Real Numbers Class 10 Worksheet Maths Chapter 1

True and False

Q1: State whether the given statement is true or false :
(i) The product of two rationals is always rational
Ans: True

(ii) The product of two irrationals is an irrational
Ans: False 

(iii) The product of a rational and an irrational is irrational
Ans: True

(iv) The sum of two rationals is always rational
Ans: True

(v) The sum of two irrationals is an irrational.
Ans: False 

Very Short Questions
Q.2. Classify the following numbers as rational or irrational:
(i) 3.1416
Ans: Rational

(ii) 3.142857
Ans: Rational

(iii) 2.040040004......
Ans: Irrational

(iv) 3.121221222...
Ans: Irrational

(v) 3√3
Ans: Irrational

Q.3. Find the prime factorization of 1152
Ans: 1152 = 2⁷ x 32

Q4: Show that the product of two numbers 60 and 84 is equal to the product of their HCF and LCM
Ans: LCM × HCF = 420×12 = 5040 Also, 60 × 84 = 5040

Q5: If p and q are two coprime numbers, then find the HCF and LCM of p and q.
Ans: HCF = 1 and LCM = pq

Short Questions
Q6: Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657)
Ans: H.C.F. (306, 657) = 9 means H.C.F. of 306 and 657 = 9
Required L.C.M. (306, 657) means required L.C.M. of 306 and 657.
For any two positive integers;
their L.C.M. =Product of the numbers / TheirH.C.F.
i.e., L.C.M. (306, 657) =306× 657/9
= 22,338. 

Q7: The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers.
Ans: Since, the product of two numbers
= Their H.C.F. × Their L.C.M.
⇒ One no. × other no. = H.C.F. × L.C.M.
⇒ Other no. =12×240 / 48 = 60.

Long Questions

Q8: Use Euclid’s Division Algorithm to show that the cube of any positive integer is either of the 9m, 9m + 1 or 9m + 8 for some integer m.
Ans: Let a and b be two positive integers such that
a is greater than b; then:
a = bq + r; where q and r are positive integers and 0 ≤ r < b.
Taking b = 3, we get:
a = 3q + r; where 0 ≤ r < 3
=> Different values of integer a are
3q, 3q + 1 or 3q + 2.
Cube of 3q = (3q)³ = 27q³ = 9(3q³) = 9m;
where m is some integer.
Cube of 3q + 1 = (3q + 1)3
= (3q)³ + 3(3q)² ×1 + 3(3q) × 12 + 13
[Q (q + b)³ = a³ + 3a²b + 3ab² + 1]
= 27q³ + 27q² + 9q + 1
= 9(3q³ + 3q² + q) + 1
= 9m + 1; where m is some integer.
Cube of 3q + 2 = (3q + 2)³
= (3q)³ + 3(3q)² × 2 + 3 × 3q × 22 + 23
= 27q³ + 54q² + 36q + 8
= 9(3q³ + 6q2 + 4q) + 8
= 9m + 8; where m is some integer.
∴ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.
Hence the required result.

Q9: Show that any positive integer which is of the form 6q + 1 or 6q + 3 or 6q + 5 is odd, where q is some integer.
Ans: If a and b are two positive integers such that a is greater than b; then according to Euclid’s division algorithm; we have
a = bq + r; where q and r are positive integers and 0 ≤ r < b.
Let b = 6, then
a = bq + r
⇒ a = 6q + r; where 0 ≤ r < 6.
When r = 0
⇒ a = 6q + 0 = 6q;
which is even integer
When r = 1 ⇒ a = 6q + 1
which is odd integer
When r = 2 ⇒ a = 6q + 2 which is even.
When r = 3 ⇒ a = 6q + 3 which is odd.
When r = 4 ⇒ a = 6q + 4 which is even.
When r = 5 ⇒ a = 6q + 5 which is odd.
This verifies that when r = 1 or 3 or 5; the integer obtained is 6q + 1 or 6q + 3 or 6q + 5 and each of these integers is a positive odd number.
Hence the required result.

Q10: Prove that
(i) √2 is irrational number
(ii) √3 is irrational number
Similarly √5, √7, √11…... are irrational numbers.
Ans: (i) Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (≠ 0) such that .√2= r/s
Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get ,√2=a/b
where a and b are coprime.
So, b √2= a.
Squaring on both sides and rearranging, we get 2b² = a². Therefore, 2 divides a². Now, by
Theorem it following that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c², that is,
b² = 2c².
This means that 2 divides b², and so 2 divides b (again using Theorem with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.

(ii) Let us assume, to contrary, that √3 is rational. That is, we can find integers a and b (≠ 0) such that √3=a/b
Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b √3= a .
Squaring on both sides, and rearranging, we get 3b² = a².
Therefore, a² is divisible by 3, and by Theorem, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b² = 9c², that is,b² = 3c².
This means that b² is divisible by 3, and so b is
also divisible by 3 (using Theorem with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.