Class 10

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.2

Question 1.
If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, then find A ∩ B and B ∩ C. Are they same ?
Solution:
A = {1, 2, 3, 4}; B = {1, 2, 3, 5, 6}
∴ A ∩ B = {1, 2, 3, 4} ∩ {1, 2, 3, 5, 6}
= {1, 2, 3}
B ∩ A = {1, 2, 3, 5, 6} ∩ {1, 2, 3, 4)
= {1, 2, 3}
Yes, A ∩ B and B ∩ A are same.

Question 2.
A = {0, 2, 4}, find A ∩ ϕ and A ∩ A. Comments. (June 15(A.P.))
Solution:
A = {0, 2, 4}
A ∩ ϕ = {0, 2, 4} ∩ ϕ
= ϕ
A ∩ A = {0, 2, 4} ∩ {0, 2, 4}
= {0, 2, 4} = A

Question 3.
If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A – B and B – A.
Solution:
A = {2, 4, 6, 8, 10);
B = {3, 6, 9, 12, 15)
A – B = {2, 4, 8, 10}
B – A = {3, 9, 12, 15}

Question 4.
If A and B are two sets such that A ⊂ B then what is A ∪ B?
Solution:
Let us consider A ⊂ B.
Set A = {1, 2, 3}
Set B = {1, 2, 3, 4, 5}
A ∪ B = {1, 2, 3} ∪ {1, 2, 3, 4, 5)
= {1, 2, 3, 4, 5}
= B
A ∪ B = B

Question 5.
If A = {x : x is a natural number}
B = {x : x is an even natural number}
C = {x : x is an odd natural number}
D = {x : x is a prime number
Find A∩B, A∩C, A∩D, B∩C, B∩D, C∩D.
Solution:
A = {1, 2, 3, 4, …….. }
B = {2, 4, 6, 8, ……….}
C = {1, 3, 5,7, ……….}
D = {2, 3, 5, 7, ……….}
A ∩ B = {1, 2, 3, 4,………. } ∩ {2, 4, 6, 8,………. }
= {2, 4, 6,…… }
= {even natural numbers)
A ∩ C = {1, 2, 3, 4,……} ∩ {1, 3, 5,……}
= {1, 3, 5,………}
= {odd natural numbers}
A ∩ D = {1, 2, 3, 4,……} ∩ {2, 3, 5, 7, ……..}
= {2, 3, ………}
= {prime natural numbers}
B ∩ C = {2, 4, 6, 8,…..} ∩ {1, 3, 5, 7, …….}
= ϕ
= Null set (Or) empty set
B ∩ D = {2, 4, 6, 8,…….. } ∩ {2, 3, 5, 7,…….. }
= {2} = {even natural number}
C ∩ D = {1, 3, 5, 7,……. } ∩ {2, 3, 5, 7,…….}
= {3, 5, 7,…….. }
= {All odd prime numbers)

Question 6.
If A = {3, 6, 9, 12, 15, 18, 21);
B = {4, 8, 12, 16, 20};
C = {2, 4, 6, 8, 10, 12, 14, 16};
D = {5, 10, 15, 20}, find
i) A – B
ii) A – C
iii) A – D
iv) B – A
v) C – A
vi) D – A
vii) B – C
viii) B – D
ix) C – B
x) D – B
Solution:
i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}
= {3, 6, 9, 15, 18, 21}

ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16)
= {3, 9, 15, 18, 21}

iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}
= {3, 6, 9, 12, 18, 21}

iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
= {4, 8, 16, 20}

v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
= {2, 4, 8, 10, 14, 16}

vi) D – A = {5, 10, 15, 20} — {3, 6, 9, 12, 15, 18, 21)
= (5, 10, 20)

vii) B – C = {4, 8, 12, 16, 20) – {2, 4, 6, 8, 10, 12, 14, 16}
= {20}

viii) B — D = {4, 8, 12, 16, 20} – {5, 10, 15, 20}
= {4, 8, 12, 16)

ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}
= {2, 6, 10, 14)

x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20}
= {5, 10, 15}

Question 7.
State whether each of the following statement is true or false. Justify your answers.
i) {2, 3, 4, 5} and {3, 6) are disjoint sets.
ii) {a, e, i, o, u) and {a, b, c, d) are disjoint sets.
iii){2, 6, 10, 14) and {3, 7, 11, 15) are disjoint sets.
iv) (2, 6, 10) and {3, 7, 11) are disjoint sets.
Answer:
Two sets are said to be disjoint sets when they have no elements in common.
i) In the given two sets. ‘3’ is common. So, they are not disjoint sets. Hence, the given statement is false.
ii) In the given two sets, ‘a’ is common. So, they are not disjoint sets. Hence, the given statement is false.
iii) There are no elements common in the given two sets. So they are disjoint sets. Hence, the given statement is true.
iv) There are no elements common in the given two sets, So they are disjoint sets. Hence. the given statement is true.

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.1

Question 1.
Which of the following are sets? Justify your answer.
(i) The collection of all the months of a year begining with the letter “J”.
Answer:
The months of a year which begin with the letter “J” are January, June and July.
The required set is {January, June, July}.
∴ The given statement is a well defined collection of objects. So, it is a set.

(ii) The collection of ten most talented writers of India.
Answer:
The given statement is not a set because the criterion for determining as most talented writers of India may vary from person to person. Thus, it is not a well defined collection.

(iii) A team of eleven best cricket batsmen of the world.
Answer:
The given statement is not a set because the criterion for determining eleven best cricket batsmen of the world may vary from person to person.

(iv) The collection of all boys in your class.
Answer:
The given statement is a set because given a student we can divide whether he/she belongs to the set or not.

(v) The collection of all even integers.
Answer:
The given statement is a set because given a number we can divide whether the number belongs to the given set or not.

Question 2.
If A = {0, 2, 4, 6}, B = {3, 5, 7} and C = {p, q, r}, then fill the appropriate symbol, ∈ or ∉ in the blanks.
(i) 0 ….. A
Answer:
0 ∈ A

(ii) 3 …… C
Answer:
3 ∉ C

(iii) 4 ….. B
Answer:
4 ∉ B

(iv) 8 ….. A
Answer:
8 ∉ A

(v) p …… C
Answer:
p ∈ C

(vi) 7 …… B
Answer:
7 ∈ B

Question 3.
Express the following statements using symbols.
(i) The element ‘x’ does not belong to ‘A’.
Answer:
x ∉ A

(ii) ‘d’ is an element of the set ‘B’.
Answer:
d ∈ B

(iii) ‘1’ belongs to the set of Natural numbers.
Answer:
1 ∈ N

(iv) ‘8’ does not belong to the set of prime numbers P.
Answer:
8 ∉ P

Question 4.
State whether the following statements are true or false. Justify your answer
(i) 5 ∉ set of prime numbers
Answer:
Not true (5 is a prime number)

(ii) S ∈ {5, 6, 7} implies 8 ∈ S.
Answer:
Not true (8 is not a member of S)

(iii) -5 ∉ W where ‘W’ is the set of whole numbers
Answer:
True (-5 is not a member of W)

(iv) \(\frac{8}{11}\) ∈ Z Where ‘Z’ is the set of integers.
Answer:
Not true (\(\frac{8}{11}\) is a rational number)

Question 5.
Write the following sets in roster form.

(i) B = {x : x is a natural number smaller than 6}
Answer:
B = {1, 2, 3, 4, 5}

(ii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}.
Answer:
C = {17, 26, 35, 44, 53, 62, 71}

(iii) D = {x : x is a prime number which is a divisor of 60}.
Answer:
D = {3, 5}

(iv) E = {x : x is an alphabet in BETTER}.
Answer:
E = {B, E, T, R}

Question 6.
Write the following sets in the set-builder form.
(i) {3, 6, 9, 12}
Answer:
A = {x: x is multiple of 3 and less than 13}

(ii) {2, 4, 8, 16, 32}
Answer:
B = {x : x is in power of 2 and x is less than 6}

(iii) {5, 25, 125, 625}
Answer:
C = {x : x is in power of 5 and x is less than 5}

(iv) {1, 4, 9, 16, 25, 100}
Answer:
D = {x : x is in square of natural number and not greater than 10)

Question 7.
Write the following sets in roster form.
(i) A = {x : x is a natural number greater than 50 but smaller than 100}
Answer:
A = {51, 52, 53, 54….., 98, 99}

(ii) B = {x : x is an integer, x = 4}
Answer:
B = {+2, -2}

(iii) D = {x : x is a letter in the word “LOYAL”}
Answer:
D = {L, O, Y, A}

Question 8.
Match the roster form with set builder form.

Answer:
i) c
ii) a
iii) d
iv) b

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.5 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.5

Question 1.
Determine the value of the following.

(i) log₂₅5
Solution:
Let log₂₅5 = x ∵ log_aN = x ⇒ a^x = N
25^x = 5; 5^2x = 5¹ ⇒ a^x = N
⇒ 2x = 1 ⇒ x = \(\frac{1}{2}\)
log₂₅5 = \(\frac{1}{2}\)

(ii) log₈₁3
Solution:
log₈₁3 = x ∵ log<sub.aN = x
81^x = 3 ⇒ (3⁴)^x = 3¹ ⇒ a^x = N
⇒ 4x = 1 ⇒ x = \(\frac{1}{4}\)
log₈₁3 = \(\frac{1}{4}\)

(iii) log₂ \(\left(\frac{1}{16}\right)\)
Solution:
log₂\(\frac{1}{16}\) = x ∵ log_aN = x ⇒ a^x = N
⇒ Then 2^x = \(\frac{1}{16}\) ⇒ 2^x = \(\frac{1}{2^4}\) = 2^-4
⇒ x = – 4
los₂(\(\frac{1}{16}\)) = -4

(iv) log₇1
Solution:
log₇1 = x ∵ log_aN = x ⇒ a^x = N
⇒ Then 7^x = 1 ⇒ 7^x – 7⁰ ⇒ x = 0
⇒ log₁a = 0

(v) log_x \(\sqrt{x}\)
Solution:
log_x \(\sqrt{x}\) ∵ log_aN = x ⇒ a^x = N
⇒ Then x^y = \(\sqrt{x}\) ⇒ x^y = x^1/2
⇒ y = \(\frac{1}{2}\)
log_x \(\sqrt{x}\) = \(\frac{1}{2}\)

(vi) log₂512
Solution:
log₂ 512 ∵ log_aN = x ⇒ a^x = N
⇒ Then 2^x = 512 ⇒ 2^x = 2⁹
⇒ x = 9
log₂512 = 9

(vii) log₁₀0.01
Solution:
log₁₀0.01 = x ∵ log_aN = x ⇒ a^x = N
Then 10^x = 0.01
⇒ 10^x = 10^-2
⇒ x = -2
log₁₀0.01 = -2

(viii) \(\log _{\frac{2}{3}}\left(\frac{8}{27}\right)\)
Solution:
\(\log _{\frac{2}{3}}\left(\frac{8}{27}\right)\) ∵ log_aN = x ⇒ a^x = N
\(\left(\frac{2}{3}\right)^x\) = \(\frac{8}{27}\)
\(\left(\frac{2}{3}\right)^2\) = \(\left(\frac{2}{3}\right)^3\) ⇒ x = 3
\(\log _{\frac{2}{3}}\left(\frac{8}{27}\right)\) = 3

(ix) \(2^{2+\log _2^3}\)
Solution:
\(2^{2+\log _2^3}\) ∵ \(a^{\log _a^m}\) = m
= 2² × \(2^{\log _2^3}\)
= 4 × 3 = 12

Question 2.
Write the following expressions as log N and find their values.

(i) log 2 + log 5
Solution:
log 2 + log 5
∵ log x + log y = log xy
log 2 + log 5 = log (2 × 5)
= log (2 × 5)
= log 10

(ii) log 16 – log 2
Solution:
log 16 – log 2
∵ log x – log y = log \(\left(\frac{x}{y}\right)\)
log 16 – log 2 = log \(\left(\frac{16}{2}\right)\) = log 8

(iii) 3 log 4
Solution:
3 log 4
∵ m log_ax = log_ax^m
3 log 4 = log 4³ = log 64

(iv) 2 log 3 – 3 log 2
Solution:
2 log 3 – 3 log 2 ∵ m log a = log a^m
log x – log y = log\(\frac{x}{y}\)

log 3² – log 2³
log \(\frac{3^2}{2^3}\) = log \(\frac{9}{8}\)

(v) log 243 + log 1
Solution:
log 243 + log 1 ∵ log x + log y = log xy
= log 243 × 1
= log 243

(vi) log 10 + 2 log 3 – log 2
Solution:
log 10 + 2 log 3 – log 2
= log 10 + log 3² – log 2
= log 10 + log 9 – log 2
= log 90 – log 2
= log \(\frac{90}{2}\) = log 45

Question 3.
Evaluate each of the following in terms of x and y, if it is given that x = log₂3 and y = log₂5
(i) log₂ 15
(ii) log₂ 7.5
(iii) log₂60
(iv) log₂6750
Solution:

Question 4.
Expand the following.

(i) log 1000
Solution:

= 3 log 2 + 3 log 5
= 3 [log 2 + log 5]

(ii) log₂\(\left(\frac{128}{625}\right)\)
Solution:
log \(\left(\frac{128}{625}\right)\)

(iii) log x²y³z⁴
Solution:
log x²y³z⁴
= log x² + log y³ + log z⁴
= 2 log x + 3 log y + 4 log z

(iv) log \(\frac{p^2 q^3}{r}\)
Solution:
log \(\frac{p^2 q^3}{r}\)
log (p²q³) – log r
= log p² + log q³ – log r
= 2 log p + 3 log q – log r

(v) \(\log \sqrt{\frac{x^3}{y^2}}\)
Solution:
log \(\sqrt{\frac{x^3}{y^2}}\) ∵ log x^m = m log x
= log \(\left(\frac{x^3}{y^2}\right)^{1 / 2}\) = \(\frac{1}{2} \log \left(\frac{x^3}{y^2}\right)\)
= \(\frac{1}{2}\)[log x³ – log y²]
= \(\frac{1}{2}\)[3 log x – 2 log y]

Question 5.
If x² + y² = 25xy, then prove that 2 log(x + y) = 3log3 + logx + logy.
Solution:
Given x² + y² = 25xy
Adding ‘2xy’ on both sides.
x² + y² + 2xy = 25xy + 2xy
(x + y)² = 27xy
Applying ‘log’ on both sides
log(x + y)² = log 27xy
2log(x + y) = log(3³ × x × y)
= log3³ + log x + log y
∴ 2log(x + y) = 31og3 + logx + logy

Question 6.
If log\(\left(\frac{x+y}{3}\right)\) = \(\frac{1}{2}\)log(x + y) = 3log3 + logx + logy.
Solution:

Squaring on both sides

Question 7.
If (2.3)^x = (0.23)^y = 1000, then find the value of \(\frac{1}{x}\) – \(\frac{1}{y}\).
Solution:
Given : (2.3)^x = (0.23)^y = 1000
(2.3)^x = 1000 = 10³
∴ 2.3 = \(10^{\frac{3}{x}}\)
Also (0.23)^y = 10³
∴ 0.23 = \(10^{\frac{3}{y}}\)
Now 0.23 = \(\frac{2.3}{10}\)

Question 8.
If 2^x+1 = 3^1-x then find the value of x.
Solution:
Given : 2^{x + 1} = 3^{1 – x}
Taking log on b.t.s
log 2^{x + 1} = log 3^{1 – x}
(x + 1) log 2 = (1 – x) log 3
x log 2 + log 2 = log 3 – x log 3
x log 2 + x log 3 = log 3 – log 2
x (log 3 + log 2) = log 3 – log 2
∴ x = \(\frac{\log 3-\log 2}{\log 3+\log 2}\) = \(\frac{\log \frac{3}{2}}{\log 6}\)

Question 9.
Is (i) log 2 rational or irrational? Justify your answer.
(ii) log 100 rational or irrational? Justify your answer.
Solution:
i) log2 is rational. Since the value of log₁₀2
= 0.3010

(ii) log 100 rational or irrational? Justify your answer.
Solution:
log 100 is rational
∴ log₁₀100 = log₁₀10²
= 2 log₁₀10 = 2

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers InText Questions

Do This

Question 1.
Find ‘q’ and ‘r’ for the following pairs of positive integers ‘a’ and ‘b’, satisfied a = bq + r
i) a = 13, b = 3
Solution:
a = 13, b = 3
a = bq + r
13 = 3(4) + 1 Here q = 4; r = 1
13 = 13

ii) a = 8, b = 80
Solution:
a = 8, b = 80
a = bq + r Here q = \(\frac{1}{10}\) and r = 0
8 = 80\(\left(\frac{1}{10}\right)\) + 0, 8 = 8

iii) a = 125; b = 5
Solution:
a = 125; b = 5
a = bq + r
q = 25, r = 0
125 = 5(25) + 0
125 = 125

iv) a = 132; b = 11
Solution:
a = 132; b = 11
a = bq + r
q = 12, r = 0
132 = 11(12) + 0
132 = 132

Question 2.
Find the HCF of the following by using Euclid division lemma (Page No. 4)
i) 50 and 70
ii) 96 and 72
iii) 300 and 550
iv) 1860 and 2615
Euclid Division Lemma
a = bq + r, q > 0 and 0 ≤ r < b
Solution:
i) 50 and 70 When 70 is divided by 50, the remainder is 20 to get 70 = 50 × 1 + 20
Now consider the division of 50, with the remainder 20 in the above and apply the division lemma to get 50 = 20 × 2 + 10
Now consider the division of 20, with the remainder 10 in apply the division lemma to get 20 = 10 × 2 + 0
The remainder = 0, when we cannot proceed further.
We conclude that the HCF of 50 and 70 is the divisor at this stage, i.e., 10
∴ So, HCF of 50 and 70 is 10.

ii) 96 and 72 When 96 is divided by 72, the remainder is 24 to get 96 = 72 × 1 + 24
Now consider the division of 72, with the remainder 24 in the above and apply the division lemma to get, 72 = 24 x 3 + 0
The remainder = 0, when we cannot proceed further.
We conclude that the HCF of 96 and 72 is the divisor at this stage, i.e., 24 so, the HCF of 96 and 72 is 24

iii) 300 and 550 When 550 is divided by 300, the remainder is 250, to get 550 = 300 × 1 + 250
Now consider the division of 300 with the remainder 250, and apply the division lemma to get 300 = 250 × 1 + 50
Now consider the division of 250 with the remainder 50, and apply the division lemma to get 250 = 50 × 5 + 0. The remainder is zero, when we cannot proceed further. We conclude that the H.C.F of 300 and 550 is the divisor at this stage i.e., 50. So, the H.C.F of 300 and 550 is 50.

iv) 1860 and 2015 When 2015 is divided by 1860, the remainder is 155, to get 2015 = 1860 × 1 + 155
Now consider the division of 1860 with the remainder 155, and apply the division lemma to get 1860 = 155 × 12 + 0
The remainder is zero, then we cannot proceed further.
We conclude that the HCF of 1860 and 2015 is the divisor at this stage i.e., 155 So, the HCF of 1860 and 2015 is 155

Think – Discuss

Question 1.
From the above questions in do this what is the nature of q and r ? (Page No. 3)
Solution:
Given a = bq + r q > 0 and r ≤ 0 r < b r lies between 0 and b

Question 2.
Can you find the HCF of 1.2 and 0.12 ? Justify your answer. (Page No. 4)
Solution:
1.2 = \(\frac{12}{10}\) and 0.12 = \(\frac{12}{100}\)
When \(\frac{12}{10}\) is divided by \(\frac{12}{100}\), then the remainder is 0.
\(\frac{12}{10}\) = \(\frac{12}{100}\) × 10 + 0
∴ yes, we can find the HCF of 1.2 and 0.12 HCF (1.2, 0.12) = 10

Textual Examples

Question 1.
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. (Page No. 5)
Solution:
Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1.
If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1.

Question 2.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer. (Page No. 5)
Solution:
Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.
That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

Think – Discuss

Question 1.
If r = 0, then what is the relationship between a, b and q in a = bq + r of Euclid division lemma? (Page No. 6)
Solution:
a = bq + 0 (∵ r = 0) ⇒ a = bq

Do This

Question 1.
Express 2310 as a product of prime factors. Also see how your friends have factorized the number. Have they done it like you ? Verify your final product with your friend’s result. Try this for 3 or 4 more numbers. What do you conclude ?
(Page No. 7)
Solution:
2310 = 2 × 1155
= 2 × 5 × (231)
= 2 × 5 × 3 × 77
= 2 × 5 × 3 × 7 × 11
= 2 × 3 × 5 × 7 × 11
Ramu, Kiran and Mohan are my friends. They factorized the number in the following way :
Ramu : 2310 = 231 × 10
= 3 × 77 × 2 × 5
= 3 × 7 × 11 × 2 × 5
Kiran : 2310 = 3 × 770
= 3 × 10 × 77
= 3 × 10 × 11 × 7
= 3 × 2 × 5 × 11 × 7
Mohan : 2310 = 7 × 330
= 7 × 3 × 110
= 7 × 3 × 11 × 10
= 7 × 3 × 11 × 2 × 5
No, but the prime factors are same.
The final product with your friends result is same.
Conclusion : The order in which the prime factors occur, the composite number is unique.

Question 2.
Find the HCF and LCM of the following given pairs of numbers by prime factorisation.
i) 120, 90
ii) 50, 60
iii) 37, 49 (Page No. 8)
H.C.F. : (Product of smallest power of each common prime factor of the numbers)
L.C.M. : (Product of greatest power of each prime factors of the numbers)
Solution:
i) 120, 90
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3¹ × 5¹
90 = 2 × 3 × 3 × 5 = 2¹ × 3² × 5¹
HCF (120, 90) = 2¹ × 3¹ × 5¹ = 30
LCM (120, 90) = 2³ × 3² × 5¹ = 360

ii) 50, 60
50 = 2 × 5 × 5 = 2¹ × 5²
60 = 2 × 2 × 3 × 5 = 2² × 3¹ × 5¹
HCF (50, 60) = 2¹ × 5¹ = 10 (Product of smallest power of each common prime factors of the numbers)
LCM (50, 60) = 2² × 5² × 3¹ = 300 (Product of greatest power of each prime factors of the numbers)

iii) 37, 49
37 is a prime number and 49 is a composite number, so the HCF(37, 49) is 1 and LCM(37, 49) is 1813.

Textual Examples

Question 1.
Consider the numbers 4ⁿ, where n is a natural number. Check whether there is any value of n for which 4ⁿ ends with the digit zero. (Page No. 5)
Solution:
For the number 4ⁿ to end with digit zero for any natural number n, it should be divisible by 5. This means that the prime factorisation of 4ⁿ should contain the prime number 5. But it is not possible because 4ⁿ = (2)²ⁿ so 2 is the only prime in the factorization of 4ⁿ. Since 5 is not present in the prime factorization, so there is no natural number n for which 4ⁿ ends with the digit zero.

Question 2.
Find the HCF and LCM of 12 and 18 by the prime factorization method. (Page No. 7)
Solution:
We have 12 = 2 × 2 × 3 = 2² × 3¹
18 = 2 × 3 × 3 = 2¹ × 3²
Note that HCF (12, 18) = 21 x 31 = 6
HCF : Product of the smallest power of each common prime factors in the numbers.
LCM (12, 18) = 2² × 3² = 36 = Product of the greatest power of each prime factors in the numbers.

Try This

Question 1.
Show that 3ⁿ × 4^m cannot end with the digit 0 or 5 for any natural numbers ‘n’ and ‘m’. (Page No. 8)
Solution:
Let the number 3ⁿ × 4^m
= 3ⁿ × (2²)^m
= 3ⁿ × 2^2m
Number 3ⁿ × 22^m to end with ‘0’ or ‘5’. It should be divisible by 2 and 5. This means that the prime factorization of 3ⁿ × 4^m should contain prime numbers. But it is not possible because 3ⁿ × 4^m = 3ⁿ × 22^m
So 2 or 3 are the only prime factors in its factorization. Since 5 is not present in the prime factorization 3ⁿ × 4^m can not end with the digits 0 or 5.

Try This

Question 1.
Show that 3ⁿ × 4^m cannot end with the digit 0 or 5 for any natural numbers ‘n’ and’m’. (Page No. 8)
Solution:
Let the number 3ⁿ × 4^m = 3ⁿ × (2²)^m
= 3ⁿ × 2^2m
Number 3ⁿ × 2^2m to end with ‘0’ or ‘5’. It should be divisible by 2 and 5. This means that the prime factorization of 3ⁿ × 4^m should contain prime numbers. But it is not possible because 3ⁿ × 4^m = 3ⁿ × 2^2m
So 2 or 3 are the only prime factors in its factorization. Since 5 is not present in the prime factorization 3ⁿ × 4^m can not end with the digits 0 or 5.

Do This

Question 1.
Write the following terminating decimals in the form of \(\frac{\mathbf{p}}{\mathbf{q}}\). q ≠ 0 and p, q are co-primes.
i)15.265
ii) 0.1255
iii) 0.4
iv) 23.34
v) 1215.8
What can you conclude about the denominators through this process? (Page No. 6)
Solution:
i) 15.265 = \(\frac{15265}{10^3}\) = \(\frac{152625}{2^3 \times 5^3}\) = \(\frac{3053}{2^3 \times 5^2}\) = \(\frac{3053}{200}\)

ii) 0.1255 = \(\frac{1255}{10^4}\) = \(\frac{1255}{2^4 \times 5^4}\) = \(\frac{3053}{200}\)

iii) 0.4 = \(\frac{4}{10}\) = \(\frac{2}{5}\)

iv) 23.34 = \(\frac{2334}{10^2}\) = \(\frac{2334}{2^2 \times 5^2}\) = \(\frac{1167}{50}\)

v) 1215.8 = \(\frac{12158}{10}\) = \(\frac{12158}{2 \times 5}\) = \(\frac{6079}{5}\)

We can conclude that the denominator in the problems have only power of 2 or power of 5 or both.

Question 2.
Write the following rational numbers in the form of \(\frac{\mathbf{p}}{\mathbf{q}}\), where q is of the form 2ⁿ5^m where n, m are non-negative integers and these write the negative Integers and then write the numbers in their decimal form.

i) \(\frac{3}{4}\)
ii) \(\frac{7}{5}\)
iii) \(\frac{51}{64}\)
iv) \(\frac{14}{25}\)
v) \(\frac{80}{100}\)
Solution:

Question 3.
Write the following rational numbers as decimals and find out the block of digits, repeating ¡n the quotient.
i) \(\frac{1}{3}\)
ii) \(\frac{2}{7}\)
iii) \(\frac{5}{11}\)
iv) \(\frac{10}{13}\) (Page No. 11)
Solution:
i) \(\frac{1}{3}\) = 0.33333 …….

The block of digits repeating in the quotient is only 3.

ii) \(\frac{2}{7}\) = 0.285714285714 ……


The block of digits repeating in the quotient is only 285714.

iii) \(\frac{5}{11}\) = 0.454545…..

The block of digits repeating in the quotient is only 45.

iv) \(\frac{10}{13}\) = 0.769230769230 …..


The block of digits repeating in the quotient is only 769230.

Textual Examples

Question 1.
Using the above theorems, without actual division, state whether the following rational numbers are terminating or non-terminating, repeating decimals.
i) \(\frac{16}{125}\)
ii) \(\frac{25}{32}\)
iii) \(\frac{100}{81}\)
iv) \(\frac{41}{75}\)
Solution:
\(\frac{16}{125}\) = \(\frac{16}{555}\) = \(\frac{16}{5^3}\)
is terminating decimal.

ii) \(\frac{25}{32}\) = \(\frac{25}{2 \times 2 \times 2 \times 2 \times 2}\) = \(\frac{25}{2^5}\)
is terminating decimal.

iii) \(\frac{100}{81}\) = \(\frac{100}{3 \times 3 \times 3 \times 3}\) = \(\frac{10}{3^4}\)
is non-terminating, repeating decimal.

iv) \(\frac{41}{75}\) = \(\frac{41}{3 \times 5 \times 5}\) = \(\frac{41}{3 \times 5^2}\)
is non-terminating, repeating decimal.

Question 2.
Write the decimal expansion of the following rational numbers without actual division.
i) \(\frac{35}{40}\)
ii) \(\frac{21}{25}\)
iii) \(\frac{7}{8}\)
Solution:

Do This

Question 1.
Verify the statement proved above for p = 2, p = 5 and for a² = 1, 4, 9, 25, 36, 49, 64 and 81. (Page No. 14)
Solution:
The statement proved already is as follow.
Let p be a prime number.
If p divides a² where ‘a’ is a positive integer then ‘p’ divides ‘a’ when p = 2 if 2 divides a (= 4, 36, 64) then p divide ‘a’ (= 2, 6, 8)
In other cases when a² = 1, 9, 25, 49 and 81 the statement is not correct.
When p = 5 if 5 divide a² then p divide ‘a’
In other cases i.e., when a²(= 1, 4, 9, 16, 36, 49, 64, 81). the statement is not correct.

Textual Examples

Question 1.
Prove that \(\sqrt{2}\) is irrational. (Page No. 14)
Solution:
Since we are using proof by contradiction, let us assume the contrary, i.e., \(\sqrt{2}\) is rational.
If it is rational, then there must exist two integers r and s (s ≠ 0) such that
\(\sqrt{2}\) = \(\frac{\mathrm{r}}{\mathrm{s}}\)
Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get \(\sqrt{2}\) = \(\frac{a}{b}\)’ w^ere a and b are co-primes.
So, b\(\sqrt{2}\) = a
On squaring both sides and rearranging, we get 2b² = a². Therefore, 2 divides a².
Now, by statement 1, it follows that if 2 divides a² it also divides a.
So, we can write a = 2c for some integer c. Substituting for a, we get 2b² = 4c², that is, b² = 2c².
This means that 2 divides b², and so 2 divides b (again using statement 1 with p = 2).
Therefore, both a and b have 2 as a common factor.
But this contradicts the fact that a and b are co-prime and have no common factors other than 1.
This contradiction has arisen because of our assumption that \(\sqrt{2}\) is rational. So, we conclude that \(\sqrt{2}\) is irrational.

Question 2.
Show that 5 – \(\sqrt{3}\) is irrational. (Page No. 14)
Solution:
Let us assume, to the contrary, that 5 – \(\sqrt{3}\) is rational.
That is, we can find co-primes a and b (b ≠ 0)
such that 5 – \(\sqrt{3}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\)
Therefore, 5 – \(\frac{\mathrm{a}}{\mathrm{b}}\) = \(\sqrt{3}\)
Rearranging this equation,
we get \(\sqrt{3}\) = 5 – \(\frac{\mathrm{a}}{\mathrm{b}}\) = \(\frac{5 b-a}{b}\)
Since a and b are integers, we get 5 – \(\frac{\mathrm{a}}{\mathrm{b}}\) is rational so \(\sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is irrational.
This contradiction has arisen because of our incorrect assumption that
5 – \(\sqrt{3}\) is rational.
So, we conclude that 5 – \(\sqrt{3}\) is irrational.

Question 3.
Show that 3\(\sqrt{2}\) is irrational.(Page No. 15)
Solution:
Let us assume, the contrary, that 3\(\sqrt{2}\) is rational.
i.e., we can find co-primes a and b (b ≠ 0) such that 3\(\sqrt{2}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\)
Rearranging, we get \(\sqrt{2}\) = \(\frac{a}{3 b}\)
Since 3, a and b are integers, \(\frac{a}{3 b}\) is rational, and so \(\sqrt{2}\) is rational.
But this contradicts the fact that \(\sqrt{2}\) is irrational.
So, we conclude that 3\(\sqrt{2}\) is irrational.

Question 4.
Prove that \(\sqrt{2}\) + \(\sqrt{3}\) is irrational. (Page No. 15)
Solution:
Let us suppose that \(\sqrt{2}\) + \(\sqrt{3}\) is rational.
Let \(\sqrt{2}\) + \(\sqrt{3}\) = \(\frac{a}{b}\), where a, b are integers and
b ≠ 0.
Therefore, \(\sqrt{2}\) = \(\frac{a}{b}\) – \(\sqrt{3}\)
Squaring on both sides, we get

∴ Since a, b are intetgers, \(\frac{a^2+b^2}{2 a b}\) is rational, and so, \(\sqrt{3}\) is rational.
This contradicts the fact that \(\sqrt{3}\) is irrational.
∴ Hence, \(\sqrt{2}\) + \(\sqrt{3}\) is irrational

Think – Discuss

Question 1.
Draw the graphs of y = 2^x; y = 4^x; y = 8^x and y = 10^x in a single graph and mention your observation. (Page No. 17)

Observations:

1. The curve never cuts the X – axis.
2. The value of y is same x = 0.
i.e., y = 2^x = 4^x = 8 = 1o^x = 1 where x = 0.
3. The value of y gets very close to zero for the values of x.
4. All curves meet Y-axis at the same point when x = 0.
Solution:

Question 2.
Write the nature of y, a and x in y = a,sup>x. Can you determine the value of x ford given y justify your answer. (Page No. 17)
Solution:
y = a^x means, the relative change is not same for all real values of x.
We are unable to determine the value of x for a given value of y in y = a^x.
For example y = 3 in y = 7^x
What should be the power to which 7 must be raised to get 3.

Question 3.
From the graph y = 2^x, y = 4^x, y = 8^x and y = 10^x you have drawn earlier have you noticed the value log 1 (any base) (Page No. 18)
Solution:
Yes, log 1 to any base is zero.

Question 4.
You know that 2¹ = 2, 4¹ = 4, 8¹ = 8 and 10¹ = 10. What do you notice about the values of \(\log _2^2\), \(\log _4^4\), \(\log _8^8\) and \(\log _10^10\)?
What can you generalise from this? (Page No. 18)
Solution:
We know that 2¹ = 2, 4¹ = 4, 8¹ = 8 and 10¹ = 10.
\(\log _2^2\) = 1, \(\log _4^4\) = 1, \(\log _8^8\) = 1 and \(\log _10^10\) = 1
∴ We generalise from this, a = N then the \(\log _a^a\) = 1

Question 5.
Does \(\log _0^10\) exist?
Solution:
No, \(\log _0^10\) does not exist.
a^x ≠ 0, a, x ∈ N

Question 6.
We know that, if 7 = 2^x then x = \(\log _2 7\). Then what is the value of \(2^{\log _2^7}\) = ? Justify your answer. Generalise the above by taking same more examples for \(a^{\log _a^N}\). (Page No. 23)
Solution:

Do THIS

Question 1.
Write the powers to which the bases to be raised in the following:

i) 7 = 2^x
ii) 10 = 5^b
iii) \(\frac{1}{81}\) = 3^c
iv) 100 = 10^z
v) \(\frac{1}{257}\) = 4^a.
Solution:
i) 7 = 2^x
We cannot determined the power of x.

ii) 10 = 5^b
2 × 5 = 5^b
We can not determined the power of b.

iii) \(\frac{1}{81}\) = 3^c
(81)^-1 = 3^c
(3⁴)^-1 = 3^c
(3)^-4 = 3^c
∴ C = -4

iv) 100 = 10^z
10² = 10^z
∴ z = 2

v) \(\frac{1}{257}\) = 4^a
We cannot determine the power of a.

Question 2.
Express the logarithms of the following Into sum of the logarithms:
i) 35 × 46
ii) 235 × 437
iii) 2437 × 3568 (Page No. 19)
Solution:
i) 35 × 46
log xy = log x + log y
∴ log₁₀ 35 × 46 = log₁₀ 35 + log₁₀ 46

ii) 235 × 437
log₁₀ 35 × 437 = log₁₀ 235 + log₁₀ 437
[∵ log xy = log x + log y]

iii) 2437 × 3568
log₁₀ 2437 × 3568 = log₁₀ 2437 + log₁₀ 3568
[∵ log xy = log x + log y]

Question 3.
Express the logarithms of the following into difference of the logarithms

i) \(\frac{23}{34}\)
ii) \(\frac{373}{275}\)
iii) 4525 ÷ 3734
iv) 5055/3303
Solution:

Question 4.
By using the formula \(\log _a^{x^n}\) = \(n \log _a^{\mathbf{x}}\) convert the following

i) \(\log _2^{7^{25}}\)
ii) \(\log _5^{8^{50}}\)
iii) \(\log 5^{23}\)
iv) log¹⁰²⁴ (Page No. 21)
Solution:

TRY THIS

Question 1.
Write the following relation in exponential form and find the values of respective variables. (Page No. 18)
i) \(\log _2^{32}\) = x
ii) \(\log _5^{625}\) = y
iii) \(\log _{10}^{10000}\) = z
iv) \(\log _7\left(\frac{1}{343}\right)\) = -a
Solution:

Question 2.
i) Find the value of \(\log _2^{32}\) (Page No. 21)
log x^m = m log x and \(\log _a^a\) = 1
Solution:
= \(\log _2^{2^5}\) = \(5 \log _2^2\) = 5 × 1 = 5

ii) Find the value of \(\log _c^{\sqrt{c}}\) (Page No. 21)
Solution:

iii) Find the value of \(\log _{10}^{0.001}\) (Page No. 21)
Solution:

iv) Find the value of (Page No. 21)
Solution:

Textual Examples

Question 1.
Expand log \(\frac{343}{125}\).
Solution:
As you know, log_a \(\frac{x}{y}\) = log_ax – log_ay
So, log \(\frac{343}{125}\) = log343 – log125
= log7³ – log5³
∴ Since, log_ax^m = m log_ax
= 3log7 – 3log5

Question 2.
Write 2log3 + 3log5 – 5log2 as a single logarithm.
Solution:
2log3 + 3log5 – 5log2
= log3² + log5³ – log2⁵
(Since in m log_ax=log_ax^m)
= log9 + log125 – log32
= log (9 × 125) – log32
(Since log_ax + log_ay = log_axy)
= log1125 – log32
= logl \(\frac{1125}{32}\) (Since log_ax – log_ay = \(\log _a \frac{x}{y}\))

Question 3.
Solve 3^x = 5^x-2 (Page No. 22)
Solution:
Taking log on both the sides
x log₁₀ 3 = (x – 2) log₁₀ 5
x log₁₀ 3 = x log₁₀ 5 – 2 log₁₀ 5
x log₁₀ 5 – 2 log₁₀ 5 = x log₁₀ 3
x(log₁₀ 5 – log₁₀ 3) = 2 log₁₀ 5

Question 4.
Find x if 2 log 5 + \(\frac{1}{2}\) log 9 – log 3 = log x. (Page No. 22)
Solution:
log x = 21og 5 + \(\frac{1}{2}\) log 9 – log 3
= log 5² + log \(9^{\frac{1}{2}}\) – log 3
= log 25 + log \(\sqrt{9}\) – log 3
= log 25 + log 3 – log 3
log x = log 25
∴ x = 25

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

Question 1.
Can the number 6ⁿ, n being a natural number, end with the digit 5 ? Give reason.
Solution:
Given number = 6ⁿ; n ∈ N
6ⁿ to be end in 5; it should be divisible by 5.
6ⁿ = (2 × 3)ⁿ
The prime factors of 6ⁿ are 2 and 3.
∴ It can’t end with the digit 5.

Question 2.
Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer.
Solution:
Given : 7 × 5 × 3 × 2 + 3
= 3(7 × 5 × 2 + 1)
= 3 × (70 + 1)
= 3 × 71
The given number has two factors namely 3 and 71.
∴ Hence it is a composite number.

Question 3.
Prove that (2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number. Also check whether (2\(\sqrt{3}\) + \(\sqrt{5}\))(2\(\sqrt{3}\) – \(\sqrt{5}\)) is rational or irrational.
Solution:
To prove : 2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number. On contrary, let us suppose that 2\(\sqrt{3}\) + \(\sqrt{5}\) be a rational number then 2\(\sqrt{3}\) + \(\sqrt{5}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\)
Squaring on both sides, we get

L.H.S = an irrational number
R.H.S. = p, q being integers, \(\frac{p^2-17 q^2}{4 q^2}\) is a rational number.
This is a contradiction to the fact that \(\sqrt{15}\) is an irrational. This is due to our assumption that 2\(\sqrt{3}\) + \(\sqrt{5}\) is a rational. Hence our assumption is wrong and 2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number.
Also,
(2\(\sqrt{3}\) + \(\sqrt{5}\)) (2\(\sqrt{3}\) – \(\sqrt{5}\)) = (2\(\sqrt{3}\))² -(\(\sqrt{5}\))²
[∵ (a + b) (a – b) = a² – b²]
= 4 × 3 – 5 = 12 – 5 = 7, a rational number.

Question 4.
If x² + y² = 6xy, prove that 2 log (x +y) = logx + logy + 3 log 2
Solution:
Given : x² + y2² = 6xy Adding both sides
⇒ x² + y² + 2xy = 6xy + 2xy (x + y)² = 8xy
Taking logarithms on both sides
log (x + y)² = log 8xy
⇒ 2 log(x + y) = log 8 + log x + log y
[∵ log x^m = mlog x]
[∵ log xy = log x + log y]
= log 2³ + log x + log y
⇒ 2log (x + y) = log x + log y + 3 log 2

Question 5.
Find the number of digits in 4²⁰¹³, if log₁₀ 2 = 0.3010.
Solution:
Given log₁₀2 = 0.3010
4²⁰¹³ = (2²)²⁰¹³ = 2⁴⁰²⁶
∴ log₁₀ 2⁴⁰²⁶ = 4026 log 2
[∵ log x^m = m log x]
= 4026 × 0.3010
= 1211.826 \(\simeq\) 1212
∴ 4²⁰¹³ has 1212 digits in its expansion.

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.2

Question 1.
Express each of the following numbers as a product of its prime factors.

(i) 140
Answer:
140

∴ 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156
Answer:
156

∴ 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13

(iii) 3825
Answer:
3825

∴ 3825 = 3 × 3 × 5 × 5 × 17
= 3² × 5² × 17

(iv) 5005
Answer:
5005

∴ 5005 = 5 × 7 × 11 × 13

(v) 7429
Answer:
7429

∴ 7429 = 17 × 19 × 23

Question 2.
Find the LCM and HCF of the following integers by the prime factorization method.
(i) 12, 15 and 21
Answer:
12, 15, 21
12 = 2 × 2 × 3, 15 = 3 × 5
21 = 3 × 7
H.C.F of 12, 15, 21 is 3.
L.C.M of 12, 15, 21 = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23, and 29
Answer:
17, 23, 29
17 = 1 × 17, 23 = 1 × 23
29 = 1 × 29
H.C.F of 17, 23 and 29 is 1.
L.C.M of 17, 23, 29 is 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
Answer:
8, 9, 25
8 = 1 × 2 × 2 × 2, 9 = 1 × 3 × 3
25 = 1 × 5 × 5
H.C.F. = 1
L.C.M = 2 × 2 × 2 × 3 × 3 × 5 × 5
= 1800

(iv) 72 and 108
Answer:
72, 108

72 = 2 × 36
= 2 × 2 × 18
= 2 × 2 × 2 × 9
= 2 × 2 × 2 × 3 × 3

108 = 2 × 54
= 2 × 2 × 27
= 2 × 2 × 3 × 9
= 2 × 2 × 3 × 3 × 3
H.C.F. = 2 × 2 × 3 × 3 = 36
L.C.M. = 2 × 2 × 2 × 3 × 3 × 3 = 216

(v) 306 and 657
Answer:
306, 657

H.C.F. = 9
L.C.M. = 9 × 34 × 73
= 22338

Question 3.
Check whether 6″ can end with the digit 0 for any natural number n.
Answer:
If the number 6ⁿ for any n, were to end with digit ‘0’ then it would be divisible by 5.
The prime factorisation of 6n would contain the prime 5.
6ⁿ = (2 × 3)ⁿ
Prime factorisation of 6ⁿ does not contain 5 as a factor, so 6ⁿ can never end with the digit 0 for any natural number.

Question 4.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
Given number is 7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= (7 × 11 + 1) × 13 distributive law
= (77 + 1) × 13
= 78 × 13
= (2 × 3 × 13) × 13
= 2 × 3 × 13²
= Product of prime factors
Hence the given number is a composite number.
Given number is
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009
= Product of prime numbers
Hence the given number is a composite number.

Question 5.
How will you show that (17 × 11 × 2)+ (17 × 11 × 5) is a composite number? Explain.
Answer:
Given number is
(17 × 11 × 2) + (17 × 11 × 5)
= 17 × 11 × (2 + 5)
= 17 × 11 × 7 = Product of primes
We know that every composite number can be expressed as a product of primes.

Question 6.
Which digit would occupy the units place of 6¹⁰⁰.
Answer:
6¹⁰⁰ = (2 × 3)¹⁰⁰
So the last digit of 6¹⁰⁰ is 6.

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.4

Question 1.
Prove that the following are irrational.

(i) \(\frac{1}{\sqrt{2}}\)
Solution:
\(\frac{1}{\sqrt{2}}\)
Let us assume to the contrary that \(\frac{1}{\sqrt{2}}\) is rational. Then there exist co-prime positive integers ‘a’ and ‘b’ such that 1 a
\(\frac{1}{\sqrt{2}}\) = \(\frac{\mathrm{a}}{\mathrm{b}}\)
\(\sqrt{2}\)a = b
\(\sqrt{2}\) = \(\frac{b}{a}\)
Here ‘a’ and ‘b’ are integers, \(\frac{b}{a}\) is rational.
∴ \(\sqrt{2}\) is rational.
This contradicts the fact that \(\sqrt{2}\) is irrational.
So our assumption is wrong.
Hence \(\frac{1}{\sqrt{2}}\) is irrational.

(ii) \(\sqrt{3}\) + \(\sqrt{5}\)
Solution:
\(\sqrt{3}\) + \(\sqrt{5}\)
Let us assume to the contrary that \(\sqrt{3}\) + \(\sqrt{5}\) is a rational number.
Then there exist co-prime positive integers ‘a’ and ‘b’ such that
\(\sqrt{3}\) + \(\sqrt{5}\) = \(\frac{a}{b}\)
\(\frac{a}{b}\) – \(\sqrt{3}\) = \(\sqrt{5}\)
S.B.S.

This contradicts the fact that \(\sqrt{3}\) is irrational.
∴ Hence, \(\sqrt{3}\) + \(\sqrt{5}\) is irrational.

(iii) 6 + \(\sqrt{2}\)
Solution:
6 + \(\sqrt{2}\)
Let us assume on the contrary that 6 + \(\sqrt{2}\) is rational. Then there exist co-prime positive integers ‘a’ and ‘b’ such that
6 + \(\sqrt{2}\) = \(\frac{a}{b}\)
⇒ \(\sqrt{2}\) = \(\frac{a-6 b}{b}\)
\(\sqrt{2}\) is rational.
\(\frac{a-6 b}{b}\) is rational.
This contradicts the fact that \(\sqrt{2}\) is irrational, so our assumption is wrong.
∴ 6 + \(\sqrt{2}\) is irrational.

(iv) \(\sqrt{5}\)
Solution:
\(\sqrt{5}\)
Let us assume, to the contrary that \(\sqrt{5}\) is irrational then there exist co-prime positive integers a and b such that
\(\sqrt{5}\) = \(\frac{a}{b}\)
\(\sqrt{5}\) b = a
S.B.S. we get
(\(\sqrt{5}\) b)² = (a)²
5b² = a² ……. (1)
5 divides a².
Hence 5 divides a.
We can write a = 5c for some integer c.
Substitute a = 5c in (1) we get
5b² = (5c)²
5b² = 25c²
b² = \(\frac{25 c^2}{5}\)
b² = 5c²
5 divides b² and 5 divide b.
‘a’ and ‘b’ have atleast as a common factor.
This contradicts the fact that ‘a’ and ‘b’ have no common factor other than 1.
So our assumption is wrong.
∵ \(\sqrt{5}\) is irrational.

(v) 3 + 2\(\sqrt{5}\)
Solution:
3 + 2\(\sqrt{5}\)
Let us assume, to the contrary that 3 + 2\(\sqrt{5}\) is rational. Then there exist co-prime positive integers ‘a’ and ’b’ such that
3 + 2\(\sqrt{5}\) = \(\frac{a}{b}\)
2\(\sqrt{5}\) = \(\frac{a}{b}\) – 3
\(\sqrt{5}\) = \(\frac{a-3 b}{2 b}\)
\(\sqrt{5}\) is rational.
\(\frac{a-3 b}{2 b}\) is rational.
This contradicts the fact that \(\sqrt{5}\) is irrational, so our assumption is wrong.
3 + 2\(\sqrt{5}\) is irrational.

Question 2.
Prove that \(\sqrt{p}\) + \(\sqrt{q}\) is an irrational, where p, q are primes.
Solution:
Let us assume to the contrary that \(\sqrt{p}\) + \(\sqrt{q}\) is rational. Then there exist co-prime positive in-tegers ‘a’ and ‘b’.

We know that square root of any prime number is irrational, we get \(\sqrt{q}\) is rational.
This contradicts the fact that \(\sqrt{q}\) is irrational.
So our assumption is wrong.
∵ \(\sqrt{q}\) is irrational.
∵ \(\sqrt{p}\) + \(\sqrt{q}\) is irrational.

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.1

Question 1.
Use Euclid’s algorithm to find the HCF of

(i) 900 and 270
Answer:
900 and 270
Euclid Division Lemma
a = bq + r, q > 0 and 0 ≤ r < b
When 900 is divided 270, the remainder is 90 to get 900 = 270 × 3 + 90
Now consider the division of 270 with the remainder 90 in the above and division algorithm to get 270 = 90 × 3 + 0 The remainder is zero, when we cannot proceed further. We conclude that the HCF of (900, 270) = 90

(ii) 196 and 38220
Answer:
196 and 38220
When 38220 is divided 196, the remainder is 0 to get 38220 = 196 × 195 + 0
The remainder is zero, when we cannot proceed further. We conclude that the HCF of (38220, 196) = 196.

(iii) 1651 and 2032
Answer:
1651 and 2032
When 2032 is divided by 1651, the remainder is 381 to get 2032 = 1651 × 1 + 381
Now consider the division of 1651 with 381 in the above and division algorithm to get 1651 = 381 × 4 + 127
Now consider the division of 381 with the remainder 127 in the above and division algorithm to get 381 = 127 × 3 + 0
The remainder is zero, when we cannot proceed further we conclude that the HCF (1651, 2032) = 127

Question 2.
Use division algorithm to show that any positive odd integer is ofthe form 6q + 1, or 6 q + 3 or 6 q + 5, where q is some integer.
Answer:
Let ‘a’ be any positive odd integer, we apply the division algorithm with a and b = 6. Since 0 ≤ r ≤ 5, the possible remainders are 0, 1, 2, 3, 4 and 5.
i. e., a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where ‘q’ is quotient, However, since ‘a’ is odd, a cannot be 6q, 6q+2, 6q+4 (since they both are divisible by 2)
∴ any odd integer in the form of 6q + 1, (or) 6q + 3 (or) 6q + 5.

Question 3.
Use division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1.
Answer:
Consider ‘a’ be the square of an integer
Applying Euclid’s division lemma with ‘a’ and ‘b’ = 3
0 ≤ r < 3 the possible remainders are 0, 1, 2 a = 3p or 3p + 1 (or) 3p + 2
Any square number is of the form 3p, 3p + 1, or 3p + 2, Where p is the quotient.

Question 4.
Use division algorithm to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m + 8.
Answer:
Let ‘a’ be the cube of a positive integer
Applying Euclid’s division lemma for ‘a’ and b = 9 a = bm + r, Where’m’ is quotient r is the remainder where 0 ≤ r < 9 ‘a’ can be of the form
9m, 9m + 1, 9m + 2 ………… or 9m + 8
The cube of any positive integer is of the form 9m, 9m + 1, 9m + 2, ….. or 9m + 8.

Question 5.
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
Answer:
Given numbers : n; n + 2 and n + 4

Case (i) : ‘n’ is divisible by 3, then n is of the form 3k. Now, n + 2 = 3k + 2 – leaves a remainder 2 when divided by 3. n + 4 = 3k + 4 = (3k + 3) + 1 = 3(k + 1) + 1 leaves a remainder 1 when divided by 3.

Case (ii) : n + 2 is divisible by 3. then n + 2 is of the form 3k.
Now n = 3k – 2 leaves a remainder 1 when divided by 3 and n + 4 = n + 2 + 2
= 3k + 2 leaves a remainder 2 when divided by 3.

Case (iii) : n + 4 is divisible by 3.
Take n + 4 = 3k
Now : n = 3k – 4 = 3(k – 1) + 3 – 4
= 3(k – 1) – 1 leaves a remainder 2 when divided by 3.
Also n + 2 = 3k-2 = 3(k- 1) + 3-2
= 3(k – 1) + 1 leaves a remainder 1 when divided by 3.
In all the above three cases only one out of n, n + 2 and n + 4 is divisible by 3.

Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1.
Write the following rational numbers in their decimal form and also state which are terminating and which are non-terminating repeating decimal form.

(i) \(\frac{3}{8}\)
Solution:
\(\frac{3}{8}\) = \(\frac{3}{2.2 .2}\) = \(\frac{3}{2^3}\) (∵ Denominator consists of only 2’s. terminating decimal)
\(\frac{3 \times 5^3}{2^3 \times 5^3}\) = \(\frac{3 \times 125}{10^3}\) = \(\frac{375}{1000}\) = 0.375

(ii) \(\frac{229}{400}\)
Solution:
\(\frac{229}{400}\) = \(\frac{229}{2.2 .2 .2 \times 5 \times 5}\) (∵ Denominator consists of only 2 ‘s.)
= \(\frac{229}{2^4 \cdot 5^2}\) (terminating decimal)
= \(\frac{229 \times 5^2}{2^4 \times 5^2 \times 5^2}\) = \(\frac{229 \times 5^2}{2^4 \times 5^4}\)
= \(\frac{229 \times 25}{(2 \times 5)^4}\) = \(\frac{5725}{10^4}\)
= \(\frac{5725}{10000}\) = 0.5725 (∵ Denominator consists of only 2’s. terminating decimal)

(iii) 4\(\frac{1}{5}\)
Solution:
4\(\frac{1}{5}\) = \(\frac{21}{5}\) (terminating decimal)
\(\frac{21}{5}\) = 4.2 (∵ Denominator consists of only 2’s.)

(iv) \(\frac{2}{11}\)
Solution:
\(\frac{2}{11}\) (∵ Denominator doesn’t contain 2s or 5’s or both. Hence it is an non-terminating, repeating decimal) (non-terminating, re-peating decimal because the denomination does not contain power of 2 or power of 5 of both 2 and 5)
\(\frac{2}{11}\) = 0.181818……..
= \(0 . \overline{18}\)

(v) \(\frac{8}{125}\)
Solution:
\(\frac{8}{125}\) = \(\frac{8}{5^3}\) (∵ Denominator does not consists of only 2’s.) (terminating decimal)
= \(\frac{8}{5^3}\) × \(\frac{2^3}{2^3}\) = \(\frac{8 \times 8}{(5 \times 2)^3}\) = \(\frac{64}{1000}\) = 0.064

Question 2.
Without performing division, state whether the following rational numbers will have a terminating decimal form or a non-terminating repeating decimal form.

(i) \(\frac{13}{3125}\)
Solution:
\(\frac{13}{3125}\). It is of the form \(\frac{p}{q}\)
\(\frac{13}{3125}\) = \(\frac{13}{5 \times 5 \times 5 \times 5 \times 5}\) = \(\frac{13}{5^5}\)
∵ q = 5⁵ which is of the form 2ⁿ5^m (n = 0; m = 5)
Given rational number has a terminating decimal expansion.

(ii) \(\frac{11}{12}\)
Solution:
\(\frac{11}{12}\) it is of the form \(\frac{p}{q}\).
\(\frac{11}{12}\) = \(\frac{11}{2 \times 2 \times 3}\) = \(\frac{11}{2^2 \times 3}\) = 0.916666.
∵ Here q = 2² × 3 which is not of the form 2ⁿ × 5^m.
∴ Given rational number has a non-termi-nating repeating decimal expansion.

(iii) \(\frac{64}{455}\)
Solution:
\(\frac{64}{455}\) It is of the form \(\frac{p}{q}\).
\(\frac{64}{455}\) = \(\frac{64}{5 \times 7 \times 13}\)
∵ q = 5 × 7 × 13 which is not of the form 2ⁿ.5^m.
∵ Given rational number has a non-terminating, repeating decimal expansion.

(iv) \(\frac{15}{1600}\)
Solution:
\(\frac{15}{1600}\) It is of the form \(\frac{p}{q}\).
\(\frac{15}{1600}\) = \(\frac{3}{320}\) = \(\frac{3}{2^6 \times 5}\)
∵ Here q = 2⁶ × 5¹ which is of the form 2ⁿ.5^m (m = 1, n = 6)
∴ Given rational number has a terminating decimal expansion.

(v) \(\frac{29}{343}\)
Solution:
\(\frac{29}{343}\) It is of the form \(\frac{p}{q}\).
\(\frac{29}{343}\) = \(\frac{29}{7 \times 7 \times 7}\) = \(\frac{29}{7^3}\)
∵ Here q = 7³ which is not of the form 2n5m.
∴ Given rational number has a non-terminating, repeating decimal expansion.

(vi) \(\frac{23}{2^3 \cdot 5^2}\)
Solution:
\(\frac{23}{2^3 \times 5^2}\) It is of the form \(\frac{p}{q}\).
∵ Here q = 2³ × 5² which is of the form of 2ⁿ5^m (n = 3, m = 2).
∴ Given rational number has a terminating decimal expansion.

(vii) \(\frac{129}{2^2 \cdot 5^7 \cdot 7^5}\)
Solution:
\(\frac{129}{2^2 \times 5^7 \times 7^5}\) It is of the form \(\frac{p}{q}\).
∵ Here q = 2² × 5⁷ × 7⁵ which is not of the form 2ⁿ5^m.
∴ Given rational number has a non-terminating, repeating decimal expansion.

(viii) \(\frac{9}{15}\) = \(\frac{3}{5}\) It is of the form \(\frac{p}{q}\).
Solution:
\(\frac{9}{15}\) = \(\frac{3}{5}\) It is of the form \(\frac{p}{q}\).
Here q = 5¹ which is of the form 2ⁿ5^m (n = 0; m = 1). .
∴ Given rational number has a terminating decimal expansion.

(ix) \(\frac{36}{100}\)
Solution:
\(\frac{36}{100}\) It is of the form \(\frac{p}{q}\).
\(\frac{36}{100}\) = \(\frac{36}{2 \times 2 \times 5 \times 5}\) = \(\frac{36}{2^2 5^2}\)
∵ Here q = 2². 5² which is of the form 2ⁿ5^m (n = 2, m = 2).
∴ Given rational number has a terminating decimal expansion.

(x) \(\frac{77}{210}\)
Solution:
\(\frac{77}{210}\) it is of the form \(\frac{p}{q}\).
\(\frac{77}{210}\) = \(\frac{11}{30}\) = \(\frac{11}{2 \times 3 \times 5}\)
∵ q = 2 × 3 × 5 which is not of the form 2ⁿ5^m
∴ Given rational number has non-terminating, repeating decimal expansion.

Question 3.
Write the following rational numbers in decimal form using Theorem 1.4.

(i) \(\frac{13}{25}\)
Solution:
\(\frac{13}{25}\) = \(\frac{13}{5 \times 5}\) = \(\frac{13 \times 2^2}{5^2 \times 2^2}\)
= \(\frac{13 \quad 4}{10^2}\) = \(\frac{52}{100}\) = 0.52

(ii) \(\frac{15}{16}\)
Solution:
\(\frac{15}{16}\) = \(\frac{15}{2 \times 2 \times 2 \times 2}\) = \(\frac{15 \times 5^4}{2^4 \times 5^4}\)
= \(\frac{15 \quad 5^4}{10^4}\) = \(\frac{9375}{10000}\) = 0.9375

(iii) \(\frac{23}{2^3 \cdot 5^2}\)
Solution:
\(\frac{23}{2^3 \times 5^2}\) = \(\frac{23 \times 5}{2^3 \times 5^2 \times 5}\)
= \(\frac{115}{10^3}\) = \(\frac{115}{1000}\) = 0.115

(iv) \(\frac{7218}{3^2 .5^2}\)
Solution:
\(\frac{7218}{3^2 \times 5^2}\) = \(\frac{3 \times 3 \times 802}{3^2 \times 5^2}\)
= \(\frac{802 \times 2^2}{5^2 \times 2^2}\) = \(\frac{3208}{100}\) = 32.08

(v) \(\frac{143}{110}\)
Solution:
\(\frac{143}{110}\) = \(\frac{11 \times 13}{11 \times 10}\) = \(\frac{13}{10}\) = 1.3

Question 4.
Express the following decimal numbers in the form of q and write the prime factors of q. What do you observe?
(i) 43.123
Solution:
43.123456789
Given decimal expansion terminates. Hence given real number is a rational.
It is in the form of \(\frac{p}{q}\)

∵ Here q = 2⁹.5⁹ ; q is of the form 2ⁿ5^m (n = 9; m = 9)

(ii) 0.120112001120001…
Solution:
0.120120012000120000 …………
Given decimal expansion is not either ter-minating or non-terminating repeating.
∵ Hence given real number is not rational.

(iii) \(43 . \overline{12}\)
Solution:
\(43 . \overline{123456789}\)
Given decimal expansion is non-terminating, repeating.
Given real number is rational and so of the \(\frac{p}{q}\)
Let x = \(43 . \overline{123456789}\)
x = \(43 . \overline{123456789}\) …… (2)
Multiplying both sides of (1) by 1000000000, we get
x = 43123456789.123456789 ……. (2)
Substracting (1) from (2) we get
999999999 x = 43123456746
x = \(\frac{43123456746}{999999999}\)
x = \(\frac{14374485582}{333333333}\)
q = 333333333 which is not of the form 2ⁿ5^m.