RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev

The document RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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                                                     Exercise - 1.5
Question: 1
Show that the following numbers are irrational.
(i) 7 √5
(ii) 6 + √2 
(iii) 3 - √5
Solution:
(i) Let us assume that 7 √5 is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
We know that √5 is an irrational number
Here we see that √5 is a rational number which is a contradiction.
(ii) Let us assume that 6+√2 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √2 is a rational number which is a contradiction as we know that √2 is an irrational number
Hence 6 + √2 is an irrational number
(iii) Let us assume that 3 - √5 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √5 is a rational number which is a contradiction as we know that √5 is an irrational number
Hence 3 - √5 is an irrational number.

Question: 2
Prove that the following numbers are irrationals.
(i) 2√7
(ii) 3/(2√5)
(iii) 4 + √2
(iv)  5√2
Solution:
(i) Let us assume that 2√7 is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev

√7 is rational number which is a contradiction
Hence 2√7 is an irrational number

(ii) Let us assume that 3/(2√5) is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
√5 is rational number which is a contradiction
Hence 3/(2√5) is irrational.

(iii) Let us assume that 3/(2√5) is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev

√2 is rational number which is a contradiction
Hence 4+ √2 is irrational.

(iv) Let us assume that 5√2 is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
√2 is rational number which is a contradiction
Hence 5√2 is irrational

Question: 3
Show that 2 - √3 is an irrational number.
Solution:
Let us assume that 2 - √3 is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √3 is a rational number which is a contradiction
Hence 2- √3 is irrational

Question: 4
Show that 3 + √2 is an irrational number.
Solution:
Let us assume that 3 + √2 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √2 is a irrational number which is a contradiction
Hence 3 + √2 is irrational

Question: 5
Prove that 4 - 5√2 is an irrational number.
Solution:
Let us assume that 4 - 5√2 is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
This contradicts the fact that √2 is an irrational number
Hence 4 – 5√2 is irrational

Question: 6
Show that 5 - 2√3 is an irrational number.
Solution:
Let us assume that 5 -2√3 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
This contradicts the fact that √3 is an irrational number
Hence 5 – 2√3 is irrational

Question: 7
Prove that 2√3 - 1 is an irrational number.
Solution:
Let us assume that 2√3 – 1 is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
This contradicts the fact that √3 is an irrational number
Hence 5 - 2√3 is irrational

Question: 8
Prove that 2 - 3√5 is an irrational number.
Solution:
Let us assume that 2 - 3√5 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
This contradicts the fact that √5 is an irrational number
Hence 2 - 3√5 is irrational

Question: 9
Prove that √5 + √3 is irrational.
Solution:
Let us assume that √5 + √3 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √3 is a rational number which is a contradiction as we know that √3 is an irrational number
Hence √5 + √3 is an irrational number

Question: 10
Prove that √3 + √4 is irrational.
Solution:
Let us assume that √3 + √4 is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √3 is a rational number which is a contradiction as we know that √3 is an irrational number
Hence √3 + √4   is an irrational number

Question: 11
Prove that for any prime positive integer p, √p is an irrational number.
Solution:
Let us assume that √p is rational. Then, there exist positive co primes a and b such that

RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
⇒ pb2 = a2
⇒ p|a2 ⇒ p|a2 
⇒ p|a ⇒ p|a
⇒ a = pc for some positive integer c ⇒ a = pc for some positive integer c

b2p = a2
⇒ b2p = p2c2 ( ∵ a = pc)
⇒ p|b2 (since p|c2p) ⇒ p|b2(since p|c2p) 
⇒ p|b ⇒ p|a and p|b
This contradicts the fact that a and b are co primes
Hence √p is irrational

Question: 12
If p, q are prime positive integers, prove that √p + √q is an irrational number.
Solution:
Let us assume that √p + √q is rational. Then, there exist positive co primes a and b such that
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here we see that √q is a rational number which is a contradiction as we know that √q is an irrational number
Hence √p + √q  is an irrational number

                                                     Exercise - 1.6

Question: 1
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
(i)  23/8
(ii) 125/441
(iii) 35/50
 (iv) 77/210
(v) RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Solution:
(i) The given number is 23/8
Here, 8 = 23 and 2 is not a factor of 23.
So, the given number is in its simplest form.
Now, 8 = 23 is of the form 2m × 5n, where m = 3 and n = 0.
So, the given number has a terminating decimal expansion.
(ii) The given number is 125/441
Here, 441 = 32 × 72 and none of 3 and 7 is a factor of 125.
So, the given number is in its simplest form.
Now, 441 = 32 × 72 is not of the form 2m × 5n
So, the given number has a non-terminating repeating decimal expansion.
(iii) The given number is 35/50 and HCF (35, 50) = 5.
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here, 7/10 is in its simplest form.
Now, 10 = 2 × 5 is of the form 2m × 5n, where in = 1 and n = 1.
So, the given number has a terminating decimal expansion.
(iv) The given number is 77/210 and HCF (77, 210) = 7.
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Here, 11/30 is in its simplest form. 30
Now, 30 = 2 × 3 × 5 is not of the form 2m × 5n.
So, the given number has a non-terminating repeating decimal expansion.
(v) The given number is RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRevClearly, none of 2, 5 and 7 is a factor of 129.
So, the given number is in its simplest form.

Question: 2
Write down the decimal expansions of the following rational numbers by writing their denominators in the form of 2m × 5n, where m, and n, are the non- negative integers.
(i)  ⅜
(ii)  13/125
(iii)  7/80
(iv) 14588/625
(v) RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
Solution.
(i) The given number is ⅜
Clearly, 8 = 23 is of the form 2m × 5n, where m = 3 and n = 0.
So, the given number has terminating decimal expansion.
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
(ii) The given number is 13/125
Clearly, 125 = 53 is of the form 2m × 5n, where m = 0 and n = 3.
So, the given number has terminating decimal expansion.
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
(iii) The given number is 7/80.
Clearly, 80 = 24 × 5 is of the form 2m × 5n, where m = 4 and n = 1.
So, the given number has terminating decimal expansion.
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
(iv) The given number is 14588/625
Clearly, 625 = 54 is of the form 2m × 5n, where m = 0 and n = 4.
So, the given number has terminating decimal expansion?
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
(v) The given number is RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRevClearly, 22 × 57 is of the form 2m × 5n, where in = 2 and n = 7.
So, the given number has terminating decimal expansion?
RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev

Question: 3
What can you say about the prime factorization of the denominators of the following rational?
(i) 43.123456789
(ii) RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
(iii) RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRev
(iv) 0.120120012000120000
Solution.
(i) Since 43.123456789 has terminating decimal expansion. So, its denominator is of the form 2m × 5n, where m, n are non-negative integers.
(ii) Since  RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRevhas non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.
(iii) Since RD Sharma Solutions: Exercise 1.5 & 1.6 - Real Numbers Notes | EduRevhas non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.
(iv) Since 0.120120012000120000 … has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.

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