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 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:**

**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.

**5. Find the smallest natural number of the form 2 ^{a}3^{b}7^{c}, 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: (2

^{a}3

^{b}7

^{c})/2 is a cube only if a – 1, b, c is all divisible by 3].

**Solution:**

Given (2^{a}3^{b}7^{c})/2 = n^{3} => 2^{(a-1)}3^{b}7^{c} = n^{3}

(2^{a}3^{b}7^{c})/3 = m^{3} => 2^{(a)}3^{(b-1)}7^{c} = m^{7}

(2^{a}3^{b}7^{c})/7 = l^{2} => 2^{(a)}3^{b}7^{(c-1)} = l^{2}

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.**

** **

number | expansion | integer part | non-repeating(impure) part | repeating(periodic) part | length |

1/3 | 0.3333…. | 0 | nil | 3 | 1 |

1/250 | 0.00400000… | 0 | 004 | 00 | 0 |

1/14 | 0.0714285714285… | 0 | 0 | 714285 | 6 |

1517/9900 | 0.15323232… | 0 | 15 | 32 | 2 |

41/6 | 6.8333333… | 6 | 8 | 3 | 1 |

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.p _{1}.p_{2}…p_{k}.* ,where a is an integer,

*p*is the non-repeating part and q

_{1}.p_{2}…p_{k}_{1}.q

_{2}…..q

_{l }is the periodic part. We consider only part r = 0.

*p*~~q~~and show that this is a rational number. The given number is simply

_{1}.p_{2}…p_{k}_{1}.q

_{2}…..q

_{l}

*a + r*. We multiply

*r*by 10

^{k}and obtain

*10 ^{k}.r = p_{1}.p_{2}…p_{k} *———————– (1)

Again multiplying this by *10 ^{k}*, we obtain

*10 ^{l}(10^{k}.r) = p_{1}.p_{2}…p_{k}q_{1}q_{2}…*———————– (2)

Observe *b = p _{1}.p_{2}…p_{k }*and

*c = p*are two numbers in the base 10.

_{1}.p_{2}…p_{k}q_{1}q_{2}…q_{l }Subtracting (2) from (1), we obtain,

*(10 ^{k+l} – 10^{k}) r = c – b *

Hence, *r = (c – b)/( (10 ^{k+l} – 10^{k}) 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

**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:**

This has a non terminal expansion.

**(ii) 123/35**

**Solution:**

**(iii) 103/625;**

**Solution:**

**(iv) 68/35;**

**Solution:**

**(v) 2013/1024;**

**Solution:**

**(VI) 2/17;**

**Solution:**

**(vii) 123/750;**

**Solution:**

**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)2.00 12**

**Solution:**

Let r = 2.00~~12~~…….

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) **0.11 2233 **

**Solution:**

Let r = 0.11~~2233~~

100r = 11.2233……

1000000r = 112233.2233……

(10^{6}r – 10^{2}r) = 112233.2233 – 11.2233 = 112222.00

999900r = 112222

r = (112222)/(999900)

**r **= (5101)/(45450)

**Construct rational numbers with period of lengths**

**(i) 10**

**Solution:**

Let r = 4.1234543213

10^{10} r = 41234543213. 1234543213

(10^{10 } – 1) r = 41234543209.0

r = (41234543209)/(10^{10}−1)

**(ii) 12**

**Solution:**

Let r = 8.431354789265

10^{12} r = 8431354789265. 431354789265

(10^{12} – 1) r = 8431354789257.0

r = (8431354789257)/(10^{12 }−1)

**(iii) 15**

**Solution:**

Let r = 2.345464748494142

10^{15} r = 2345464748494142. 345464748494142

(10^{15 } – 1) r = 2345464748494140.0

r = (2345464748494140.0)/(10^{15}−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

10^{6} r = 142857. 142857

(10^{6 } – 1) r = 142857.0

r = 142857/999999

We observe that r = 1/7

The two rational numbers are not the same.

**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 ***r _{1} *

**and**

*r*

_{2}**such that**

*r*

_{1}< r_{2}**, we can find an irrational number**

*i*

_{1}**such that**

*r*

_{1}<i_{1 }< r_{2 }**. 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 ***i _{1}*

**and**

*i*

_{2}**such that**

*i*

_{1}< i_{2}**, we can find a rational number**

*r*

_{1}**such that**

*i*

_{1}< r_{1 }< i_{2}**. 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 *r _{1 }= 1.010010001000*…,

*r*

_{2}= 1.011222222…, r_{3}= 1.0122222222… These are three rational numbers and1.0100100010000… < *r _{1} < r_{2} < r_{3 }<* 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,

Therefore, √𝟐 = 1.414213

Let √3,

√3 = 1.7320508

We consider r_{1} = 1.422222……..

r_{2} = 1.433333…….

r_{3} = 1.44444……..

r_{4} = 1.5555……..

r_{5} = 1.666666……

r_{6} = 1.7111111……

These are rational numbers between 2 and 3

√2< r_{1}< r_{2} < r_{3} < r_{4} < r_{5} < r_{6} < √3

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

**Solution:**

Therefore, √3 = 1.7320508

Then, √5,

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:**

√2^{8} = 2^{1/8} is an irrational number.

The 8th power is (2^{1/8} )^{8} = 2^{1/8 x 8} = 2^{1} = 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 r_{1} = 0.12222…..

r_{2} = 0.122333…..

r_{3} = 0.122444……

r_{4} = 0.122555……

r_{5} = 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:**

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 *a *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 *c *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 *b * 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 *c *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 * a *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 *a *and *b *are real numbers such that * a < b * and let *c * 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 *b * by another real number *a. *We define this by *a + (-b), where -b *is the additive inverse of *b * and we denote this as *a – b. *Similarly, if *b ≠ 0 *is a real number and *a *is some other real number, we define the division of *a *by *b *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.

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 ***n ***and positive real number ***a, ***there is a unique real number ***b ***such that *** b ^{n} = a. *

**If**

*n*

**is an odd positive integer, then for any real number**

*a*

**there is a unique real number**

*b*

**such that**

*b*^{n}= 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 *a *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:**

**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 AB*²* + BC*²* = CA*². *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 OA*²* + AK*²* = OK*²* Thus OK*²* = 1*²* + 1*² = 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.

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 OL*²* = OB*²* + BL*². *But OB = √2 so that OB = 2. Hence OL*²* = 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.

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 OS*²* = ON*² + *NS*² = (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.

**Example: Represent √7 on the number line.**

**Solution: **

Let* O* be the origin on the line *l*. Let *A *be on the line such that *OA = 1. *Draw *AB *perpendicular to *OA *at *A* such that * AB = 1. *Then *O**B² = OA² + AB² = 1**²* + 1*²* = 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 *OD**² = OC² + CD²* = 2 + 1 = 3. Thus OD = √3. Draw an arc with *O *as centre and *OD *as radius to cut *l *in *E. *Then *OE = OD = *√3. Draw *FE *perpendicular to *l* at *E *such that *EF = 2 *and join *OF. *Now,

*OF**² = OE² + EF² = 3 + 4 = 7.*

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

## 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 *A *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

Hence OP = √8. With O as centre and OP as radius, cut *l *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 *A *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

Hence OP = √10. With O as centre and OP as radius, cut *l *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 *A *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

Hence OP = √13. With O as centre and OP as radius, cut *l *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 *l *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

Hence AC = √5. With A as centre and AC as radius, cut *l *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 *l *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

Hence AC = 2√3. With A as centre and AC as radius, cut *l *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.*