CBSE Class 10 Mathematics Real Numbers Worksheet Set 07

Very Short Answer Type Questions 

Question. Find the LCM (306, 1314), if HCF (306, 1314) = 18.
Answer: 22338

Question. Find the LCM of 336 and 54 by prime factorisation method.
Answer: 3024

Question. The HCF and LCM of two numbers are 9 and 90 respectively. If one number is 18, find the other.
Answer: 45

Question. Show that every positive even integer is of the form \( 2q \) and that every positive odd integer is of the form \( 2q + 1 \), where \( q \) is some integer.
Answer: According to Euclid’s division lemma, \( a = bq + r \). Let \( b = 2 \). Then \( r \) can be 0 or 1. If \( r = 0, a = 2q \) (even). If \( r = 1, a = 2q + 1 \) (odd).

Question. In the adjoining factor tree (I), find the numbers \( m, n \):
(Tree I shows: \( m \) splits into 2 and another branch; that branch splits into 2 and \( n \); \( n \) splits into 2 and another branch; that branch splits into 2 and 5.)
Answer: \( m = 160, n = 40 \)

Question. Use Euclid’s division algorithm to find H.C.F. of 870 and 225.
Answer: 15

Question. Find the LCM and HCF of 120 and 144 by fundamental theorem of Arithmetic.
Answer: HCF = 24, LCM = 720

Question. In the adjoining factor tree (II), find the numbers \( m, n \):
(Tree II shows: \( m \) splits into 2 and another branch; that branch splits into 3 and \( n \); \( n \) splits into 5 and another branch; that branch splits into 5 and 2.)
Answer: \( m = 300, n = 50 \)

Question. Without actually performing the long division, state whether the \( \frac{543}{225} \) has a terminating decimal expansion or non terminating recurring decimal expansion.
Answer: Terminating

Question. Use Euclid’s Lemma to show that the square of any positive integer is of form \( 4m \) or \( 4m + 1 \) for some integer \( m \).
Answer: Let \( a \) be any positive integer. By Euclid's division lemma, \( a = 2q + r \), where \( r = 0, 1 \).
Case 1: \( a = 2q \implies a^2 = 4q^2 = 4m \).
Case 2: \( a = 2q + 1 \implies a^2 = (2q + 1)^2 = 4q^2 + 4q + 1 = 4(q^2 + q) + 1 = 4m + 1 \).

Short Answer Type Questions 

Question. Prove that \( 3 + \sqrt{7} \) is an irrational number.
Answer: Standard proof by contradiction.

Question. Prove that \( 2\sqrt{3} - 4 \) is an irrational number.
Answer: Standard proof by contradiction.

Question. A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
Answer: 60 litres

Question. Show that any positive even integer is of the form \( 6m, 6m + 2 \) or \( 6m + 4 \), where \( m \) is some integer.
Answer: According to Euclid's Lemma, \( a = 6m + r \) where \( r \in \{0, 1, 2, 3, 4, 5\} \). For even integers, \( r \) must be 0, 2, or 4.

Question. Find the HCF and LCM of 306 and 54. Verify that HCF \( \times \) LCM = Product of the two numbers.
Answer: HCF = 18, LCM = 918. Verification: \( 18 \times 918 = 16524 \); \( 306 \times 54 = 16524 \).

Question. Find the LCM and HCF of 15, 18, 45 by the prime factorisation method.
Answer: LCM = 90, HCF = 3

Question. Prove that \( \frac{1}{2 + \sqrt{3}} \) is an irrational number.
Answer: Standard proof by contradiction after rationalizing the denominator.

Question. Show that \( 9^n \) can’t end with 2 for any integer \( n \).
Answer: Prime factors of \( 9^n \) are \( (3^2)^n = 3^{2n} \). Since it doesn't contain factors 2 and 5, it cannot end in 0, and the powers of 9 only end in 9 or 1.

Question. Show that the square of any positive odd integer is of the form \( 8m + 1 \), for some integer \( m \).
Answer: Let the odd integer be \( a = 4q + 1 \) or \( 4q + 3 \).
If \( a = 4q + 1, a^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1 = 8m + 1 \).
If \( a = 4q + 3, a^2 = 16q^2 + 24q + 9 = 8(2q^2 + 3q + 1) + 1 = 8m + 1 \).

Question. Prove that \( \sqrt{3} + \sqrt{5} \) is an irrational number.
Answer: Standard proof by contradiction.

Question. Prove that \( 2\sqrt{3} - 7 \) is an irrational.
Answer: Standard proof by contradiction.

Question. Show that any positive odd integer is of the form \( 6p + 1, 6p + 3 \) or \( 6p + 5 \) where \( p \) is some integer.
Answer: According to Euclid's Lemma, \( a = 6p + r \) where \( r \in \{0, 1, 2, 3, 4, 5\} \). For odd integers, \( r \) must be 1, 3, or 5.

Question. Prove that \( (5 - \sqrt{3}) \) is an irrational number.
Answer: Standard proof by contradiction.

VALUE BASED QUESTIONS

Question. A charitable trust donates 28 books (each on different topic) of Maths, 16 books (each on different topic) of Science and 12 (each on different topic) of Social Science to the poor students (each student is given maximum number of books (equal number of books) and only one subject of their interest.
(a) Find the number of books each student got.
(b) Find the total number of students.
(c) How it helps our society?
Answer: (a) Number of books given to each student = HCF of 28, 16 and 12 = 4
(b) Number of students \( = \frac{28}{4} + \frac{16}{4} + \frac{12}{4} = 7 + 4 + 3 = 14 \).
Values:

• Helping poor students will encourage the society toward better education.
• Donating books will increase the interest of poor students towards study.
• It will stop the cause like child labour etc. and dropping students from schools.

Question. Rahul, Navin and Vinay has piece of lands in the form of circles of diameters 10m, 12 m and 16 m. With the help of ropes each of them calculates the ratio of the circumference to length of diameters of their respective fields.
(a) Calculate the HCF of their ratios.
(b) What values are show.
Answer: (a) Their HCF is 1. The HCF of their ratios will be 1 because ratio of circumference to diameter is known as \( \pi \) and value of \( \pi \) varies circles to circle.
Values:

• Concept of \( \pi \) i.e. definition of \( \pi \).
• Reason why \( \pi \) is an irrational number.
• HCF of two or more irrational number is always 1.
• Each irrational number is not \( \pi \).

Question. 6 retired teachers, 8 retired doctors and 10 retired defence officers are willing to render their services to a village. Each of doctor, teacher and defence officer serves the different persons in that village.
(a) Find least number of persons served.
(b) Suggest the value of services.
Answer: (a) Least no of persons served = LCM of number of teacher, doctors and defence officers = 120.
Values:

• We can give our services to our nation even after retirement which is a patriotism.
• Serving in villages will stop the pressure in urban areas.
• Self satisfaction and motivation comes after such services.
• It will bridge the gaps between rural and urban areas.

Question. A class of 20 boys and 15 girls is divided into \( n \) groups so that each group has \( x \) boys and \( y \) girls. Find \( x, y \) and \( n \). What values are referred in a class.
Answer: HCF of 20 and 15 = 5. So the 5 students are in each group so \( n = \frac{20 + 15}{5} = 7 \). Hence \( x = 4, y = 3 \) and \( n = 7 \).
Values:

• Promote coeducation.
• Promote and help to educate girl child.
• Role of activity in groups.
• Increasing healthy friendly environment at school level.

Question. Write the HCF of two co-prime numbers
Answer: 1

Question. We can not find LCM of two positive integers by using Euclid’s division algorithm. (True / False)
Answer: True

Question. Product of two integers is equal to the _______ of their HCF and _______.
Answer: Product, LCM

Question. If \( a \) and \( b \) are two co-prime numbers. What is their LCM?
Answer: \( \text{LCM} = a \times b \)

Question. \( \pi \) is a/an _______ number.
Answer: Irrational

Question. From given factor tree find value of \( a + b + c - d \)
(Tree shows: \( d \) splits into 2 and \( c \); \( c \) splits into 1 and \( b \); \( b \) splits into 3 and \( a \); \( a \) splits into 10 and 9.)
Answer: 90

Question. Explain why \( 2 \times 3 \times 7 + 7 \) is a composite number?
Answer: Resulting number is 49 which is clearly \( 7 \times 7 \) so it is composite number.

Question. Show that: \( (3 + 2\sqrt{2})(3 - 2\sqrt{2}) \) is not an irrational number.
Answer: \( (3)^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \) Which is rational i.e. not irrational.

Question. Find the largest number that divides 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.
Answer: Required number = 17

Question. Find H.C.F. of 510 and 92 by using Euclid’s division algorithm.
Answer: HCF = 2

Question. Without performing long division check which of following are terminating and which are non-terminating:
(i) 48/375
(ii) 6/27
(iii) 11/550
Answer: (i) Terminating (ii) Non-Terminating (iii) Terminating

Question. Explain why \( (42)^n \) cannot end with zero. Also find the unit’s place digit in the expansion of \( (42)^n \) if \( n = 17 \).
Answer: No it can not end with zero because prime factors of base i.e. \( 42 = 2 \times 3 \times 7 \). There is no number 5. Hence it cannot end with zero. Units digit in the expansion \( (42)^{17} \) will be 2.

Question. Can two numbers have 18 as their HCF and 380 as their LCM. Give reason.
Answer: No, because 380 is not divisible by 18.

Question. The circumferences of the fore and wheels of a carriage are \( 6\frac{3}{14} \text{ m} \) and \( 8\frac{1}{18} \text{ m} \) respectively. At any given moment, a chalk mark is put on the point of contact of each wheel with the ground. Find the distance travelled by the carriage so that both the chalk marks are again on the ground at the same time.
OR
Two persons A and B walk round a circle whose diameter is 1.4 km. A walks at a speed of 165 metres per minute while B walks at a speed of 110 metres per minute. If they both start at the same point and walk in the same direction, at what interval of time would they both be at the same starting point again?
Answer: 217.5 m because required distance = LCM of \( 6\frac{3}{14} \) and \( 8\frac{1}{18} \)
\( \implies \) LCM of \( \frac{87}{14}, \frac{145}{18} \)
\( \implies \) \( \frac{\text{LCM of 87 and 145}}{\text{HCF of 14 and 18}} = \frac{435}{2} = 217.5 \)
OR
Circumference of circular path \( = 2 \times \frac{22}{7} \times 700 \text{ m} = 4400 \text{ m} \)
Time taken by A to complete one round \( = \frac{4400}{165} \) minutes
Time taken by B to complete one round \( = \frac{4400}{110} \) minutes
Now, required time \( = \text{LCM of } \frac{4400}{165} \text{ and } \frac{4400}{110} \)
\( = \frac{4400}{55} = 80 \text{ minutes} \)

Question. Prove that \( \sqrt{3} \) is an irrational number.
Answer: To prove \( \sqrt{3} \) is an irrational number we will use contradiction method i.e. first we will assume the opposite of that what we want to prove and then at the end we will prove that what we assume is wrong and finally we will come in position to prove the required result.
Now, if possible let \( \sqrt{3} \) is a rational number so as per definition of rational numbers we can say that \( \sqrt{3} = \frac{p}{q} \) {where \( p \) & \( q \) are co-primes and \( q \neq 0 \)} ...(1)
On squaring both sides of equation number (1) we get \( (\sqrt{3})^2 = \left(\frac{p}{q}\right)^2 \)

\( \implies 3 = \frac{p^2}{q^2} \)

\( \implies p^2 = 3q^2 \) ...(2)
It is very clear that R.H.S. of equation number (2) is \( 3q^2 \), so we can say this result (\( 3q^2 \)) is clear cut multiple of 3 [as \( 3q^2 = 3 \times q^2 \)]
As R.H.S. of the equation number (2) is multiple of 3. So L.H.S. i.e. \( p^2 \) will also be multiple of 3, because LHS and RHS of any equation are same [equal] from above discussion it become clear that \( p^2 \) is multiple of 3.
So
\( \implies p \) is multiple of 3 ...(3)
From equation number (3) we are saying that \( p \) is multiple of 3 so we can say that \( p = 3m \) [where \( m \) is any integer]
Now on replacing '\( p \)' by \( 3m \) in equation number (2) we get
\( (3m)^2 = 3q^2 \)

\( \implies 3m \times 3m = 3q^2 \)

\( \implies 3m^2 = q^2 \) ...(4)
Now clearly LHS of equation number (4) is multiple of 3 because LHS is \( 3m^2 \) which can further be written as : \( 3 \times m^2 \).
So RHS i.e. '\( q^2 \)' of equation number (4) will also be multiple of 3 but \( q^2 \) can be multiple of 3 only if \( q \) is multiple of 3. ...(5)
Now from equation number (3) and (5) we can conclude that '\( p \)' and '\( q \)' both are multiple of 3.
But in the beginning i.e. is equation number (1) we have assumed [as per definition of rational numbers] that \( p \) and \( q \) are co-prime i.e. \( p \) and \( q \) have no common factor other than 1.
But right now we have shown that '\( p \)' and '\( q \)' have 3 as common factor so it contradict our supposition that \( \sqrt{3} \) is rational.
Hence our supposition that \( \sqrt{3} \) is rational is wrong.
Hence, we can conclude from this that \( \sqrt{3} \) is an irrational number.

Question. Prove that \( \sqrt{2} \) is an irrational number by contradiction method.
Answer: We assume that \( \sqrt{2} \) is rational.

\( \implies \sqrt{2} = \frac{p}{q} \) (where \( p \) and \( q \) are coprimes, \( q \neq 0 \))

\( \implies 2 = \frac{p^2}{q^2} \) (Squaring both sides)

\( \implies p^2 = 2q^2 \) ...(1)

\( \implies 2q^2 = p^2 \)

\( \implies 2 \) divides \( p^2 \)

\( \implies 2 \) divides \( p \) ...(2) (By Concept 3)
Let \( p = 2m \)

\( \implies p^2 = 4m^2 \)
Putting the value of \( p^2 \) in (1), we have
\( 4m^2 = 2q^2 \)

\( \implies 2m^2 = q^2 \)

\( \implies 2 \) divides \( q^2 \)

\( \implies 2 \) divides \( q \) (By Concept 3)
Thus, 2 divides \( p \) and \( q \) [From (2)]
It means 2 is a common factor of \( p \) and \( q \). This contradicts the assumption as there is no common factor of \( p \) and \( q \).
Thus, \( \sqrt{2} \) is not rational.
Hence, \( \sqrt{2} \) is irrational. Proved.

Question. Prove that \( \sqrt{5} \) is an irrational number by contradiction method.
Answer: Suppose \( \sqrt{5} \) represents a rational number. Then \( \sqrt{5} \) can be expressed in the form \( \frac{p}{q} \), where \( p, q \) are integers and have no common factor, \( q \neq 0 \).
\( \sqrt{5} = \frac{p}{q} \)
Squaring both sides, we get

\( \implies 5 = \frac{p^2}{q^2} \)

\( \implies p^2 = 5q^2 \) ...(1)

\( \implies 5 \) divides \( p^2 \)

\( \implies 5 \) divides \( p \). ...(2) (Concept 3)
Let \( p = 5m \)

\( \implies p^2 = 25m^2 \)
Putting the value of \( p^2 \) in (1), we get
\( 25m^2 = 5q^2 \)

\( \implies 5m^2 = q^2 \)

\( \implies 5 \) divides \( q^2 \)

\( \implies 5 \) divides \( q \). ...(3) (Concept 3)
Thus from (2), 5 divides \( p \) and from (3), 5 also divides \( q \). It means 5 is a common factor of \( p \) and \( q \). This contradicts the supposition so there is no common factor of \( p \) and \( q \).
Hence, \( \sqrt{5} \) is an irrational number. Proved.

Question. Show that \( \sqrt{7} \) is an irrational number. (Without using contradiction method)
Answer: Let us find out square root of 7 by division method.
(Calculation shown: \( \sqrt{7} \approx 2.6457513... \))

\( \implies \sqrt{7} = 2.6457513... \) is non terminating and non recurring decimal.
Hence, \( \sqrt{7} \) is an irrational number. Proved.

Question. Prove that \( \sqrt{7} \) is an irrational number by contradiction method.
Answer: Suppose \( \sqrt{7} \) represents a rational number. Then \( \sqrt{7} \) can be expressed in the form \( \frac{p}{q} \), where \( p, q \) are coprimes (\( q \neq 0 \)).
\( \sqrt{7} = \frac{p}{q} \)
Squaring both the sides, we get
\( 7 = \frac{p^2}{q^2} \)

\( \implies 7q^2 = p^2 \) ...(1)

\( \implies 7 \) divides \( p^2 \)

\( \implies 7 \) divides \( p \) ...(2) (Concept 3)
Let \( p = 7m \)

\( \implies p^2 = 49m^2 \)
Putting the value of \( p^2 \) in (1), we get
\( 7q^2 = 49m^2 \)

\( \implies q^2 = 7m^2 \)

\( \implies 7 \) divides \( q^2 \)

\( \implies 7 \) divides \( q \) ...(3) (Concept 3)
Thus from (2) \( p \) is a multiple of 7 and from (3), \( q \) is also a multiple of 7. It means 7 is a common factor of \( p \) and \( q \). This contradicts the supposition so there is no common factor of \( p \) and \( q \). Hence, \( \sqrt{7} \) is an irrational number. Proved.

Question. Prove that \( 3 + 2\sqrt{5} \) is irrational.
Answer: Let us assume, to the contrary, that \( 3 + 2\sqrt{5} \) is a rational number.
Now, let \( 3 + 2\sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are coprime and \( b \neq 0 \).
So, \( 2\sqrt{5} = \frac{a}{b} - 3 \)
or \( \sqrt{5} = \frac{a}{2b} - \frac{3}{2} \)
Since \( a \) and \( b \) are integers, therefore \( \frac{a}{2b} - \frac{3}{2} \) is a rational number
\( \therefore \sqrt{5} \) is a rational number.
But \( \sqrt{5} \) is an irrational number.
This shows that our assumption is incorrect.
So, \( 3 + 2\sqrt{5} \) is an irrational number. Proved.

Question. Prove that the following are irrationals : (i) \( \frac{1}{\sqrt{2}} \) (ii) \( 7\sqrt{5} \) (iii) \( 6 + \sqrt{2} \)
Answer: (i) Let us assume, to the contrary, that \( \frac{1}{\sqrt{2}} \) is rational. That is, we can find co-prime integers \( p \) and \( q \) (\( q \neq 0 \)) such that
\( \frac{1}{\sqrt{2}} = \frac{p}{q} \) or \( \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{p}{q} \) or \( \frac{\sqrt{2}}{2} = \frac{p}{q} \)
or \( \sqrt{2} = \frac{2p}{q} \)
Since \( p \) and \( q \) are integers so \( \frac{2p}{q} \) is rational, and so \( \sqrt{2} \) is rational.
But this contradicts the fact that \( \sqrt{2} \) is irrational.
So, we conclude that \( \sqrt{2} \) is an irrational. Proved.
(ii) Let us assume, to the contrary, that \( 7\sqrt{5} \) is rational. That is, we can find co-prime integers \( p \) and \( q \) (\( q \neq 0 \)) such that \( 7\sqrt{5} = \frac{p}{q} \).
So, \( \sqrt{5} = \frac{p}{7q} \).
Since \( p \) and \( q \) are integers, \( \frac{p}{7q} \) is rational and so is \( \sqrt{5} \).
But this contradicts the fact that \( \sqrt{5} \) is irrational. So, we conclude that \( 7\sqrt{5} \) is an irrational. Proved.
(iii) Let us assume, to the contrary, that \( 6 + \sqrt{2} \) is rational. That is, we can find integers \( p \) and \( q \) (\( q \neq 0 \)) such that
\( 6 + \sqrt{2} = \frac{p}{q} \) or \( 6 - \frac{p}{q} = -\sqrt{2} \)
or \( \sqrt{2} = 6 - \frac{p}{q} \)
Since \( p \) and \( q \) are integers, we get \( 6 - \frac{p}{q} \) is rational, and so \( \sqrt{2} \) is rational.
But this contradicts the fact that \( \sqrt{2} \) is irrational.
So, we conclude that \( 6 + \sqrt{2} \) is an irrational. Proved.

Question. Show that \( \sqrt[3]{6} \) is not a rational number.
Answer: Let us assume that \( \sqrt[3]{6} \) is a rational number.
Let \( \sqrt[3]{6} = \frac{p}{q} \), (where \( p \) and \( q \) are coprimes, \( q \neq 0 \))
\( \frac{p^3}{q^3} = 6 \) ...(1) (Taking cube on both sides)
Now \( 1^3 = 1 \quad 2^3 = 8 \)
Thus \( 1 < 6 < 8 \)

\( \implies 1 < \frac{p^3}{q^3} < 8 \)

\( \implies 1 < \frac{p}{q} < 2 \)
From (1) \( \frac{p^3}{q^3} = 6 \)

\( \implies 6q^2 = \frac{p^3}{q} \) ...(2)
As \( q \) is an integer \( \implies 6q^2 \) is an integer ...(3)
Since \( p \) and \( q \) have no common factor.

\( \implies p^3 \) and \( q \) will have no common factor.

\( \implies \frac{p^3}{q} \) is a fraction ...(4)
From (2), (3) and (4)
Integer = fraction
Thus our assumption is wrong.
Hence \( \sqrt[3]{6} \) is not a rational number. Proved.

Question. Prove that \( \sqrt{n} \) is not a rational number, if \( n \) is not a perfect square.
Answer: Suppose \( \sqrt{n} \) is a rational number.
\( \sqrt{n} = \frac{p}{q} \) (\( p \) and \( q \) have no common factor, coprimes, and \( q \neq 0 \))

\( \implies n = \frac{p^2}{q^2} \)

\( \implies p^2 = nq^2 \) (Squaring both sides) ...(1)

\( \implies n \) divides \( p^2 \)

\( \implies n \) divides \( p \) ...(2) (Concept 3)
Let \( p = nm \)

\( \implies p^2 = n^2 m^2 \)
Putting the value of \( p^2 \) in (1), we get
\( n^2 m^2 = nq^2 \)

\( \implies q^2 = nm^2 \)

\( \implies n \) divides \( q^2 \).

\( \implies n \) divides \( q \). ...(3) (Concept 3)
From (2), \( n \) divides \( p \) and from (3) \( n \) divides \( q \). It means \( n \) is a factor of both \( p \) and \( q \). This contradicts the assumption that \( p \) and \( q \) have no common factor. So our supposition is wrong.
Hence, \( \sqrt{n} \) cannot be a rational number. Proved.

Question. Prove that \( \sqrt{2} + \sqrt{3} \) is an irrational number.
Answer: Assume that \( \sqrt{2} + \sqrt{3} \) is a rational number 'a'.
\( \therefore a = \sqrt{2} + \sqrt{3} \)

\( \implies a^2 = (\sqrt{2} + \sqrt{3})^2 \) (Squaring both sides)

\( \implies a^2 = 2 + 3 + 2\sqrt{2}\sqrt{3} \)

\( \implies a^2 = 5 + 2\sqrt{6} \)

\( \implies a^2 - 5 = 2\sqrt{6} \)

\( \implies \frac{a^2 - 5}{2} = \sqrt{6} \) ...(1)
Now \( \frac{a^2 - 5}{2} \) is a rational number as '\( a \)' is a rational number (by assumption)
From (1), \( \sqrt{6} \) is a rational number. [\( \therefore \) L.H.S. is rational so R.H.S. will be rational]
But \( \sqrt{6} \) is an irrational number. (Example 6)
Thus, there is a contradiction.
Hence, our assumption is wrong.
Therefore, \( \sqrt{2} + \sqrt{3} \) is an irrational number. Proved.

Question. In the following equations find which variables \( x, y, z \) etc. represent rational or irrational numbers.
(i) \( x^2 = 5 \)
(ii) \( y^2 = 9 \)
(iii) \( z^2 = 0.04 \)
(iv) \( u^2 = \frac{17}{4} \)
(v) \( v^2 = 3 \)
(vi) \( w^3 = 27 \)
(vii) \( t^2 = 0.4 \)
Answer: (i) \( x^2 = 5 \implies x = \sqrt{5} = 2.2360679 .... \) which is an irrational number.
(ii) \( y^2 = 9 \implies y = 3 \) which is a rational number.
(iii) \( z^2 = 0.04 \implies z = 0.2 \) which is a rational number.
(iv) \( u^2 = \frac{17}{4} \implies u = \frac{\sqrt{17}}{2} = 2.0615528 .... \) which is an irrational number.
(v) \( v^2 = 3 \implies v = \sqrt{3} = 1.7320508 ...... \) which is an irrational number.
(vi) \( w^3 = 27 \implies w = 3 \) which is a rational number.
(vii) \( t^2 = 0.4 \implies t = 0.63245 ...... \) which is an irrational number.

VALUE BASED QUESTION

Question. Why \( \pi \) is an irrational numbers.
Answer: We know that the ratio of circumference of a circle to length of its corresponding diameter is known as \( \pi \) (pie).
i.e. \( \pi = \frac{c}{d} \)
Now, read this to understand why \( \pi \) is an irrational number.
If we take circumference of a circle as an integer (natural number) then we find length of its corresponding diameter comes in decimal [not an integer (natural number)] so ratio \( \frac{c}{d} \) is not in the form of \( \frac{p}{q} \), i.e. \( p \) and \( q \) both are not integers. Similarly if we take diameter of circle as an integer (natural number) then circumference of its corresponding circle comes out to be in the decimal form so again ratio \( \frac{c}{d} = \pi \) does not comes in the form of \( \frac{p}{q} \) because diameter, i.e. '\( q \)' is an integer and '\( p \)' is in decimal form. So it is clear that for any circle if length of diameter is natural number then circumference is the decimal form or if circumference is natural number then diameter is a decimal number. So we can say that the ratio \( \frac{c}{d} \) can never come in the form of \( \frac{p}{q} \) where \( p \) and \( q \) both integer and \( q \neq 0 \). So it is clear that the ratio \( \frac{c}{d} \), i.e. \( \pi \) can never be rational hence \( \pi \) is always an irrational number.

Question. Identify the following as rational or irrational.
(i) \( \sqrt{4} \)
(ii) \( 3\sqrt{18} \)
(iii) \( \sqrt{1.44} \)
(iv) \( \sqrt{\frac{9}{27}} \)
(v) \( -\sqrt{0.64} \)
(vi) \( \sqrt{100} \)
(vii) \( (\sqrt{3} + 1)^2 \)
(viii) \( (\sqrt{5} + 1)(\sqrt{5} - 1) \)
(ix) \( (\sqrt{3} + \sqrt{4})(\sqrt{5} + \sqrt{6}) \)
(x) \( \frac{6}{2\sqrt{3}} \)
Answer: (i) Rational Number (ii) Irrational Number (iii) Rational Number (iv) Irrational Number (v) Rational Number (vi) Rational Number (vii) Irrational Number (viii) Rational Number (ix) Irrational Number (x) Irrational Number.