Real Numbers – Full Chapter – Class IX

We have studied Rational Numbers and their properties. Every rational number is of the form p/q, where p is an integer and q is a natural number; like 3/5, 1/3, -7/11. We also know how to add two rational numbers and how to multiply two rational numbers. In this chapter let us study Real Numbers in brief…

For example,

Real Numbers


Real Numbers Exercise 1.2.1

1. Prove that fraction (m(n+1)+1)/(m(n+1)-n) is irreducible for every positive integer’s m and n.

Solution:

Real Numbers

Real Numbers


2. Do there exist 10 distinct integers such that the sum of any 9 of them is a perfect square?

Solution:
Yes, for example 61, 56, 53, 48, 37, 13, -2, -28, -59, -108. Infect one can find infinitely many such sets.


3. Prove that any integer a ≥ 7 is the sum of two relatively prime integers. (we say two integers m and n are relatively if their HCF (i.e. GCD) is 1)

Solution:
When a is odd
a = p + q where either p is even or q is even
If p is even the q is odd and hence p and q are relatively prime to each other. Similarly vice versa. i. e. GCD (p, q) = 1
Thus any integer n ≥ 7 is the sum of two relatively prime integers.


Real Numbers

Real Numbers


5. Find the smallest natural number of the form 2a3b7c, such that the half of the number is the cube of an integer, one third of the number is the seventh power of an integer and one seventh of the number is square of an integer.
[Hint: (2a3b7c)/2 is a cube only if a – 1, b, c is all divisible by 3].

Solution:

Given (2a3b7c)/2 = n3 => 2(a-1)3b7c = n3

(2a3b7c)/3 = m3 => 2(a)3(b-1)7c = m7

(2a3b7c)/7 = l2 => 2(a)3b7(c-1) = l2

By trial and error method we find that

a = 28 b = 36 and c = 21

Furthermore a – 1, b, c are divisible by 3


Now let us study real numbers,

Real numbers are those which comprises of both rational numbers and irrational numbers.

The expansion of a rational number, after carrying out the division, consists of:

(1) integer part (the integer which you get before the decimal point);

(2)some finite number of digits which is not a repeating part and which occur soon after the decimal point;

(3) a string of digit which repeats itself and which start repeating after the non-repeating part.

 

numberexpansioninteger partnon-repeating(impure) partrepeating(periodic) partlength
1/30.3333….0nil31
1/2500.00400000…0004000
1/140.0714285714285…007142856
1517/99000.15323232…015322
41/66.8333333…6831

The repeating part of a rational number is called the periodic part or simply  the period of the rational number. The non-repeating or nonrecurring part is some times called the impure part of the rational number. Thus a rational number has three parts: integer part: impure part: periodic part. The number of digits in the period is called its length.

There may not be any impure part for a rational number: like 1/3 = 0.3333333… in such case we say the rational number is purely periodic. If the periodic part starts after the non-repeating part, we say the expansion is eventually periodic. Thus 0.333333… is denoted by 0.3


Theorem 1: Any eventually periodic decimal number corresponds to a rational number.

Proof:

Let us take an eventually periodic decimal expansion. It is of the form a.p­­1.p­2…p­k.q1.q2…..ql ,where a is an integer, p­­1.p­2…p­k is the non-repeating part and q1.q2…..ql is the periodic part. We consider only part r = 0. p­­1.p­2…p­k q1.q2…..ql and show that this is a rational number. The given number is simply a + r. We multiply r by 10k and obtain

10k.r = p­­1.p­2…p­k q1.q2…..ql ———————– (1)

Again multiplying this by 10k, we obtain

10l(10k.r) = p­­1.p­2…p­kq1q2ql . q1.q2…..ql ———————– (2)

Observe b = p­­1.p­2…p­k and c = p­­1.p­2…p­kq1q2…ql are two numbers in the base 10.

Subtracting (2) from (1), we obtain,

(10k+l – 10k) r = c – b

Hence, r = (c – b)/( (10k+l – 10k) r.

This shows that r is a rational number. Finally the given number is a+r is also rational since a is an integer and r is rational number.


A number is rational if and only if it has a decimal expansion which is either terminating or eventually periodic.

A number is rational if and only if it has a decimal expansion which is eventually periodic.


Rational Numbers – Exercise 1.2.2

  1. Write down the decimal expansion of the following number:

(i) 13/31;

(ii) 123/35;

(iii) 103/625;

(iv) 68/35;

(v) 2013/1024;

(VI) 2/17;

(vii) 123/750;

Which of them have terminating expansion? What are the periods?

(i) 13/31

Solution:

Real Numbers Exercise 1.2.2

This has a non terminal expansion.


(ii) 123/35

Solution:

Real Numbers Exercise 1.2.2


(iii) 103/625;

Solution:

Real Numbers Exercise 1.2.2

(iv) 68/35;

Solution:

Real Numbers Exercise 1.2.2


(v) 2013/1024;

Solution:

Real Numbers Exercise 1.2.2

(VI) 2/17;

Solution:

Real Numbers Exercise 1.2.2


(vii) 123/750;

Solution:

Real Numbers Exercise 1.2.2


  1. Find the rational number which is represented by the following decimal numbers:

(i) 0.999999

Solution:

Let, r = 0.9999 …….

Then, 10r  = 9.9999 ……

10r – r = 9.0000

9r = 9

r = 9/9 = 1

(ii)2.0012

Solution:
Let r = 2.0012…….
100r = 200.1212 ……
10000 r = 20012.1212
10000 r – 100r = 19812.00
9900r = 19812
r = 19812/9900
r = 𝟏𝟔𝟓𝟏/𝟖𝟐𝟓


(iii) 2013.13

Solution:

Let r = 2013.1313…….

100r = 201313.13……

100r – r = 199300.00

99r = 199300

r = 𝟏𝟗𝟗𝟑𝟎𝟎/𝟗𝟗


(iv) 0.112233

Solution:

Let r = 0.112233

100r = 11.2233……

1000000r = 112233.2233……

(106r – 102r) = 112233.2233 – 11.2233 = 112222.00

999900r = 112222

r = (112222)/(999900)

r = (5101)/(45450)


  1. Construct rational numbers with period of lengths

(i) 10

Solution:

Let r = 4.1234543213

1010 r  = 41234543213. 1234543213

(1010  – 1) r = 41234543209.0

r = (41234543209)/(1010−1)


(ii) 12

Solution:

Let r = 8.431354789265

1012  r = 8431354789265. 431354789265

(1012  – 1) r = 8431354789257.0

r = (8431354789257)/(1012 −1)


(iii) 15

Solution:

Let r = 2.345464748494142

1015  r = 2345464748494142. 345464748494142

(1015  – 1) r = 2345464748494140.0

r = (2345464748494140.0)/(1015−1)


  1. Find the rational numbers whose decimal expansions are: (i) 0.142857 (ii) 0.142857. Are these two same?

Solution:

(i) 0.142857

We have r = 0.142857 = (142857)/(1000000)

r = (142857)/(1000000)

(ii) 0.142857

Let r = 0.142857

106  r = 142857. 142857

(106  – 1) r = 142857.0

r  = 142857/999999

We observe that r = 1/7

The two rational numbers are not the same.


  1. Write 1 as an infinite decimal.

Solution:

We can write 1as infinite decimal as 0.9999…… both 1.00 and 0.9999….. are rational numbers which are one and the same. The first one terminates and the second one is periodic hence infinite. i.e., 1 = 0.9


1.2.3 Irrational numbers

A number having an infinite decimal expansion which is not eventually periodic is called an irrational number. Some examples for irrational numbers: √2, √3, pi, 7√5, In fact, √p is irrational for any prime number p.


For any two rational numbers r1 and r2 such that r1 < r2, we can find an irrational number i1 such that r1 <i1 < r2 . in fact we can find infinitely many rational numbers between any two rational numbers. This property is true for any two distinct rational numbers.

For example:

Let us find 3 rational numbers between ½ and 2/3

Observe ,

½ = 0.5 = 0.4999999…, and

2/3 = 0.666666…

Consider the number a = 0.50500500050000… This is larger than 0.5 and is irrational. But this is smaller than 0.6666…, Similarly, the number b= 0.60600600060000…is an irrational number and is smaller than 0.66666… Consider another number c = 0.56566566656666… This is an irrational and lies between a and b. (we can write it as a < c <b)


For any two irrational numbers i1 and i2 such that i1 < i2 , we can find a rational  number r1 such that i1 < r1 < i2. In fact we can find infinitely many rational numbers between any two irrational numbers. This property is true for any two distinct irrational numbers.

For example:

Let us find 3 rational numbers between 1.0100100010000… and 1.012303003000…

Observe these two are irrational numbers. Consider the numbers r1 = 1.010010001000…, r2 = 1.011222222…, r3 = 1.0122222222… These are three rational numbers and

1.0100100010000… < r1 < r2 < r3 < 1.012303003000…


Real Numbers – Exercise 1.2.3

1. Find 5 irrational numbers between 4 and 5.

Solution:
The numbers a = 4.101001000….
b = 4.02002000…..
c = 4.303003000……
d = 4.404004000……
e = 4.05005000…..
Are 5 irrational numbers between 4 and 5 since 4 < a < b < c < d < e < 5


2. Find 6 rational numbers between √𝟐 and √𝟑

Solution:

Let √2,

Real Numbers 1.2.3
Therefore, √𝟐 = 1.414213

Let √3,

Real Numbers 1.2.3

√3 = 1.7320508

We consider r1 = 1.422222……..

2 = 1.433333…….

r3 = 1.44444……..

r4 = 1.5555……..

r5 = 1.666666……

r6 = 1.7111111……

These are rational numbers between 2 and 3

√2< r1< r2 < r3 < r4 < r5 < r6 < √3


3. Write 5 irrational numbers between √𝟑 and √𝟓

Solution:

Real Numbers - Exercise 1.2.3

Therefore, √3 = 1.7320508

Then, √5,

Real Numbers - Exercise 1.2.3

Therefore, √5 = 2.236067

Consider a = 1.7404004000……
b = 1.7505005000……
c = 1.8202002000……
d = 1.9303003000……
e = 2.101001000… are 5 irrational numbers between √3 and √5


4. Prove that √2 + √5 is irrational.

Solution:

If possible, let √2 + √5 be rational
Then (√2 + √5) is rational
═> (√2 + √5) is rational
═> 2 + 5 + 2√2 .√5
═> 7 + 2 √10 is rational
But (7 + 2 √10 ) being the sum of a rational and an irrational. These we arrive at a contradiction to the assumption that ( √2 + √5 ) is rational.
Hence is ( √2 + √5 ) is irrational


5. Give an example of an irrational number such that its 8-th power is a rational number.

Solution:

√28 = 21/8 is an irrational number.

The 8th power is (21/8 )8 = 21/8 x 8 = 21 = 2

2 is a rational number.


6. Why is 0.111222333444……., where each number appears 3 times in a row. Irrational?

Solution: 
0.111222333444….., where each number appears 3 times in a row is a non- terminating, non-recurring decimal expansion. Hence it is irrational.


7. Find 3 rational numbers between 0.1122334455….. and 0.1234567…. here each number appears 2 times in the first expansion and each number appears only once in the second expansion.

Solution:

Consider r1 = 0.12222…..

r2 = 0.122333…..

r3 = 0.122444……

r4 = 0.122555……

r5 = 0.122666……

This is 5 rational numbers between 0.1122334455….. and 0.1234567……


8. Find the 3 irrational numbers between 2/3 and ¾ using their decimal expansion.

Solution:

Real Numbers - Exercise 1.2.3

Consider a = 0.6707007000….

b = 0.6808008000….

c = 0.6909009000…

d = 0.707007000….

e = 0.7202002000….

These are 5 irrational numbers and 2/3 < a < b < c < d < e < 3/4


1.2.4 Real Number System

The collection of all rational numbers and irrational numbers together is the set of all Real numbers. This is denoted by ℜ. Thus every real number is either a rational number or an irrational number. We can put two binary operations  on ℜ which are called addition (denoted by +) and multiplication (denoted by . or x). With these operations ℜ has all the nice properties we have seen in Q, the set of all rational numbers. With the background we have, it is not possible to properly define these operations and prove the properties of these operations. We take them for granted and record all these properties:

(1) closure property: 

The set ℜ of real numbers is closed under addition and multiplication. Thus we can add any two real numbers and we get another real number; for example 2 and √2 can be added to get  (2 + 2√2) which again is a real number; for example (√5 x ∛3) is a real number. More generally, for any two real numbers and b, both  a + b  and a x b  are real numbers.

(2) Commutative property: 

Just like rational numbers, addition and multiplication on real numbers also have commutative

For example we have,

∛7 + 1.234 = 1.234 + ∛7 ; √2 x ∛1.75 = ∛1.75 x √2.

(3) Associative Property: 

Suppose you have 3 real numbers, say a, b and c,  then you have

(a + b) + c = (a + b) + c;

(a x b) x c = a x (b x c)

Thus, for exampple,

(2 + √3) + π = 2 + (√3 + π);

(0.1234567891011… x 1.8) x √5 = 0.1234567891011… x (1.8 x √5)

(4) Distributive Property: 

In real number system, the multiplication is distributive over addition. If a, b and are three real number, then

a x (b + c) = (a x b) + (a x c);

For example,

π x (√17 + 1/17) = (π x √17) +(π  x 1/17)

(5) Additive and Multiplicative Identities: 

In real number system, the multiplication is distributive over addition, such that,

a + 0 = a, for all real numbers a.

a x 1 = a, for all real numbers a.

Thus we see that, for example, (√(1+5√2)) + 0 = (√(1+5√2)) and (π x √8) x 1 = (π x √8). We say that 0 is the additive identity and 1 is the multiplicative identity.

(6) Additive inverse and multiplicative inverse: 

For every real number a, there exists a unique real number denoted by (-a) such that, a + (-a) = 0 = (-a) + a. We say that (-a)  is the additive inverse of a. Thus for example the additive inverse of  (1 + √5) is -(1 + √5) since,  (1 + √5) + [-(1 + √5)] = 0.

Similarly, if  is anon-zero real number then there exists a unique real number denoted by b-1 such that b x (b-1) = 1 = (b-1) x b. This is called multiplicative inverse of b. We also denote b-1 as 1/b. The most important thing here is that the multiplicative of a real number b exists if and only if b ≠ 0. Thus if we take the real number √13, then its multiplicative inverse  1/√13 exists as real number and √13 x (1/√13) = 1.

(7) Cancellation laws: 

Suppose a,b and are real numbers such that a x c = b x c. If c ≠ 0, then a = b. This means we can cancel non-zero quantity on both the sides  of an equality. This is a very useful property while solving equations. Similarly, if a + c = b + c, then cancelling c on both the sides we get  a = b. This is again useful in solving equations, since x + a = b gives  x = b – a.

(8) Ordering property: 

Given any two real numbers and b, we can compare the,: either  a = b or a < b or b < a,  and exactly one of these three relation holds. Thus we can order the real numbers.

In practice, comparing two real numbers may be harder. The rational numbers and irrational numbers are inseparably intertwined in this system: between any two real numbers, we can find infinitely many rational and irrational numbers.

Suppose and are real numbers such that  a < b  and let  be a positive real number, i.e.,  c > 0. Then a x c < b x c. Thus we can multiply an inequality by a positive real number without affecting it. If d > 0, then  a x d > b x d; thus the inequality changes when it is multiplied by a negative number. For example: we have 7 < 10. If we multiply both sides by -6, 7 x (-6) = -42 and 10 x (-6) = -60. We know -42 > -60. Thus 7 x (-60) > 10 x (-6), and the inequality is reversed. Hence it is very important to check the sign og the number before we multiply an inequality by it.

Here we define subtraction and division  using addition and multiplication respectively. Suppose we want to subtract a real number  by another real number a. We define this by a + (-b), where -b is the additive inverse of  and we denote this as a – b. Similarly, if b  ≠ 0 is a real number and is some other real number, we define the division of by to be a x (b-1) where (b-1) is multiplicative inverse of b. We denote this by a/b.

We know that Q contains Z, the set of all integers. Now ℜ contains Q and irrational numbers. We can put all these pictorially.

Real Numbers

Thus, we see that,

N ⊂ W ⊂ Z ⊂ Q ⊂ R, J ⊂ R and R = Q U J

We know there is no rational number p/q such that (p/q)² = 2. However, we have included √2 in the real number system R and (√2)² = 2.

Given any positive integer and positive real number a, there is a unique real number such that  bn = a. If is an odd positive integer, then for any real number there is a unique real number such that bn = a.


Example: For any real number a, prove that a x 0 = 0.

Solution:

For any real number a, by the additive identity property of 0. Using distributive property

a x 0 = a x (0 + 0) = a x 0 + a x 0.

We can cancel a x 0 on both sides and get a x 0 = 0.

Example: Prove that (-1) x (-1) = 1

Solution:

Since 1 is the multiplicative identity, we know that 1 x (-1) = (-1).  Using distributive property,

0 = (1 – 1) x (-1) = 1 x (-1) + (-1) x (-1)

Thus, (-1) x (-1) + (-1) = 0. Adding 1 on both sides, we get,

(-1) x (-1) = 1


Example : Prove that the square of a real number is always non-negative.

Solution:

Suppose a is a real number. Then either a > 0 or a = 0 or a < 0 and only one of these is true.

Suppose a > 0. Then a x a > 0 x a = 0, since we are multiplying by a positive real number. Hence a² > 0. 

If  a = 0. then a² > 0.

Suppose a < 0. Then a x a < 0 x a, as the inequality changes when multiplied by a negative number. Again we obtain a² > 0.

We conclude that a² > 0 when is either positive or negative; and a² = 0 when a = 0. Thus a² ≥ 0 for any real number a.


Real Number  – Exercise 1.2.4

1. Give an example to each of the property of real numbers associativity and computability of addition and multiplication; distributivity of multiplication over addition.

Solution:
Associative property
(a) Let a = 2, b = 5 and c = 8
Then a + (b + c) = 2 + (5 + 8)
= 2 + 13
= 15
a + (b + c) = (2 + 5) + 8
= 7 + 8
= 15
a + (b + c) = (a + b) + c
This is associative property of addition
(b) Now a. (b. c) = 2. (5 . 8)
= 2.40
= 80
(a. b). c = (2.5).8
= 10.8
= 80
This is associative property of multiplication.

Commutative property:
Let a = 3 and b = 7
(a) a + b = 3 + 7 = 10
b + a = 7 + 3 = 10
a + b = b + a
This is commutative property of addition.
(b) a. b = 3.7 = 21
b. a = 7.3 = 21
∴ a. b = b .a
This is commutative property of multiplication.
Distribution of multiplication over addition.
Let a = 4, b = 6 and c = 9
a (b + c) = 4(6 + 9) = 4 x 15 = 60 …… (1)
ab + bc = 4.6 + 4.9
= 24 + 36
= 60 …… (2)
From (1) and (2) we see that
∴ a (b + c) = ab + bc
This is left distribution law.
Also

(b + c) a = (6 + 9). 4 = 15.4 = 60 ……. (3)
ba + ca = 6.4 + 9.4 = 24 + 36 = 60 …….. (4)
From (3) and (4)
(b + c) a = ba + ca
This is right distributive law.


2. What are the properties of R used in the following?
(i) 8 x 7 = 7 x 8

Solution:
Commutative property of multiplication.
(ii) n + (𝛑 + c) = (n + 𝛑) + c

Solution:
Associative property of addition
(iii) 0 + 0 = 0

Solution:
Identity element property with respect to addition. 0 is the identity element
(iv) 𝛑 x 1 = 𝛑
Solution:

Identity element property with respect to multiplication.

(v) √2 (1 + √2 ) = 2 + √2
Solution:

Distribution property of multiplication of over addition.


3. Find the additive inverse of each of the following:
(a) √𝟓

(b) 1 + 𝛑

(c) 7 + 2(1/4)

(d) –𝟑/√𝝅

(e) (–3 + √𝟑)²
Solution:
(a) Additive inverse of √5 is – √5
(b) Additive inverse of (1 + 𝛑) is –(1 + 𝛑)
(c) Additive inverse of 7 + 2(1/4) is -(7+ 2(1/4))

(d) Additive inverse of (−3/√𝜋) is (3/√𝜋)
(e) Additive inverse of (–3 + √𝟑)² is -(–3 + √𝟑)²


4. Find the multiplicative inverse of each of the following:

Real Numbers Exercise 1.2.4


Representation Of Irrational numbers on Number Line

We can use number line to represent irrational numbers as we represent rational numbers on number line.

Pythagoras’ theorem:

One of the classical result in geometry dating back to Greek period is Pythagoras’ theorem. Given a right angle triangle ABC with ∠B = 90°,this theorem states that AB² + BC² = CA². This is very useful in constructing newer lengths as we see now.

Take an infinite line and select a point on it which is called origin. This is the point used to represent 0. Now let us take a unit length on it, say OA = 1. Draw a perpendicular AK to the line at K such that AK = 1 Join OK. Pythagoras’ theorem says that OA² + AK² = OK² Thus OK² = 1² + 1² = 1 + 1 = 2. We obtain OK = √2. With O as centre and OK as radius, draw an arc cutting the line at B. Then OB = OK = √2. Thus √2 is represented by the point B on the line. Real Numbers

Suppose we want to represent √3 on the number line. We draw BL perpendicular to the line at B such that BL = 1. Join OL. Using Pythagoras theorem, we get OL² = OB² + BL². But OB = √2 so that OB = 2. Hence OL² = 3. and this gives OL = √3. Now draw an arc with O as centre and OL as radius to cut the line in C. Then we get OC = √3. Real Numbers

We can continue this process in steps; knowing 2, we can represent √5, since 5 = √(2²+1²); knowing √5, we can represent √6 and so on. More generally, suppose we have represented √n on the number line by a point N. we can use this information to represent √(n+1) on the number line.

Draw NS perpendicular to the line at N such that NS = 1. Join OS. The Pythagoras’ theorem gives OS² = ON² + NS² = (n + 1). Hence OS = √(n+1). With O as the centre and OS as radius, draw an arc cutting the line at R. Then OR = OS = √(n+1). Hence R represents √(n+1) on the number line. Real Numbers

Example: Represent √7 on the number line.

Solution: 

Let O be the origin on the line l. Let be on the line such that OA = 1. Draw AB perpendicular to OA at A such that  AB = 1. Then OB² = OA² + AB² = 1² + 1² = 2. Thus OB  = √2. With O as centre and OB as radius draw an arc cutting the line at C. Then OC = OB = √2. Again draw CD perpendicular to l such that CD = 1. Then OD² = OC² + CD² = 2 + 1 = 3. Thus OD = √3. Draw an arc with O as centre and OD as radius to cut in E. Then OE = OD = √3. Draw FE perpendicular to l at such that EF = 2 and join OF. Now,

OF² = OE² + EF² = 3 + 4 = 7.

Hence OF  = √7. With O as centre and OF as radius, cut at G. Then OG = OF = √7. Thus G represents √7 on the line l.

Real Numbers


 

Real Numbers Exercise 1.2.5 – Class IX

1. Represent √𝟖, √𝟏𝟎 and √𝟏𝟑 on the number line.

Solution:

Representation of √8:

Let O be the origin on the line l. Let be on the line such that OA = 2. Draw AP perpendicular to OA at A such that  AP = 2. 
In Δ AOP, by Pythagoras’ theorem, we have, OP² = AP² + OA²
OP² = 2² + 2² = 4 + 4
OP² = 8
OP = √8

Real Numbers exercise 1.2.5

Hence OP  = √8. With O as centre and OP as radius, cut at C. Then OC = OP = √8. Thus C represnts √8 on the line l.
Therefore, OC represents √8.

Representation of √10:

Let O be the origin on the line l. Let be on the line such that OA = 1. Draw AP perpendicular to OA at A such that  AP = 3. 
In Δ AOP, by Pythagoras’ theorem, we have, OP² = AP² + OA²
OP² = 3² + 1² = 9 + 1 = 10
OP² = 10
OP = √10

Real Numbers exercise 1.2.5

 

Hence OP  = √10. With O as centre and OP as radius, cut at C. Then OC = OP = √10. Thus C represnts √10 on the line l.
Therefore, OC represents √10.

Representation of √13:

Let O be the origin on the line l. Let be on the line such that OA = 3. Draw AP perpendicular to OA at A such that  AP = 2. 
In Δ AOP, by Pythagoras’ theorem, we have, OP² = AP² + OA²
OP² = 2² + 3² = 4 + 9 = 13
OP² = 13
OP = √13

Real Numbers exercise 1.2.5

Hence OP  = √13. With O as centre and OP as radius, cut at C. Then OC = OP = √13. Thus C represnts √13 on the line l.
Therefore, OC represents √13.


2. Represent 1 +√5, 2 + 2√3 on the number line.

Solution:

Representation of 1 + √5

Let O be apoint on the number line and A be the origin on the line l. Let OA = 1 and B be another point on the number line such that AB = 1 on the number line l.  Draw BC perpendicular to AB at B such that  BC = 2. 
In Δ ABC, by Pythagoras’ theorem, we have, AC² = AB² + BC²
AC² = 1² + 2² = 1 + 4 = 5
AC² = 5
AC = √5

Real Numbers exercise 1.2.5

Hence AC  = √5. With A as centre and AC as radius, cut at D. Then AC = AD = √5. Thus D represnts √5 on the line l. Since OA = 1. Thus OD = OA + AD = 1 + √5
Therefore, OD represents 1 + √5 on the number line l.

Representation of 2 + 2√3

Let O be apoint on the number line and A be the origin on the line l. Let OA = 2 and B be another point on the number line such that AB = √6 on the number line l.  Draw BC perpendicular to AB at B such that  BC = √6. 
In Δ ABC, by Pythagoras’ theorem, we have, AC² = AB² + BC²
AC² = (√6)² + (√6)² = 6 + 6 = 12
AC² = 12
AC = √(12) = √(4×3) = 2√3

Real Numbers exercise 1.2.5

Hence AC  = 2√3. With A as centre and AC as radius, cut at D. Then AC = AD = 2√3. Thus D represnts 2√3 on the line l. Since OA = 2. Thus OD = OA + AD = 2 + 2√3
Therefore, OD represents 2 + 2√3 on the number line l.


Square Root