Practice Questions

1. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3*m* or 3*m* + 1 for some integer *m*.

[Hint: Let *x* be any positive integer then it is of the form 3*q*, 3*q* + 1 or 3*q* *+ *2. Now square each of these and show that they can be rewritten in the form 3*m* or 3*m + *1.]

Ans. Let *a* be any positive integer and *b* = 3.

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

And *r* = 0, 1, 2 because 0 ≤ *r* < 3

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

Or,

Where *k*1, *k*2, and *k*3 are some positive integers

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

2. Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Ans. (i) 140 = 2 ´ 2 ´ 5 ´ 7 = 22´ 5 ´ 7

(ii) 156 = 2 ´ 2 ´ 3 ´ 13 = 22´ 3 ´ 13

(iii) 3825 = 3 ´ 3 ´ 5 ´ 5 ´ 17 = 32´ 52´ 17

(iv) 5005 = 5 ´ 7 ´ 11 ´ 13

(v) 7429 = 17 ´ 19 ´ 23

3. Given that HCF (306, 657) = 9, find LCM (306, 657).

Ans. HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

\ LCM × HCF =306 × 657

LCM = 22338

4. Check whether 6*n* can end with the digit 0 for any natural number *n*.

Ans. If any number ends with the digit 0, it should be divisible by 10.

In other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6*n =* (2 ×3)n

It can be observed that 5 is not in the prime factorisation of 6*n*.

Hence, for any value of *n*, 6*n* will not be divisible by 5.

Therefore, 6*n* cannot end with the digit 0 for any natural number *n*.

5. Prove that (3+2) is irrational.

Ans. We will prove this by contradiction.

Let us suppose that (3+2) is rational.

It means that we have co-prime integers *a* and *b* (*b*≠0) such that

⇒

⇒

⇒ * … *(1)

*a* and *b* are integers.

It means L.H.S of (1) is rational but we know that is irrational. It is not possible. Therefore, our supposition is wrong. (3+2) cannot be rational.

Hence, (3+2) is irrational.

Practice Questions

1. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3*m* or 3*m* + 1 for some integer *m*.

[Hint: Let *x* be any positive integer then it is of the form 3*q*, 3*q* + 1 or 3*q* *+ *2. Now square each of these and show that they can be rewritten in the form 3*m* or 3*m + *1.]

Ans. Let *a* be any positive integer and *b* = 3.

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

And *r* = 0, 1, 2 because 0 ≤ *r* < 3

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

Or,

Where *k*1, *k*2, and *k*3 are some positive integers

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

2. Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Ans. (i) 140 = 2 ´ 2 ´ 5 ´ 7 = 22´ 5 ´ 7

(ii) 156 = 2 ´ 2 ´ 3 ´ 13 = 22´ 3 ´ 13

(iii) 3825 = 3 ´ 3 ´ 5 ´ 5 ´ 17 = 32´ 52´ 17

(iv) 5005 = 5 ´ 7 ´ 11 ´ 13

(v) 7429 = 17 ´ 19 ´ 23

3. Given that HCF (306, 657) = 9, find LCM (306, 657).

Ans. HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

\ LCM × HCF =306 × 657

LCM = 22338

4. Check whether 6*n* can end with the digit 0 for any natural number *n*.

Ans. If any number ends with the digit 0, it should be divisible by 10.

In other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6*n =* (2 ×3)n

It can be observed that 5 is not in the prime factorisation of 6*n*.

Hence, for any value of *n*, 6*n* will not be divisible by 5.

Therefore, 6*n* cannot end with the digit 0 for any natural number *n*.

5. Prove that (3+2) is irrational.

Ans. We will prove this by contradiction.

Let us suppose that (3+2) is rational.

It means that we have co-prime integers *a* and *b* (*b*≠0) such that

⇒

⇒

⇒ * … *(1)

*a* and *b* are integers.

It means L.H.S of (1) is rational but we know that is irrational. It is not possible. Therefore, our supposition is wrong. (3+2) cannot be rational.

Hence, (3+2) is irrational.

6. is a

(a) Composite number

(b) Whole number

(c) Prime number

(d) None of these

Ans. (a)and(b)both

7. For what least value of ‘n’ a natural number, is divisible by 8?

(a) 0

(b) -1

(c) 1

(d) No value of ‘n’ is possible

Ans. (c) 1

8. The sum of a rational and an irrational number is

(a) Rational

(b) Irrational

(c) Both (a) & (c)

(d) Either (a) or (b)

Ans. (b) Irrational

9. HCF of two numbers is 113, their LCM is 56952. If one number is 904, then other number is:

(a) 7719

(b) 7119

(c) 7791

(d) 7911

Ans. (b) 7119

10. A lemma is an axiom used for proving

(a) other statement

(b) no statement

(c) contradictory statement

(d) none of these

Ans. a) other statement

11. If HCF of two numbers is 1, the two numbers are called relatively ________ or ________.

(a) prime, co-prime

(b) composite, prime

(c) Both (a) and (b)

(d) None of these

Ans. (a) prime, co-prime

12. is

(a) a terminating decimal number

(b) a rational number

(c) an irrational number

(d) Both (a) and (b)

Ans. (b) a rational number

13. 2.13113111311113……is

(a) a rational number

(b) a non-terminating decimal number

(c) an irrational number

(d) Both (a) & (c)

Ans. (c) an irrational number

14. The smallest composite number is:

(a) 1

(b) 2

(c) 3

(d) 4

Ans. (c) 3

15. is

(a) an integer

(b) an irrational number

(c) a rational number

(d) None of these

Ans. (c) a rational number

16. is

(a) a rational number

(b) an irrational number

(c) both (a) & (b)

(d) neither rational nor irrational

Ans. (b) an irrational number

17. (2+) is

(a) a rational number

(b) an irrational number

(c) an integer

(d) not real number

Ans. (b) an irrational number