take the left red subdivision in zoom level 3, it will represent 2.333. It is fairly accurate. We know that 0.003 is a small quantity. But we are able to mark it by 'zooming in'. It is like zooming in on Google Maps. At higher zoom levels, we are able to see smaller details. But in our case of number line, there is no limit. We can mark any quantity, however small it may be. All we need to do, is to zoom in to the required level. 
So we can conclude that:
• Any natural number can be represented on a number line
• Any fraction can be represented on a number line

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Now we try to do the reverse: We are given a number line with a mark  on it. We want to represent that mark as a number or a fraction. 
Consider the mark P in fig.22.3(a). It is exactly at 4

Fig.22.3

So we write: The mark P represents number 4. 
Consider mark Q. It is exactly at 6.5 
So we write: The mark Q represents number 6.5, which is equal to ¹³⁄₂
Consider mark R. We have seen that any recurring decimal can be represented on a number line. If it is given that R is 5.333... , can we represent it as a fraction? 
The answer is yes. The procedure is as follows:
• Let x = 5.333...
• 10x = 10 × 5.333... = 53.333... = 48 + 5.333... = 48 + x (∵ x = 2.333..)
⇒ 10x = 48 + x ⇒ 9x = 48 ⇒ x = ⁴⁸⁄₉ = 5³⁄₉ = 5¹⁄₃ = ¹⁶⁄₃

So, even if the given mark on the number line is a recurring decimal, we can convert it into a fraction. Let us see a few more examples:
■ Express 1.272727... as a fraction:
1. We can write: 1.272727... = 1.2̅7
2. A line is drawn above 2 and 7. That means the block '27' repeats for ever
3. Let x = 1.272727... Since two digits are repeating, we will multiply by 100
• 100x = 100 × 1.272727... = 127.272727... = 126 + 1.272727... = 126 + x (∵ x = 1.272727..)
⇒ 100x = 126 + x ⇒ 99x = 126 ⇒ x = ¹²⁶⁄₉₉ = 1³⁄₁₁

■ Express 0.2353535... as a fraction:
1. We can write: 0.2353535... = 0.23̅5
2. A line is drawn above 3 and 5. That means the block '35' repeats for ever
3. Let x = 0.2353535... Since two digits are repeating, we will multiply by 100
• 100x = 100 × 0.2353535... = 23.535353... =  23.3 + 0.2353535... = 23.3 + x (∵ x = 0.2353535...)
⇒ 100x = 23.3 + x ⇒ 99x = 23.3 ⇒ x = ²³³⁄₉₉₀

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■ Based on the above discussion, we can write the following:
1. Any natural number can be represented on a number line
2. Any fraction can be represented on a number line
    ♦ The fractions  like ¹⁄₂, ¹⁄₅, ³⁄₅ etc., that give terminating decimals
    ♦ The fractions like ¹⁄₃, ¹⁄₇, ³⁄₅ etc., that give non-terminating, recurring decimals
■ Reverse of the above is also possible: 
If a point is marked on a number line, it can be represented by any one of the two categories given below:
1. A natural number
2. A fraction
    ♦ The fraction may be one  like ¹⁄₂, ¹⁄₅, ³⁄₅ etc., that give terminating decimals
    ♦ The fraction may be one like ¹⁄₃, ¹⁄₇, ³⁄₅ etc., that give non-terminating, recurring decimals

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■ Any number that can be expressed as a fraction ^p⁄_q where q is not equal to zero is called a rational number. 

• Simple natural numbers like 2, 5 etc., can be written as ²⁄₁ and ⁵⁄₁. So they are rational numbers. 
• Decimals like 0.2, 0.35, 0.5 etc., can be easily written in the form of fractions (with non-zero denominators). So they are rational numbers.
• Non-terminating recurring decimals like 0.333..., 1.2̅7, 0.23̅5, etc., can be written in the form of fractions (with non-zero denominators). So they are rational numbers.

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In this section we saw the details about rational numbers. In the next section we will see irrational numbers.

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