Prove that 2 - 3 5 is an irrational number.

Let us assume on the contrary that 2 - 3 5 is a rational number. Then, there exist co-prime positive integers a and b such that 2 - 3 5 = a b ; b ≠ 0 , where r is a rational number. ⇒ 2 - a b = 3 5 . ⇒ 2 b - a 3 b = 5 . Since r is a rational number, then 2 b - a 3 b is a rational number, Then, 5 is also a rational number. But we know that, 5 is an irrational number. So, our supposition is wrong. Hence, 2 - 3 5 is an irrational number. (proved)

Prove that 2 - 3 5 is an irrational number.

Let us assume on the contrary that 2 - 3 5 is a rational number. Then, there exist co-prime positive integers a and b such that 2 - 3 5 = a b ; b ≠ 0 , where r is a rational number. ⇒ 2 - a b = 3 5 . ⇒ 2 b - a 3 b = 5 . Since r is a rational number, then 2 b - a 3 b is a rational number, Then, 5 is also a rational number. But we know that, 5 is an irrational number. So, our supposition is wrong. Hence, 2 - 3 5 is an irrational number. (proved)

Prove that 5 + 3 is irrational.

Let us assume that 5 + 3 is a rational number. Then, there exist co-prime positive integers a and b such that, 5 + 3 = a b ; b ≠ 0 . Squaring on both sides, 5 + 3 + 2 15 = a b 2 ⇒ 2 15 = a 2 b 2 - 8 ⇒ 15 = a 2 - 8 b 2 2 b 2 Since, a 2 - 8 b 2 2 b 2 is a rational number, then 15 is also a rational number. But, we know that 15 is an irrational number. So, our assumption is wrong. Hence, 5 + 3 is irrational number. (Proved)

Prove that 2 + 3 is an irrational number.

Let us assume on the contrary that 2 + 3 is a rational number. Then, there exist co-prime positive integers a and b such that 2 + 3 = a b ; b ≠ 0 ⇒ a b - 2 = 3 ⇒ a b - 2 2 = ( 3 ) 2 [Squaring both sides] ⇒ a 2 b 2 - 2 a b 2 + 2 = 3 . ⇒ a 2 b 2 - 1 = 2 a b 2 . ⇒ a 2 - b 2 b 2 = 2 a b 2 . ⇒ a 2 - b 2 2 a b = 2 which is a contradiction as the L.H.S. is a rational number while 2 is irrational. Hence, our assumption is wrong. Thus, 2 + 3 is an irrational number. (proved)

Given that 2 is irrational, prove that ( 5 + 3 2 ) is an irrational number.

Let us assume 5 + 3 2 is a rational number. Then, there exist co-prime positive integers a and b such that, 5 + 3 2 = a b ; b ≠ 0 . ⇒ 2 = a - 5 b 3 b As a - 5 b 3 b is a rational number. But, 2 is irrational number. So, our assumption is wrong. Hence, 5 + 3 2 is an irrational number. (Proved)

Prove that for any prime positive integer p , p is an irrational number.

Suppose p is a rational number. Let p = a b , where a , b are integers and b ≠ 0 and both have no common factors. Squaring both sides, we get ⇒ p = a 2 b 2 ⇒ b 2 = a 2 p So, if p | a 2 ⇒ p | a , i.e,. p is a factor of a , say a = p x .....(i) ⇒ b 2 = p 2 x 2 p ⇒ b 2 = p x 2 b 2 p = x 2 Similarly if p | b 2 ⇒ p | b , hence p is a factor of b .......(ii) From (i) and (ii), p is a common factor of both a and b , So, our supposition is wrong. Hence, p is an irrational number, for any prime positive integer. (proved)

If p , q are prime positive integers, prove that p + q is an irrational number.

Let us assume on the contrary that p + q is a rational number. Then, there exist co-prime positive integers a and b such that p + q = a b ; b ≠ 0 ⇒ a b - p = q ⇒ a b - p 2 = ( q ) 2 [Squaring both sides] ⇒ a 2 b 2 - 2 a b p + p = q . ⇒ a 2 b 2 + p - q = 2 a b p . ⇒ a 2 + p - q b 2 b 2 = 2 a b p . ⇒ a 2 - p - q b 2 2 a b = p . which is a contradiction as the L.H.S. is a rational number while p is irrational because p is prime. Hence, our assumption is wrong. Thus, p + q is an irrational number. (proved)

Without actually performing the long division, state whether the rational number 35 50 will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Given rational number is 35 50 Now, expressing denominator as a product of primes, we get: 35 50 = 7 10 = 7 2 1 × 5 1 So, the denominator of 35 50 is of the form 2 m × 5 n , where m , n are non-negative integers. Hence, 35 50 has terminating decimal expansion.

E M B I B E

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.5>Q 8

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Prove that $2 - 3 \sqrt{5}$ is an irrational number.

Important Questions on Real Numbers

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.5>Q 9

Prove that

\sqrt{5} + \sqrt{3}

is irrational.

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.5>Q 10

Prove that

\sqrt{2} + \sqrt{3}

is an irrational number.

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.5>Q 11

Given that

\sqrt{2}

is irrational, prove that

(5 + 3 \sqrt{2})

is an irrational number.

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.5>Q 12

Prove that for any prime positive integer

p, \sqrt{p}

is an irrational number.

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.5>Q 13

If

p, q

are prime positive integers, prove that

\sqrt{p} + \sqrt{q}

is an irrational number.

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.6>Q 1

Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

$\frac{23}{8}$

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.6>Q 1

Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

$\frac{125}{441}$

MEDIUM

10th CBSE

IMPORTANT

MATHEMATICS CLASS X>Chapter 1 - Real Numbers>EXERCISE 1.6>Q 1

Without actually performing the long division, state whether the rational number