ML Aggarwal Solutions Class 9 Math 1st Chapter Rational and Irrational Numbers Exercise 1.4
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions First Chapter Rational and Irrational Numbers Exercise 1.4. APC Solution Class 9 Exercise 1.4.
(1) Simplify the following:
(i) √45 – 3 √20 + 4 √5
Solution:
√5 × √9 – 3√4 × √5 + 4√5
= √5 (3 – 3 × 2 + 4)
= √5 (3 – 6 + 4)
= √5 × 1
= √5
(ii) 3√3 + 2√27 + 7/√3
Solution:
3√3 + 2√27 + 7/√3
= 3√3 + 2√9 √3 + 7/√3
= √3 (3 + 6 + 7/√3×√3)
= √3 (9 + 7/3)
= √3 (27+7/3)
= √3 × 34/3
= 34√3/3
(iii) 6√5 × 2√5
Solution:
6√5 × 2√5
= 6 × 2 × 5
= 60
(iv) 8√15+ 2√3
Solution:
8√15 + 2√3
= 8√5 × √3/2×√3
= 4√5
(v) √24/8 + √54/9
Solution:
√24/8 + √54/9
= √6×√4/8 + √6×√9/9
= √6 (2/8 + 3/9)
= √6 (1/4 + 1/3)
= √6 (3+4/12)
= 7√6/12
(vi) 3/√8 + 1/√2
Solution:
3/√8 + 1/√2
= 3/√4×√2 + 1/√2
= 3/2√2 + 1/√2
= 3+2/2√2
= 5/2√2
= 5×√2/2√2×√2
= 5√2/4
(2) Simplify the following:
(i) (5 + √7) (2 + √5)
Solution:
(5 + √7) (2 + √5)
= 10 + 5√5 + 2√7 + √7 √5
= 10 + 5√5 + 2√7 + √35
(ii) (5 + √5) (5 – √5)
Solution:
(5 + √5) (5 – √5)
= 53 – (√5)2 [a2 – b2 = (a+b) (a-b)]
= 25 – 5
= 20
(iii) (√5 + √2)2
Solution:
(√5 + √2)2
= (√5) + 2√5 × √2 + (√2)2
= 5 + 2√10 + 4
= 9 + 2√10
(iv) (√3 – √7)2
Solution:
(√3 – √7)2
= (√3)2 – 2 × √3 × √7 + (√7)2
= 3 – 2√21 + 7
= 10 – 2√27
(v) (√2 + √3) (√5 + √7)
Solution:
(√2 + √3) (√5 + √7)
= √2 × √5 + √2 × √7 + √3 × √5 + √3 × √7
= √10 + √14 + √15 + √21
(vi) (4 + √5) (√3 – √7)
Solution:
(4 + √5) (√3 – √7)
= 4√3 – 4√7 + √5 × √3 – √5 × √7
= 4√3 – 4√7 + √15 – √35
(3) If √2 = 1.414, then find the value of:
(i) √8 + √50 + √72 + √98
(ii) 3√32 – 2√50+ 4√128 – 20√18
Solution:
Given, √2 = 1.414
(i) √8 + √50 + √72 + √98
= √4 × √2 + √25 × √2 + √36 × √2 + √49 × √2
= 2√2 + 5√2 + 6√2 + 7√2
= √2 (2 + 5 + 6 + 7)
= 1.414 × 20
= 28.28
(ii) 3√32 – 2√50 + 4√128 – 20√18
= 3√16 × √2 – 2√25 × √2 + 4 × √64 × √2 – 20 × √9 × √2
= √2 (3×4 – 2×5 + 4×8 – 20×3)
= √2 (12 – 10 + 32 – 60)
= 26√2
= 26 × 1.414
= -36.764
(4) If√3 = 1.732, then find the value of:
(i) √27 + √75 + √108 – √243
(ii) 5√12 – 3√48 + 6√75 + 7√108
Solution:
Given, √3 = 1.732
(i) √27 – √75 + √108 – √243
= √9 × √3 – √3 × √25 + √3 × √36 – √3 × √81
= √3 (3 – 5 + 6 – 9)
= 1.732x – 5
= 8.66
(ii) 5√12 – 3√48 + 6√75 + 7√108
= 5√3 × √4 – 3 ×√3 × √16 + 6√3 × √25 + 7√3 × √36
= √3 (5×2 – 3×4 + 6×5 + 7×6)
= 1.732 (10 – 12 + 30 + 42)
= 1.732 × 70
= 121.24
(5) State which of the following numbers are irrational:
(i) √4/9, – 3/70, √7/25, √16/5
(ii) – √2/49, 3/200, √25/3, – √49/16
Solution:
(i) √4/9 = √4/√9 = 2/3 is a rational number.
-3/70 is a rational number.
√7/25 = √7/√25 = √7/5 is an irrational number.
= 2√7/10 = 0.2√7
√16/5 = √16/√5 = 4/√5 = 4√5/5 = 8√5/10 = 0.8√5 is an irrational number,
(ii) – √2/49 = – √2/√49 -√2/7 is an irrational number.
3/200 is a rational number.
v25/3 = √25/√3 = 5/√3 is an irrational number.
= √49/16 = -√49/√16 = -7/4 is a rational number.
(6) State which of the following numbers will change into non-terminating non-recurring decimals.
(i) -3√2
(ii) √256/81
(iii) √27×16
(iii) √5/36
Solution:
(i) -3√2 is an irrational number because √2 is non-recurring and non-terminating decimal.
(ii) √256/81 = 16/9 is a rational number
(iii) √27×16 = √27 × √16
= 4√9 × √3
= 4 × 3 × √3
12√3 is an irrational number since √3 is irrational.
(iv) √5/36 = √5/6 is an irrational number since √5 is an irrational number.
(7) State which of the following numbers are irrational:
(i) 3 – √7/25
(ii) – 2/3 + 3√2
(iii) 3/√3
(iv) – 2/7 3√5
(v) (2 – √3) (2 + √3)
(vi) (3 + √5)2
(vii) (2/5 √7)2
(viii) (3 – √6)2
Solution:
(i) 3 – √7/25 = 3 – √7/5 = 15-√7/5 is an irrational number since √7 is an irrational number.
(ii) – 2/3 + 3√2 = 33√2 – 2/3 is an irrational numbers
Since 3√2 is an irrational number.
(iii) 3/√3 = 3√3/3 = √3 [Multiplying √3 an denominator and numerator]
∴ √3 is an irrational number
∴ 3/√3 is also irrational
(iv) – 2/7 3√5 is an irrational number since 5√5 is an irrational number.
(v) (2 – √3) (2 + √3)
= 22 – (√3)2 [∵ a2 – b2 = (a + b) (a – b)]
= 4 – 3
= 1 is a rational number.
(vi) (3 + √5)2 = 32 + 2 × 3 × √5 + (√5)2
= 9 + 6√5 + 5
= 14 + 6√5 is an irrational number
since √5 is an irrational number.
(vii) (2/5 √7)2 = 22/5 × (√7)2 = 4/25 × 7
= 42/25 is a rational number.
(viii) (3 – √6)2 = 32 – 2 × 3 × √6 + (√6)2
= 9 – 6√6 + 6
= 15 – 6√6 is an irrational number since √6 is an irrational number.
(8) Prove that the following numbers are irrational:
(i) 3√2
(ii) 3√5
(iii) 4√5
Solution:
(i) 3√2
Suppose 3√2 is a rational number, then, 3√2 = p/q where p, q are both integers, q ≠ 0 and p, q have no common factors.
∴ 2 = (p/q)3 => p3 = 2q3 —– (i)
As 2 divides 2q3, so 2 divides p3 but 2 is prime
∴ 2 divides P.
Let, P = 2k, where k is some integer.
Substituting the value of k in equation (i) we get,
(2k)3 = 2q3
=> 8k3 = 2q3
=> 4k3 = q3
As, 2 divides 4k3 so 2 divides q3 but 2 is prime.
∴ 2 divides q3
∴ 2 divides q
Thus, p & q has a common factor 2. This contradicts that p & q have no common factors.
Hence our initial presumption is wrong.
∴ 3√2 is on irrational number.
(ii) 3√3
Let, 3√3 be a rational number then,
3√3 = p/q where q≠0 and p, q have no common factor.
Or, (3√3)3 = p3/q3
Or, p3 = 3q3 —- (i)
As 3 divides 3q3, so 3 divides p3
∴ 3 divides p
Let, p = 3k where k is an integer.
Putting the value of p in equation (i) we get,
(3k)3 = 3q3
Or, 27k3 = 3q3
Or, q3 = qk3
As 3 divides qk3 so 3 divides q3
∴ 3 divides q
Thus, p & q has a women factor 3. The contradicts our initial assumption of that p & q has no common factors.
∴ 3√3 is an irrational number.
(iii) 4√5
Let 4√5 be a rational number, then
4√5 = p/q where q≠0, p & q has no common factors between them.
Or, (4√5)4 = p4/q4
Or, 5q4 = p4 —– (i)
As 5 divides 5q4 so 5 divides q.
∴ 5 divides p4
∴ 5 divides p
Let, p = 5k where k is an integer.
Putting the value of p in equation (i) we get,
5q4 = (5k)4
Or, 5q4 = 625 k4
Or, q4 = 125 k4
As 5 divides 125k4, so 5 divides k
∴ 5 divides q4
∴ 5 divides q
Thus, we have 5 as a common factor of p & q. Which contradicts p & q has no common factor.
∴ Our initial assumption is wrong.
4√5 is an irrational number.
(9) Find the greatest and the smallest real number among the following real number:
(i) 2√3, 3/√2, -√7, √15
(ii) -3√2, 9/√5, -4 4/3√5, 3/2 √3
Solution:
(i) 2√3, 3/√2, -√7, √15
2√3 = √4 × √3 = √12
3/√2 = √9/√2 = √9/2 = √4.5
∴ √15 > √12 > √4.5 > – √7
∴√5 is the greatest real number
– √7 is the smallest real number.
(ii) -3√2, 9/√5, -4, 4/3√5, 3/2√3
-3√2 = – √9 × √2 = – √18
9/√5 = √81/5 = √182/10 = √16.2
4/3 √5 = √16×√5/√9 = √80/9 = √8.88
3/2 √3 = √9×√3/√4 = √27/4 = √6.75
∴ √16.2 > √8.88 > √6.75 > – √18
∴ 9/√5 is the greatest real number
-3√2 is the smallest real number
(10) Write the following numbers in ascending order:
(i) 3√2, 2√3, √15, 4
(ii) 3√2, 2√8, 4, √50, 4√3
Solution:
(i) 3√2, 2√3, √15, 4
3√2 = √9 × √2 = √18
2√3 = √4 × √3 = √12
4 = √16
∴ √18 > √16 > √15 > √12
Or, 3√2 > 4> √15 > 2√3
In ascending order => 2√3, √15, 4, 3√2
(ii) 3√2, 2√8, 4, √50, 4√3
3√2 = √9 × √2 = √18
2√8 = √4 × √8 = √32
4 = √16
4√3 = √16 × √3 = √48
∴ √50 > √48 > √32 > √18 > √16
Or, √50 > 4√3 > 2√8 >3√2 > 4
In ascending order = 4, 3√2, 2√8, 4√3, √50
(11) Write the following real numbers in descending order:
(i) 9/√2, 3/2 √5, 4√3, 3 √6/5
(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7
Solution:
(i) 9/√2, 3/2 √5, 4√3, 3 √6/5
9/√2 = √81/2 = √40.5
3/2 √5 = √9×√5/√2 = √45/2 = √22.5
4√3 = √16×√3 = √48
3√6/5 = √9×√6/√5 = √54/5 = √108/10 = √10.5
∴ √48 > √40.5 > √22.5 > √10.5
Or, 4√3 > 9/√2 > 3/2 √5 > 3 √6/5
In descending order => 4√3, 9/√2, 3/2√5, 3√6/5
(ii) 5/√3, 7/3 √2, – √3, 3√5, 2√7
5/√3 = √25/3 = √8.33
7/3 √2 = √49×√2/√9 = √98/9 = √10.88
– √3 = – √3
3√5 = √9 × √5 = √45
2√7 = √4 × √7 = √28
∴ √45 > √28 > √10.88 > √8.33 > – √3
∴ In descending order.
= 3√5, 2√7, 7/3 √2, 5/√3, – √3
(12) Arrange the following numbers in ascending order: 3√2, √3, 6√5.
Solution:
3√2, √3, 6√5
LCM, of 3, 2 & 6 is 6.
∴ 3√2 = (2)1/3 = (22)1/6 = 41/6
√3 = 31/2 = (33)1/6 = 271/6
6√5 = 51/6
As, 27 > 5 > 4 so, 271/4> 51/6> 41/6
∴ √3 > 2√5 > 3√2
∴ In ascending order, 3√2, 6√5, √3