**Excercise 1.1**

- Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Solution:**

(i) Check Greater Value 225 > 135 ,Always greater value divide with small number

Divide 225 by 135 get 1 quotient and 90 remainder as

225 = 135× 1 + 90

Divide 135 by 90 get 1 quotient and 45 remainder as

135 = 90×1 + 45

Divide 90 by 45 get 2 quotient and NO remainder as

90 = 45 × 2 + 0

As there is no remainder and divisor is 45,so HCF is 45

(ii) 38220 > 196

Divide 38220 by 196 then we get quotient 195 and no remainder

38220 = 196 × 195 + 0

So no remainder, 196 is divisor therefore HCF is 196

(iii) 867 > 255

Divide 867 by 255 we get quotient 3 and remainder 102 ,so

867 = 255 × 3 + 102

Now, divide 255 by 102,then we get quotient is 2 and remainder 51 as

255 = 102 × 2 + 51

Divide 102 by 51, then we get quotient 2 and no remainder,so

102 = 51 × 2 + 0

As there is no remainder ,so divisor is 51 is HCF

2. Show that any positive odd integer is of the form 6q+1, or 6q + 5, where q is some integer.

**Sol:** We know Euclid’s Algorithm is

a = bq + r , Given equation is : 6q + 1,

so b(Divisor) =6, Now we get equation that 6q +r ,r is remainder and q(Quotient) is equal to or more than 0, r = 1,2,3,4,5 because 0 ≤ r ≤ 6

So possible form is

6q + 0, 6 is divisible by 2,so it is a even number

6q+ 1, 6 is divisible by 2 but 1 is not divisible by 2, so it is odd number

Therefore 6q + 5, 6 is divisible by 2 but 5 is not divisible by 2, so it is odd number

3. An army contingent of 616 members is to march behind an army band of 32 members is a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

**Sol:** By HCF (616,32) we can find maximum number of column in which they can march. (HCF means Highest Common Fector)

By Euclid’s Algorithm to find HCF

616 = 32 × 19 + 8

35 = 8 × 4 + 0

The HCF(616,32) is 8.

Therefore, they can march in 8 column each.

4. Use Euclid’s division lemma to show that the square of any integer is either of form 3m or 3m+1 for some integer m.[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

**Sol:**

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a² = (3q)² or (3q + 1)² or (3q + 2)²

a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 4

= 3 × (3q²) or 3(3q² + 2q) + 1 or 3(3q² + 4q + 1) + 1

= 3k₁ or 3k₂ + 1 or 3k₃ + 1

Where k₁, k₂, k₃ are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1

- Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9
*m*, 9*m*+ 1 or 9*m*+ 8.

**Sol:**

Let a be any positive integer and b = 3

*a* = 3*q* + *r*, where *q* ≥ 0 and 0 ≤ *r* < 3

∴ *a* = 3q or 3*q* + 1 or 3*q* + 2

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When *a* = 3*q*,

*a*^{3} = (3*q*)^{3} = 27*q*^{3} = 9(3*q*)^{3} = 9*m*,

Where *m* is an integer such that *m* = 3*q*^{3}

Case 2: When *a *= 3q + 1,

*a*^{3} = (3*q* +1)^{3}

*a*^{3}= 27*q*^{3} + 27*q*^{2} + 9*q* + 1

*a*^{3} = 9(3*q*^{3} + 3*q*^{2} + *q*) + 1

*a*^{3} = 9*m* + 1

Where *m* is an integer such that *m* = (3*q*^{3} + 3*q*^{2} + *q*)

Case 3: When *a* = 3*q* + 2,

*a*^{3} = (3*q* +2)^{3}

*a*^{3}= 27*q*^{3} + 54*q*^{2} + 36*q* + 8

*a*^{3} = 9(3*q*^{3} + 6*q*^{2} + 4q) + 8

*a*^{3} = 9*m* + 8

Where *m* is an integer such that *m* = (3*q*^{3} + 6*q*^{2} + 4*q*)

Therefore, the cube of any positive integer is of the form 9*m*, 9*m* + 1,

or 9*m* + 8.

**Excercise 1.2**

- Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

**Sol:**

(i) 140 = 2 × 2 × 5 × 7 = 2^{2} × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2^{2} × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

- Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

**Sol:**

(i) 26 = 2 × 13

91 =7 × 13

HCF = 13 (HCF is Higher Common Factor. Choose values those exist both sides)

LCM =2 × 7 × 13 =182 (LCF is Least common Factor)

Product of two numbers 26 × 91 = 2366

Product of HCF and LCM 13 × 182 = 2366

Hence, product of two numbers = product of HCF × LCM

Other way to find LCM is:-

LCM :- 13 × 2 × 7 = 182 ( But Our first priority should by divide by 2 then 3 and so on other prime numbers)

(ii) 510 = 2 × 3 × 5 × 17

92 =2 × 2 × 23

HCF = 2

LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers 510 × 92 = 46920

Product of HCF and LCM 2 × 23460 = 46920

Hence, product of two numbers = product of HCF × LCM

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024

Product of two numbers 336 × 54 =18144

Product of HCF and LCM 6 × 3024 = 18144

Hence, product of two numbers = product of HCF × LCM.

- Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

**Sol:**

(i) 12 = 2 × 2 × 3

15 =3 × 5

21 =3 × 7

HCF = 3

LCM = 2 × 2 × 3 × 5 × 7 = 420

**Excercise 1.3**

- Prove that √5 is irrational.

**Sol:**

Let take √5 as rational number

If *a* and *b* are two co prime number and* b* is not equal to 0.

We can write √5 = *a*/*b*

Multiply by b both side we get

*b*√5 = *a*

To remove root, Squaring on both sides, we get

5*b*^{2} = *a*^{2} … **(i)**

Therefore, 5 divides *a*^{2} (i.e b₂ = a₂ / 5) and according to theorem of rational number, for any prime number *p* which is divides *a*^{2} then it will divide *a* also.

That means 5 will divide *a*. So we can write

*a* = 5*c*

Putting value of *a* in equation **(i)** we get

5*b*^{2} = (5*c*)^{2}

5*b*^{2} = 25*c*^{2}

Divide by 25 we get

*b*2/5 = *c*^{2}

Similarly, we get that *b* will divide by 5

and we have already get that *a* is divide by 5

but *a* and *b* are co prime number. so it contradicts.

Hence √5 is not a rational number, it is irrational.

- Prove that 3 + 2√5 is irrational.

**Sol:**

Let take that 3 + 2√5 is a rational number.

So we can write this number as

3 + 2√5 = *a*/*b*

Here a and b are two co prime number and b is not equal to 0

Subtract 3 both sides we get

2√5 = *a*/*b* – 3

2√5 = (*a*-3*b*)/*b*

Now divide by 2, we get

√5 = (*a*-3*b*)/2*b*

Here *a* and *b* are integer so (*a*-3*b*)/2*b* is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.

Hence, 3 + 2√5 is a irrational number.

- Prove that the following are irrationals:

(i) 1/√2 (ii) 7√5 (iii) 6 + √2

**Sol:**

(i) Let take that 1/√2 is a rational number.

So we can write this number as

1/√2 = *a*/*b*

Here *a *and *b* are two co prime number and *b* is not equal to 0

Multiply by √2 both sides we get

1 = (*a*√2)/*b*

Now multiply by *b*

*b* = *a*√2

divide by a we get

*b*/*a* = √2

Here *a* and *b* are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.

Hence, 1/√2 is a irrational number

(ii) Let take that 7√5 is a rational number.

So we can write this number as

7√5 = *a*/*b*

Here *a* and *b* are two co prime number and *b* is not equal to 0

Divide by 7 we get

√5 = *a*/(7*b*)

Here *a* and *b* are integer so *a*/7*b* is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts. Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.

So we can write this number as

6 + √2 = *a*/*b*

Here a and b are two co prime number and b is not equal to 0

Subtract 6 both side we get

√2 = *a*/*b* – 6

√2 = (*a*-6*b*)/*b*

Here *a* and *b* are integer so (a-6*b*)/*b* is a rational number so √2 should be a rational number. But √2 is a irrational number so it contradicts.

Hence, 6 + √2 is a irrational number.

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