**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 = 3^{2} + 1^{2}

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,

OB^{2} = OA^{2} + AB^{2}

=> 3^{2} + 2^{2} = 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 = 4^{2} + 1^{2}

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 = 13/5×5^{4} = 13/5^{5}

Therefore 13/3125 is non-terminating decimal.

(ii) 17/8 = 17/2^{3} = 17×5^{3}/2^{3}×5^{3} = 2125/10000 = 2.125

Therefore 17/8 is terminating decimal.

(iii) 7/80 = 7/8 x 2 x 5 = 7/2^{4 }x 5

= 7 x 5^{3} / 2^{4} x 5^{4}

= 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 = 2×17×37/5^{4} = 2×17×37×2^{4}/5^{4}×2^{4}

= 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 = 47/2^{2}×5^{3}

= 47×2/2^{3}×5^{3}

= 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) 2 ^{2}×7/5^{4}**

**(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 = p^{2}/q^{2}

Or, p^{2} = 23 q^{2} where p & q ≠ 0 —- (i)

As 23 divides q^{2} so 23 divides p^{2} but 23 is prime

∴ 23 divides p.

Now, Let, p = 23m

∴ Putting the value of p in equation (i) we get.

(23m)^{2} = 23q^{2}

Or, 529m^{2} = 23q^{2}

Or, q^{2} = 23m^{2}

As 23 divides m^{2} so 23 divides q^{2} 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 = √15^{2} = 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 = 3709158/2^{5}×5^{6} 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 = 3^{2} × 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 = 3^{2} × 2^{2} × 5^{2} × 11 × 101

Hence, the prime factors of 9 are 3, 2, 5, 11, 101

One rational number between 2 & 3

∴ 2^{4} = 4 & 3^{2} = 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.