CBSE Class 10 Maths HOTs Real Numbers Set D

Very Short Answer Type Questions

Question. The LCM of two numbers is 182 and their HCF is 13. If one of the number is 26, find the other.
Answer: We know that \( \text{HCF} (a, b) \times \text{LCM} (a, b) = a \times b \)
So, \( 13 \times 182 = 26 \times b \)
\( \Rightarrow b = \frac{13 \times 182}{26} = 91 \)
Thus, the other number is 91.

Question. Given that HCF (135, 225) = 45, find the LCM (135, 225).
Answer: We know that
\( \text{LCM} \times \text{HCF} = \text{Product of two numbers} \)
\( \therefore \text{LCM} (135, 225) = \frac{\text{Product of 135 and 225}}{\text{HCF}(135, 225)} \)
\( = \frac{135 \times 225}{45} \)
\( = 675 \)

Question. After how many decimal places will the decimal representation of the rational number \( \frac{229}{2^2 \times 5^7} \) terminate ?
Answer: Here,
\( \frac{229}{2^2 \times 5^7} = \frac{229 \times 2^5}{2^2 \times 5^7 \times 2^5} = \frac{229 \times 2^5}{(10)^7} \)
Hence, the given rational number will terminate after 7 decimal places.

Question. Are the smallest prime and the smallest composite numbers co-prime? Justify.
Answer: No.
We know that,
Smallest prime number is 2 and smallest composite number is 4.
HCF of (2, 4) = 2
Since, there is a common factor 2.
So, they are not co-prime.

Question. The HCF of two numbers \( a \) and \( b \) is 5 and their LCM is 200. Find the product \( ab \).
Answer: Given, HCF \( (a, b) = 5 \)
LCM \( (a, b) = 200 \)
\( \text{HCF} \times \text{LCM} = \text{Product of the numbers} \)
\( \Rightarrow a \times b = 5 \times 200 \)
\( \Rightarrow ab = 1000 \)
Hence, the product of \( ab \) is 100.

Question. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
Answer: No.
We know that:
“The HCF of any two numbers must be a factor of the LCM of those numbers.”
So, two numbers cannot have their HCF 18 and LCM 380, as 18 does not divide 380.

Question. Find a rational number between \( \sqrt{2} \) and \( \sqrt{7} \).
Answer: \( \sqrt{2} = 1.414 \)
and \( \sqrt{7} = 2.6 \)
Let the rational number be x.
\( \therefore \sqrt{2} < x < \sqrt{7} \)
or \( 1.4 < x < 2.6 \)
Hence, any rational number like 1.5, 2.0, 2.5, can be the answer.

Question. Write the number of zeroes in the end of a number whose prime factorization is \( 2^2 \times 5^3 \times 3^2 \times 17 \).
Answer: Given, \( 2^2 \times 5^2 \times 5 \times 3^2 \times 17 \)
\( = (2 \times 5)^2 \times 5 \times 3^2 \times 17 \)
[\( \because \) on multiplying \( 2 \times 5 \) we get 10]
\( = (10)^2 \times 5 \times 3^2 \times 17 \)
The power of 10 in the given expression is 2.
Hence, the number of zeroes in the end will be = 2.

Question. If the HCF of (336, 54) = 6, find the LCM (336, 54).
Answer: The HCF of (336, 54) = 6.
We know that:
\( \text{LCM} \times \text{HCF} = \text{Product of two numbers} \)
\( \Rightarrow \text{LCM} = \frac{336 \times 54}{6} \)
\( = 336 \times 9 = 3024 \)
Hence, the LCM of the two numbers is 3024.

Question. Prove that the number \( 4^n \), \( n \) being a natural number, can never end with the digit 0.
Answer: If \( 4^n \) ends with 0, then it must have 5 as a factor
But, \( (4)^n = (2^2)^n = 2^{2n} \), i.e., the only prime factor of \( 4^n \) is 2.
Also, we know from the Fundamental Theorem of Arithmetic that the prime factorisation of each number is unique.
\( \therefore 4^n \) can never end with 0.

Question. Find a rational number betwen \( \sqrt{2} \) and \( \sqrt{3} \).
Answer: Rational number between \( \sqrt{2} \) (1.41 approx) and \( \sqrt{3} \) (1.73 approx) can be 1.5, 1.6, 1.63 etc.
So, a required rational number may be 1.5.

Question. Write one rational and one irrational number lying between 0.25 and 0.32.
Answer: Rational number = 0.30
Irrational number = 0.3010203040...
Or any other correct rational and irrational number.

Question. Write the exponent of 3 in the prime factorization of 144.
Answer: Prime factorization of \( 144 = 2^4 \times 3^2 \)
So, exponent of 3 = 2.

Question. A merchant has 120 litres and 180 litres of two kinds of oil. He wants to sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest volume of such a tin?
Answer: In order to find volume of such a tin, we need to find the largest number which exactly divides 120 and 180 which is nothing but the HCF (120, 180).
\( 120 = 2 \times 2 \times 2 \times 3 \times 5 \)
\( 180 = 2 \times 2 \times 3 \times 3 \times 5 \)
H.C.F. (120, 180) \( = 2 \times 2 \times 3 \times 5 = 60 \)
Hence, the greatest volume of each tin is 60 litres.

Question. Three bells toll at intervals of 12 minutes, 15 minutes and 18 minutes respectively, if they start tolling together, after what time will they next toll together?
Answer: The required time is the LCM of 12, 15 and 18.
\( 12 = 2 \times 2 \times 3 \)
\( 15 = 3 \times 5 \)
\( 18 = 2 \times 3 \times 3 \)
LCM \( = 2^2 \times 3^2 \times 5 = 180 \)
So, next time the bells will ring together after 180 minutes or 3 hours.

Question. Without actually performing the long division, write the decimal expansion of \( \frac{11725}{2^3 \times 5^4} \).
Answer: \( \frac{11725}{2^3 \times 5^4} = \frac{11725 \div 5}{2^3 \times 5^3} = \frac{2345}{(10)^3} = 2.345 \)

Question. Write any two irrational numbers whose product is a rational number.
Answer: Consider two irrationals as, \( 5 - 2\sqrt{2} \) and \( 5 + 2\sqrt{2} \)
Here,
\( (5 - 2\sqrt{2})(5 + 2\sqrt{2}) = 5^2 - (2\sqrt{2})^2 \)
\( = 25 - 8 = 17 \) (a rational number)

Short Answer Type Questions

Question. Check whether \( 12^n \) can end with the digit 0 for any natural number \( n \).
Answer: Let, if possible, \( 12^n \) have a value which ends with the digit 0.
\( \Rightarrow 10 \) is a factor of \( 12^n \)
\( \Rightarrow 5 \) is a prime factor of \( 12^n \)
i.e., \( 12^n = 5 \times q \), where \( q \) is some natural number
\( \Rightarrow (2^2 \times 3)^n = 5 \times q \)
or \( 2^{2n} \times 3^n = 5 \times q \)
The assumption, 5 is a prime factor of \( 2^{2n} \times 3^n \), is not possible because \( 2^{2n} \times 3^n \) can have only 2 and 3 as prime factors.
Hence, our assumption is wrong and \( 12^n \) cannot end with the digit 0.

Question. The product of the LCM and HCF of two natural numbers is 24. The difference of two numbers is 2. Find the numbers.
Answer: Let the natural numbers be \( p \) and \( q \).
According to question,
\( p \times q = 24 \) ...(i)
and \( p - q = 2 \)
\( p = 2 + q \) ...(ii)
From (i) and (ii)
\( (q + 2) \times q = 24 \)
\( q^2 + 2q - 24 = 0 \)
\( q^2 + 6q - 4q - 24 = 0 \)
\( (q + 6) (q - 4) = 0 \)
\( q = - 6, 4 \)
\( q = 4 \)
[Since –6 is not a natural number]
So, the numbers are 4, 6.

Question. Two alarm clocks ring their alarms at regular intervals of 72 seconds and 50 seconds if they first beep together at 12 noon, at what time will they beep again for the first time?
Answer: Here, we need to find the LCM of 72 and 50.
\( 72 = 2 \times 2 \times 2 \times 3 \times 3 \)
\( 50 = 2 \times 5 \times 5 \)
LCM of 72 and 50 \( = 2^3 \times 3^2 \times 5^2 = 1800 \)
So, 1800 sec = 30 min
Hence, alarm clocks will beep again for the first time at 12.30 pm.

Question. Find the HCF of 612 and 1314 using prime factorisation.
Answer: Prime factors of 612 and 1314.
\( 612 = 2 \times 2 \times 3 \times 3 \times 17 \)
\( 1314 = 2 \times 3 \times 3 \times 73 \)
HCF (612, 1314) \( = 2 \times 3 \times 3 \)
\( = 18 \)
Hence, the HCF of 612 and 1314 is 18.

Question. Write the smallest number which is divisible by both 306 and 657.
Answer: Given numbers are 306 and 657.
The smallest number divisible by 306 and 657
\( = \text{LCM}(306, 657) \)
prime factors of 306 \( = 2 \times 3 \times 3 \times 17 \)
prime factors of 657 \( = 3 \times 3 \times 73 \)
The LCM of (306, 657) \( = 2 \times 3 \times 3 \times 17 \times 73 \)
\( = 22338 \)
Hence, the smallest number divisible by 306 and 657 is 22,338.

Question. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of \( q \), when this number is expressed in the form \( \frac{p}{q} \)? Give reasons.
Answer: As 327.7081 is a terminating decimal number, the denominator of the rational number must be of the form \( 2^m \times 5^n \).
Thus, \( 327.7081 = \frac{3277081}{10000} \)
\( = \frac{3277081}{10^4} \)
\( = \frac{3277081}{2^4 \times 5^4} = \frac{p}{q} \)
So, the prime factors of \( q \) are 2 and 5.
Here, \( q \) is of the form \( 2^m \times 5^n \), where \( m \) and \( n \) are natural numbers. The prime factors of \( p \) and \( q \) will be either 2 or 5 or both.

Question. Using prime factorisation method, find the HCF and LCM of 210 and 175.
Answer: The prime factorisations of 210 and 175 are:
\( 210 = 2 \times 3 \times 5 \times 7 \)
\( 175 = 5 \times 5 \times 7 \)
So, HCF (210, 175) \( = 5 \times 7 = 35 \); and
LCM (210, 175) \( = 2 \times 3 \times 5 \times 7 \times 5 = 1050 \)

Question. Find the two numbers which on multiplication with \( \sqrt{360} \) gives a rational number. Are these numbers rational or irrational?
Answer: \( \sqrt{360} = \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 5} \)
\( = 6\sqrt{10} \)
If we multiply \( 6\sqrt{10} \) with \( \sqrt{10} \) and 1.
We get,
\( 6\sqrt{10} \times \sqrt{10} \times 1 = 60 \)
Hence, numbers are \( \sqrt{10} \) and 1.
Where, 1 is a rational number and \( \sqrt{10} \) is an irrational number.

Question. Prove that \( \sqrt{5} \) is an irrational number.
Answer: Let us assume, to the contrary, that \( \sqrt{5} \) is a rational number and its simplest form is \( \frac{a}{b} \), where a and b are integers having no common factor other than 1 and \( b \neq 0 \).
Now, \( \sqrt{5} = \frac{a}{b} \)
\( \Rightarrow 5 = \frac{a^2}{b^2} \)
\( \Rightarrow 5b^2 = a^2 \) ...(i)
\( \Rightarrow a^2 \) is divisible by 5 [\(\because 5b^2\) is divisible by 5]
\( \Rightarrow a \) is divisible by 5 [\(\because\) 5 is a prime number and divides \( a^2 \Rightarrow 5 \) divides \( a \)]
Let \( a = 5c \), for some integer ‘c’
On substituting \( a = 5c \) in (i), we get
\( 5b^2 = (5c)^2 \)
\( \Rightarrow 5b^2 = 25c^2 \)
\( \Rightarrow b^2 = 5c^2 \)
\( \Rightarrow b^2 \) is divisible by 5 [\(\because 5c^2\) is divisible by 5]
\( \Rightarrow b \) is divisible by 5
Since \( a \) and \( b \) are both divisible by 5, 5 is common factor of \( a \) and \( b \).
But this contradicts the fact that \( a \) and \( b \) have no common factor other than 1.
This contradiction has arisen because of our incorrect assumption that \( \sqrt{5} \) is a rational number.
Hence, \( \sqrt{5} \) is irrational.

Question. Prove that \( 2 + 5\sqrt{3} \) is an irrational number, given that \( \sqrt{3} \) is an irrational number.
Answer: Let us assume that \( 2 + 5\sqrt{3} \) is a rational number.
Therefore: \( 2 + 5\sqrt{3} = \frac{a}{b} \)
(where, ‘\( a \)’ and ‘\( b \)’ are co-primes)
\( \Rightarrow 5\sqrt{3} = \frac{a}{b} - 2 \)
\( \Rightarrow 5\sqrt{3} = \frac{a - 2b}{b} \)
\( \Rightarrow \sqrt{3} = \frac{a - 2b}{5b} \)
Therefore, \( \frac{a - 2b}{5b} \) is in the form of \( \frac{a}{b} \) which is a rational number.
But, this contradicts the fact that \( \sqrt{3} \) is an irrational number.
Therefore, our assumption is wrong and \( 2 + 5\sqrt{3} \) is an irrational number.

Question. Prove that \( \sqrt{2} \) is an irrational number.
Answer: Let us assume \( \sqrt{2} \) be a rational number and its simplest form be \( \frac{a}{b} \), \( a \) and \( b \) as coprimes.
So, \( \sqrt{2} = \frac{a}{b} \)
\( \Rightarrow a^2 = 2b^2 \)
Thus, \( a^2 \) is a multiple of 2.
\( \Rightarrow a \) is a multiple of 2. ...(i)
Let \( a = 2m \) for some integer m.
Then, \( b^2 = 2m^2 \)
Thus, \( b^2 \) is a multiple of 2.
\( \Rightarrow b \) is a multiple of 2. ...(ii)
From (i) and (ii,) 2 is a common factor of \( a \) and \( b \).
This contradicts the fact that \( a \) and \( b \) are co-primes.
Hence, \( \sqrt{2} \) is an irrational number.

Question. Find HCF and LCM of 404 and 96 and verify that HCF \( \times \) LCM = Product of the two given numbers.
Answer: Given, numbers are 404 and 96.
Prime factorisation of both the number.
\( 404 = 2 \times 2 \times 101 \)
\( 96 = 2^5 \times 3 \)
HCF of 404 and 96 \( = 2^2 = 4 \)
LCM of 404 and 96 \( = 2^5 \times 3 \times 101 \)
\( = 9696 \)
Now, HCF \( \times \) LCM \( = 4 \times 9696 \)
\( = 38784 \) ...(i)
Product of two numbers \( = 404 \times 96 \)
\( = 38784 \) ...(ii)
From (i) and (ii) we get
HCF \( \times \) LCM = Product of two numbers.
Hence, verified.

Question. Write the denominator of rational number \( \frac{257}{5000} \) in the form \( 2^m \times 5^n \), where \( m, n \) are non-negative integers. Then, write its decimal expansion without actual division.
Answer: Denominator of the rational number \( \frac{257}{5000} \) is 5000.
Now, \( 5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 \)
\( = (2)^3 \times (5)^4 \)
which is of the type \( 2^m \times 5^n \), where \( m = 3 \) and \( n = 4 \) are non–negative integers.
Simplifying the given fraction:
\( \Rightarrow \frac{257}{5000} = \frac{257}{5^4 \times 2^3} \)
\( = \frac{257}{5^4 \times 2^3} \times \frac{2}{2} \)
\( = \frac{514}{5^4 \times 2^4} = \frac{514}{10^4} \)
\( = 0.0514 \)
So, 0.0514 is the required decimal expansion of the rational number \( \frac{257}{5000} \).

Question. Find if \( \frac{987}{10500} \) will have terminating or non–terminating (repeating) decimal expansion. Give reasons for your answer.
Answer: Yes, it will have a terminating decimal expansion.
Simplified denominator has factor in the form of \( 2^m \times 5^n \).
So, this is a terminating decimal.
Now, \( \frac{987}{10500} = \frac{3 \times 7 \times 47}{2 \times 2 \times 3 \times 5 \times 5 \times 5 \times 7} \)
\( = \frac{47}{5^3 \times 2^2} \times \frac{2}{2} = \frac{94}{5^3 \times 2^3} \)
\( = \frac{94}{1000} \)
\( = 0.094 \)
And we know, if p/q is a rational number, such that the prime factorization of \( q \) is of the form \( 2^m \times 5^n \) where \( n \) and \( m \) are non–negative integers, than x has a decimal expansion which terminates.
Hence, it terminates.

Question. On a morning walk, three people step off together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk, so that each can covers the same distance in complete steps?
Answer: We know that the LCM is the product of the greatest power of each prime factor of the numbers.
We have to find the LCM of 40, 42 and 45 to get the required minimum distance.
For this, we find prime factorisation,
\( 40 = 2 \times 2 \times 2 \times 5 \)
\( 42 = 2 \times 3 \times 7 \)
\( 45 = 3 \times 3 \times 5 \)
LCM (40, 42, 45) \( = 2 \times 3 \times 5 \times 2 \times 2 \times 3 \times 7 \)
\( = 2520 \)
Hence, each person should walk a minimum distance of 2520 cm, so that each of them can cover the same distance in complete steps.

Question. Using prime factorisation, find HCF and LCM of 18, 45 and 60. Check if HCF \( \times \) LCM = product of the numbers.
Answer: Here,
\( 18 = 2 \times 3^2 \)
\( 45 = 3^2 \times 5 \)
and \( 60 = 2^2 \times 3 \times 5 \)
So, HCF (18, 45, 60) = 3; and
LCM (18, 45, 60)\( = 2^2 \times 3^2 \times 5 = 180 \)
Clearly, HCF \( \times \) LCM \( = 3 \times 180 = 540 \)
whereas, product of numbers
\( = 18 \times 45 \times 60 = 48600 \)
Hence, HCF \( \times \) LCM \( \neq \) Product of numbers.

Question. Show that the square of any positive odd integer is of the form 8m + 1, for some integer \( m \).
Answer: Let \( a \) be any positive integer.
So, it is of the form \( 2q + 1 \), for some integer \( q \)
i.e. \( a = 2q + 1 \)
\( \Rightarrow a^2 = (2q + 1)^2 = 4q^2 + 4q + 1 \)
\( = 4q (q + 1) + 1 \)
Now, \( q (q + 1) \) is either 0 or even. So, it is \( 2m \), where \( m \) is a whole number
\( \therefore a^2 = (2q + 1)^2 \)
\( = 4 \times 2m + 1 \) i.e. \( 8m + 1 \)

Question. Prove that \( \sqrt{p} + \sqrt{q} \) is irrational, where \( p \) and \( q \) are primes.
Answer: Let us suppose that \( \sqrt{p} + \sqrt{q} \) is rational.
Let \( \sqrt{p} + \sqrt{q} = a \), where \( a \) is a rational number,
\( \Rightarrow \sqrt{p} = a - \sqrt{q} \)
On squaring both sides, we get
\( \Rightarrow p = a^2 + q - 2a\sqrt{q} \)
[Using \( (a - b)^2 = a^2 + b^2 - 2ab \)]
\( \Rightarrow \sqrt{q} = \frac{a^2 + q - p}{2a} \)
Therefore, the above statement is a contradiction as the right hand side is a rational number, while the left hand side \( \sqrt{q} \) is irrational, since \( p \) and \( q \) are prime numbers.
So, our assumption is wrong. Hence, \( \sqrt{p} + \sqrt{q} \) is irrational.

Question. Prove that \( 3 + 2\sqrt{5} \) is irrational number.
Answer: Let \( 3 + 2\sqrt{5} \) be a rational number.
So we can write this number as
\( 3 + 2\sqrt{5} = \frac{a}{b} \)
Here \( a \) and \( b \) are two co-prime integers and \( b \neq 0 \).
Subtracting 3 from both sides, we get
\( 2\sqrt{5} = \frac{a}{b} - 3 \)
\( 2\sqrt{5} = \frac{(a - 3b)}{b} \)
On dividing both sides by 2, we get
\( \sqrt{5} = \frac{(a - 3b)}{2b} \)
Here \( a \) and \( b \) are integers, so \( \frac{(a - 3b)}{2b} \) is a rational number which implies \( \sqrt{5} \) should be a rational number, but \( \sqrt{5} \) is an irrational number so it is a contradiction.
Hence, \( 3 + 2\sqrt{5} \) is an irrational number.

Question. Show that \( 5 + 2\sqrt{7} \) is an irrational number, where \( \sqrt{7} \) is given to be an irrational number.
Answer: Let us assume, on the contrary, that \( 5 + 2\sqrt{7} \) is a rational number.
i.e., \( 5 + 2\sqrt{7} = \frac{a}{b} \),
where ‘\( a \)’ and ‘\( b \)’ are co-prime numbers.
\( \Rightarrow 2\sqrt{7} = \frac{a}{b} - 5 \)
\( \Rightarrow \sqrt{7} = \frac{a - 5b}{2b} \)
Since \( \frac{a - 5b}{2b} \) is a rational number, \( \sqrt{7} \) is also a rational number,
which is contradiction to the given results.
Hence, \( 5 + 2\sqrt{7} \) is irrational.

Long Short Answer Type Questions

Question. Show that the square of any positive integer cannot be of the form (5q + 2) or (5q + 3) for any integer q.
Answer: Let ‘\( a \)’ be any positive integer. Then, it is of the form \( 5p \), or \( 5p + 1 \) or \( 5p + 2 \) or \( 5p + 3 \) or \( 5p + 4 \)
Case 1 When \( a = 5p \)
\( \Rightarrow a^2 = 25p^2 = 5(5p^2) = 5q \), where \( q = 5p^2 \)
Case 2 When \( a = 5p + 1 \)
\( \Rightarrow a^2 = 25p^2 + 10p + 1 = 5(5p^2 + 2p) + 1 = 5q + 1 \), where \( q = 5p^2 + 2p \).
Case 3 When \( a = 5p + 2 \)
\( \Rightarrow a^2 = 25p^2 + 20p + 4 = 5(5p^2 + 4p) + 4 = 5q + 4 \), where \( q = 5p^2 + 4p \).
Case 4 When \( a = 5p + 3 \)
\( \Rightarrow a^2 = 25p^2 + 30p + 9 = 5(5p^2 + 6p + 1) + 4 = 5q + 4 \), where \( q = 5p^2 + 6p \).
Case 5 When \( a = 5p + 4 \)
\( \Rightarrow a^2 = 25p^2 + 40p + 16 = 5(5p^2 + 8p + 3) + 1 = 5q + 1 \), where \( q = 5p^2 + 8p + \)
Thus, square of any positive integer cannot be of the form \( 5q + 2 \) or \( 5q + 3 \), for any integer \( n \).

Question. Prove that one of every three consecutive positive integers is divisible by 3.
Answer: Let \( n \), \( n + 1 \) and \( n + 2 \) be three consecutive positive integers
Also, we know that a positive integer n is of the form \( 3q \), \( 3q + 1 \) or \( 3q + 2 \)
Case I: When \( n = 3q \)
Here n is clearly divisible by 3.
But (\( n + 1 \)) and (\( n + 2 \)) are not divisible by 3.
[When (\( n + 1 \)) is divided by 3, remainder is 1 and when (\( n + 2 \)) is divided by 3, the remainder is 2]
Case II: When \( n = 3q + 1 \)
Here \( n + 2 = 3q + 3 = 3(q + 1) \)
Clearly, it is divisible by 3.
But \( n \) and (\( n + 1 \)) are not divisible by 3.
Case III: when \( n = 3q + 2 \)
Here, \( n + 1 = 3q + 3 = 3(q + 1) \)
clearly, (\( n + 1 \)) is divisible by 3
But \( n \) and (\( n + 2 \)) are not divisible by 3
Hence, one of every three consecutive positive integers is divisible by 3.

Question. Prove that \( \sqrt{n} \) is not a rational number, if n is not perfect square.
Answer: Let \( \sqrt{n} \) be a rational number.
\( \therefore \sqrt{n} = \frac{p}{q} \), where \( p \) and \( q \) are co-prime integers, \( q \neq 0 \).
On squaring both sides, we get
\( \Rightarrow n = \frac{p^2}{q^2} \)
\( p^2 = nq^2 \) ...(i)
\( \Rightarrow n \) divides \( p^2 \)
[Let p be a prime number. If p divided \( a^2 \) then p divides \( a \), where a is a positive integer]
\( \Rightarrow n \) divides \( p \) ...(ii)
Let \( p = nm \), where \( m \) is any integer.
\( \Rightarrow p^2 = n^2 m^2 \)
(i) \( \Rightarrow n^2 m^2 = nq^2 \)
\( \Rightarrow q^2 = nm^2 \)
\( \Rightarrow n \) divides \( q^2 \)
\( \Rightarrow n \) divides \( q \) ...(iii)
[Let p be a prime number. If p divided \( a^2 \) then p divides \( a \), where a is a positive integer]
From (ii) and (iii), \( n \) is a common factor of both \( p \) and \( q \) which contradicts the assumption that \( p \) and \( q \) are co-prime integer.
So, our supposition is wrong, \( \sqrt{n} \) is an irrational number.

Question. The decimal expansions of some real numbers are given below. In each case, decide whether they are rational or not. If they are rational, write it in the form \( \frac{p}{q} \). What can you say about the prime factors of q?
(A) 0.140140014000140000 ...
(B) \( 0.\overline{16} \)
Answer: (A) We have, 0.140140014000140000... a non-terminating and non-repeating decimal expansion. So it is irrational. It cannot be written in the form of \( \frac{p}{q} \).
(B) We have, \( 0.\overline{16} \) a non-terminating but repeating decimal expansion. So it is rational.
Let \( x = 0.\overline{16} \)
Then, \( x = 0.1616... \) ...(i)
\( 100x = 16.1616 \) ...(ii)
On subtracting (i) from (ii), we get
\( 100x - x = 16.1616 - 0.1616 \)
\( \Rightarrow 99x = 16 \Rightarrow x = \frac{16}{99} \)
The denominator (\( q \)) has factors other than 2 or 5.