ML Aggarwal Solutions Class 9 Math 1st Chapter Rational and Irrational Numbers Exercise 1.3
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions First Chapter Rational and Irrational Numbers Exercise 1.3. APC Solution Class 9 Exercise 1.3.
(1) Locate √10 and √17 on the number line.
Solution:
We write 10 as sum of two square
10 = 9+1 = 32 + 12
Let, h be the number line. If point 0 represents.
0 and point A represents 3, then length of segment OA = But at A draw AC⊥OA. From AC, Cut off AB = 1 unit. We observe that OAB is a right-angled triangle at A. By Pythagoras theorem we get,
OB2 = OA2 + AB2
=> 32 + 22 = 13
=> OB = √10
With 0 at centre we draw an arc are of a circle to which neat the number line at P. Hence point P represents √10 on number line.
2nd Part:
We write 17 as sum of two Square
17 = 16+1 = 42 + 12
Here point P represent √10 on number line.
(2) Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:
(i) 36/100
(ii) 4 1/8
(iii) 2/9
(iv) 2/11
(v) 3/13
(vi) 329/400
Solution:
(3) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) 13/3125
(ii) 17/8
(iii) 23/75
(iv) 6/15
(v) 1258/625
(vi) 77/210
Solution:
(i) 13/3125 = 13/5×625 = 13/5×54 = 13/55
Therefore 13/3125 is non-terminating decimal.
(ii) 17/8 = 17/23 = 17×53/23×53 = 2125/10000 = 2.125
Therefore 17/8 is terminating decimal.
(iii) 7/80 = 7/8 x 2 x 5 = 7/24 x 5
= 7 x 53 / 24 x 54
= 875 / 10000
= 0.0875
(iv) 6/15 = 2×3/5×3
= 2/5
= 2×2/5×2
= 0.4
Therefore 6/15 is terminating decimal.
(v) 1258/625 = 2×17×37/54 = 2×17×37×24/54×24
= 20128/10000 = 2.0128
∴ It is a terminating decimal.
(vi) 77/210 = 11/30 = 11/3×10 is in the form of p/q and p and q = 11/3×2×5 are co prime and q has 3 prime factors therefore it is a near terminating decimal.
(4) Without actually performing the long division, find if 987/10500 will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.
Solution:
987/10500 = 3×329/5×7×3×100 = 3×7×47/5×7×3×2×5×2×5 = 47/22×53
= 47×2/23×53
= 94/1000
= 0.094
∴ It is a terminating decimal. Since 987 is divisible 10500 in decimal form.
(5) Write the decimal expansions of the following numbers. Which have terminating decimal expansions.
(i) 17/8
(ii) 13/3125
(iii) 7/80
(iv) 6/13
(v) 22×7/54
(vi) 237/1500
Solution:
(7) Write the decimal expansion of 1/7. Hence, write the decimal expansion of 2/7, 3/7, 4/7, 5/7 and 6/7
Solution:
(8) Express the following numbers in the form P/q, where p and q are both integers and q ≠ 0:
(i) 0.3
(ii) 5.2
(iii) 0.404040
(iv) 0-47
(v) 0.134
(vi) 0.001
(8) Express the following numbers in the form P/q, where p and q are both integers and q ≠ 0:
(i) 0.3
(ii) 5.2
(iii) 0.404040
(iv) 0-47
(v) 0.134
(vi) 0.001
Solution:
(i) 0.3 = 0.3333 …
Let, x = 0.3333 —– (i)
Multiplying both sides with 10 we get.
10x = 3.333 —– (ii)
Substracting equation (i) from equation (ii) we get.
9x = 3
Or, x = 3/9 = 1/3 within in the p/q form of 0.
(ii) 5.2 = 5.2222 …..
Let, x = 5.2222 —- (i)
Multiplying equation both sides with 10 we get
10x = 52.222 —– (ii)
Substracting equation (i) from equation (ii) we get.
9x = 47
Or, x = 47/4 which is the p/q from of 5.2
(iii) 0.404040 ……
Let, x = 0.404040 —– (i)
Multiplying both sides with 100 we get.
100x = 40.4040 —- (ii)
Substracting equation (i) from equation (ii) we get,
99x = 40
Or, x = 40/99 is the p/q from of 0.404040
(iv) 0.47 = 0.474747 …..
Let, x = 0.474747 —- (i)
Multiplying both sides with 100 we get,
100x = 47.4447 —– (ii)
Substracting equation (i) from equation (ii) we get,
99x = 47
Or, x = 47/99 which is the p/q from of 0.47
(v) 0.134 = 001343434 …..
Let, x = 0.1343434 ….. (i)
Multiplying equation (i) by 10 we get,
10x = 1.343434 ….. (ii)
Multiplying equation (ii) by 100 we get.
1000x = 134.3434 …… (iii)
Substracting equation (ii) from equation (iii) we get,
990x = 133
Or, x = 133/990 is the p/q form of 0.134
(vi) 0.001 = 0.001001001 …..
Let, x = 0.001001001 …. (i)
Multiplying equation (i) by 1000 we get,
1000x = 0001.001001 ….. (ii)
Substracting equation (i) from equation (ii) we get,
999x = 1
Or, x = 1/999 is the p/q from of 0.001
(9) Classify the following numbers as rational or irrational:
(i) √23
(ii) √225
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001….
(vi) 345.0456
Solution:
(i) √23
Let, √23 = p/q be a rational number.
∴ Squaring both sides.
23 = p2/q2
Or, p2 = 23 q2 where p & q ≠ 0 —- (i)
As 23 divides q2 so 23 divides p2 but 23 is prime
∴ 23 divides p.
Now, Let, p = 23m
∴ Putting the value of p in equation (i) we get.
(23m)2 = 23q2
Or, 529m2 = 23q2
Or, q2 = 23m2
As 23 divides m2 so 23 divides q2 but 23 is prime
∴ 23 divides q.
Hence, p & q has one common prime factor 23 where contradicts that p & q has no common factor.
Hence our initial assumption is wrong.
∴ √23 is an irrational number.
(ii) √225 = √15×√15 = √152 = 15
∵ √225 amounts to a positive integer.
∴ √225 is a rational number.
(iii) 0.3796 = 3796/10000 is a p/q from of 0.3796.
Hence, it is a rational number.
(iv) 7.478478 …..
Let, x = 7.478478 —— (i)
Multiplying both sides with 1000 we get.
1000x = 7478.478 —– (ii)
Substracting equation (i) from equation (ii) we get,
999x = 7471
Or, x = 7471/999 is the p/q from of 7.478478 …
Hence 7.478478 is a rational number.
(v) 1.101001000100001….
The above decimal is non-terminating and non-recurring decimal number. Therefore we can conclude that 1.101001000100001…. is an irrational number.
(vi) 345.0456 = 345.0456456456 ….
Let, x = 345.0456456456 —– (i)
Multiplying equation (i) with 10 we get,
10x = 3450.456456 —- (ii)
Multiplying equation (ii) with 1000 we get.
10000x = 3450.456456456 —– (iii)
Subtracting equation (ii) from (iii) we get,
9990x = 3447006
Or, x = 3447006/9990 is the p/q from of 3450456
∴ 345.0456 is a rational number.
(10) The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form p/q, where p, q are integers, q ≠ 0 and p, q are co-prime, then what can you say about the prime factors of q?
(i) 37.09158
(ii) 423.04567
(iii) 8.9010010001…
(iv) 2.3476817681….
Solution:
(i) 37.09158 = 3709158/100000 = 3709158/25×56 is the p/q form of 37.09158. Therefore it is a rational number. q has two prime factors 2 & 5 where prime factorization.
(ii) 423.04567 = 423.0456704567 ….
Let, x = 423.0456704567 —– (i)
Multiplying 100000 on both sides.
100000x = 42304567.04567 —– (ii)
Subtracting equation (i) from equation (ii) we get,
99999x = 42304144
Or, x = 42304144/99999 is the p/q from of 423.04567
Hence, it is a rational number.
Now, The denominator 9 can be further simplified,
99999 = 9×11111 = 32 × 41 × 271
(iii) 8.9010010001….
The above number is a non-terminating & non-recurring decimal number hence it is an irrational number.
(iv) 2.3476817681 ……
Let, x = 2.3476817681 —– (i)
Multiplying equation (i) with 1000 we get,
1000x = 234.76817681 —– (ii)
Multiplying equation (ii) with 100000 we get.
10000000x = 2347681.7681 —- (iii)
Substracting equation (ii) from equation (iii) we get,
999900x = 2347447
Or, x = 2347447/999900 is the p/q form of 2.3476817681 ….
∴ 2.3496817681… is a rational number.
Here the denominator 9 can be further simplified into prime factors.
999900 = 9×100×1111 = 32 × 22 × 52 × 11 × 101
Hence, the prime factors of 9 are 3, 2, 5, 11, 101
One rational number between 2 & 3
∴ 24 = 4 & 32 = 9
Any number between 4 & 9 without having perfect surface can be used as irrational number.
∵ 4 < 5 < 9
∴ √4 < √5 < √9
∴ 2 <√5 <3
∴ √5 is the desired irrational number.
(11) Insert an irrational number between the following:
(i) 1/3 and 1/2
(ii) – 2/5 and 1/2
(iii) 0 and 0.1
Solution:
(13) Write two irrational numbers between 4/9 and 7/11.
Solution:
Any non-recurring and non-terminating decimal number between 0.44 & 0.63 can be our irrational number.
One such number is 0.54321
Another number is 0.58124
(14) Find a rational number between √2 and √3
Solution:
√2 & √3
∴ Now, (√2)2 = 2 & (√3)2 = 3
Now, Find any rational number between 2 & 3 which, is a perfect square of a rational number.
One such number is 2.25 also, √2.25 = 1.5
1.5 is our rational number between √2 & √3
(15) Find two rational numbers between 2√3 and √15.
Solution:
2√3 & √15
∴ (2√3)2 = 4×3 = 12 & (√15)2 = 15
Now, find any rational number between 12 & 15 that is a perfect square of a rational number.
One such number is 12.25 also √12.25 = 3.5
∴ 3.5 is our desired rational number.
Another number is 12.96 also √12.96 = 3.6
(16) Insert an irrational number between √5 and √7.
Solution:
√5 & √7
∴ (√5)2 = 5 & (√7)2 = 7
∵ 5 < 6 < 7
∴ √5 < √16 < √7
∴ √6 is our desired irrational number.
(17) Insert two irrational numbers between √3 and √7.
Solution:
√3 & √7
∴ (√3)2 = 3 & (√7)2 = 7
∵ Insert any no. that is not a perfect square.
∴ 3 < 5 < 6 < 7
∴ √3 <√5 < √6 < √7
∴ √5 & √6 are our desired irrational number.
