Rational & Irrational Numbers Notes for RRB ALP Maths Trade Exam. These Notes will help you score good marks in the upcoming RRB ALP Maths & Physics Trade Exam.

## Introduction: Rational & Irrational Numbers

1. Euclid’s Division lemma:- Given Positive integers a and b there exist unique integers q and r satisfying a=bq +r, where 0≤r<b, where a, b, q and r are respectively called as dividend, divisor, quotient and remainder.

2. Euclid’s division Algorithm:- To obtain the HCF of two positive integers say c and d, with c>0, follow the steps below:

- Step I: Apply Euclid’s division lemma, to c and d, so we find whole numbers, q and r such that c =dq +r, 0≤r<d
- Step II: If r=0, d is the HCF of c and d. If r is not equal to 0, apply the division lemma to d and r.
- Step III: Continue the process until the remainder is zero. The divisor at this stage will be the required HCF

3. The Fundamental theorem of Arithmetic:- Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. Ex.: 24= 2x2x2x3 = 3x2x2x2

**Theorem:** Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form of p/q where p & q are co-prime and the prime factorization of q in the form of 2^{n}.5^{m} , where n, m are non-negative integers. Ex. 7/10 = 7/2x5 = 0.7

**Theorem:** Let x = p/q be a rational number such that the prime factorization of q is not of the form of 2^{n}.5^{m}, where n, m are non-negative integers. Then x has a decimal expansion which is non terminating repeating (recurring). Ex. 7/6 = 7/2x3 = 1.1666...

**Theorem:** For any two positive integers a and b, HCF (a,b) X LCM (a,b)=a X b

Ex.: 4 & 6; HCF (4,6) = 2, LCM (4,6) = 12; HCF X LCM = 2 X 12 =24 Ans. : a X b = 24

## Questions for Practice

**Q1. Use Euclid’s division algorithm to find the HCF of :**

(i) 135 and 225

**Solution:**

We know that,

= 225 > 135

Applying Euclid’s division algorithm

225 = 135 × 1+90

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90 × 1+45

Here remainder = 45,

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence ,

HCF of (135, 225) = 45

**(ii) 196 and 38220 **

**Solution:**

We know that,

38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0

Remainder = 0

Hence, HCF of (196, 38220) = 196

**(iii) 867 and 255**

**Solution:**

We know that,

867>255

So, Applying Euclid’s division algorithm

867 = 255×3+102

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

**Hence,**** (HCF 0f 867 and 255) = 51**

**Q2. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?**

**Solution:**

By using, Euclid’s division algorithm

616 = 32×19+8

Remainder ≠ 0

So, again Applying Euclid’s division algorithm

32 = 8×4+0

HCF of (616, 32) is 8.

So ,

They can march in 8 columns each.

**Q3. ****Express each number as a product of its prime factors:**

**(i) 140**

**Solution:**

140 = 2×2×5×7 = 2^{2}×5×7

**(ii) 156**

**Solution:**

156 = 2×2×3×13 = 2^{2}×3×13

**(iii) 3825**

**Solution:**

3825 = 3×3×5×5×17 = 3^{2}×5^{2}×17

**(iv) 5005**

**Solution:**

5005 = 5×7×11×13

**(v) 7429**

**Solution:**

7429 = 17×19×23

**Q4. ****Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = ****product of the two numbers.**

**(i) ****26 and 91**

**Solution:**

26 = 2×13

91 = 7×13

HCF =13

LCM = 2×17×13 =182

Product of the two numbers = 26×91 = 2366

HCF×LCM = 13×182 = 2366

Hence, product of two numbers = HCF×LCM

**(ii) ****510 and 92**

**Solution:**

510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

Product of the two numbers = 510×92 = 46920

HCF×LCM = 2×23460

HCF×LCM = 46920

Hence, product of two numbers = HCF×LCM

**(iii) ****336 and 54**

**Solution:**

**336 = 2**×2×2×2×3×7

336 = 2^{4}×3×7

54 = 2×3×3×3

54 =2×3^{3}

HCF = 2×3 = 6

LCM = 2^{4}×3^{3}×7 = 3024

Product of the two numbers = 336×54 = 18144

HCF×LCM = 6×3024 = 18144

Hence, product of two numbers = HCF×LCM

**Q5. ****Find the LCM and HCF of the following integers by applying the prime factorisation ****method.**

**(i) ****12, 15 and 21**

**Solution:**

12 = 2×2×3 = 2^{2}×3

15 = 3×5

21 = 3×7

HCF = 3

LCM = 2^{2}×3×5×7 = 420

**(ii) ****17, 23 and 29**

**Solution:**

17 = 1×17

23 = 1×23

29 = 1×29

HCF = 1

LCM = 17×23×29 = 11339

**(iii) ****8, 9 and 25**

**Solution:**

8=2×2×2

9 = 3×3

25 = 5×5

HCF = 1

LCM = 2×2×2×3×3×5×5 = 1800

**Q6. ****Given that HCF (306, 657) = 9, find LCM (306, 657).**

**Solution:**

HCF of (306, 657) = 9

We know that,

LCM×HCF = product of two numbers

LCM×HCF = 306×657

LCM =

LCM = 22338

**Q7. ****Check whether 6 n can end with the digit 0 for any natural number n.**

**Solution: **

If any number has last digit 0,

Then, it should be divisible by 10

Factors of 10 = 2×5

So, Value 6n should be divisible by 2 or 5

Prime factorisation of 6^{n} = (2×3)^{n}

Hence, 6n is divisible by 2 but not by 5.

It can not end with 0.

**Q8. ****There is a circular path around a sports field. Sonia takes 18 minutes to drive one round ****of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the ****same point and at the same time, and go in the same direction. After how many minutes ****will they meet again at the starting point?**

**Solution: **

They will meet again after LCM of both values at starting point.

18 = 2×3×3

And

12 = 2×2×3

LCM of 12 and 18 = 2×2×3×3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

**Q9. 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)

**Solution:**

Factorize the denominator we get,

3125 = 5×5×5×5×5 = 5^{5}

The denominator is of the form 5^{m}

Hence, the decimal expansion of is terminating.

(ii)

**Solution:**

Factorize the denominator we get,

8 = 2 × 2 ×2 = 2^{3}

The denominator is of the form 2^{m}

Hence, the decimal expansion of is terminating.

(iii)

**Solution:**

Factorize the denominator we get,

455 = 5×7×13

Since, the denominator is not in the form of 2^{m }× 5^{n}, and it also contains 7 and 13 as its factors,

Its decimal expansion will be non-terminating repeating.

(iv)

**Solution:**

Factorize the denominator we get,

1600 = 2^{6}×5^{2}

The denominator is in the form 2^{m }× 5^{n}

Hence, the decimal expansion of is terminating.

(v)

**Solution:**

Factorize the denominator we get,

343 = 7^{3}

Since the denominator is not in the form of 2^{m }× 5^{n}, it has 7 as its factors.

So, the decimal expansion of non-terminating repeating.

(vi)

Solution:

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(vii)

**Solution:**

Since, the denominator is not in the form of 2^{m }× 5^{n}, as it has 7 in denominator.

So, the decimal expansion of is non-terminating repeating.

(viii)

**Solution:**

The denominator is in the form 5^{n}

Hence, the decimal expansion of is terminating.

(ix)

**Solution:**

Factorize the denominator we get,

10 = 2×5

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(x)

**Solution:**

Factorize the denominator we get,

30 = 2×3×5

Since the denominator is not in the form of 2^{m }× 5^{n}, as it has 3 in denominator.

So, the decimal expansion of is a non-terminating repeating.

**Q 10.** **Write down the decimal expansions of those rational numbers in which have terminating decimal expansions.**

**Solution:**

**(i)**

** **

** **

** **

**(ii) **

** **

** **

**(i) **

** **

** **

** **

**(viii) **

** **

** **

** **

** **

**(ix) **

** **

**Q 11. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form what can you say about the prime factors of q?**

**Solution: **

(i) **43.123456789**

Since this number has a terminating decimal expansion, it is a rational number of the form and q is of the form 2^{m}×5^{n} . That is, the prime factor of q will be 2 or 5 or both.

(ii) **0.120120012000120000...**

The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) **43.123456789**

^{m}×5

^{n}that is, the prime factors of q will also have a factor other than 2 or 5.

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