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Real Numbers

Question
CBSEENMA10006108
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Prove that square root of 5 is irrational.

Solution
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So, we can find coprime integers a and b (≠ 0) such that
                 space space space space space space space space space space space space square root of 5 equals straight a over straight b space space space space rightwards double arrow space space space space space space square root of 5 space straight b equals straight a

Squaring on both sides, we get

5b2 = a2

Therefore, 5 divides a2

Therefore, 5 divides a

So, we can write

a = 5c for some integer c.

Substituting for a, we get

5b2 = 25c2

⇒    b2 = 5c2

This means that 5 divides b2, and so 5 divides b.

Therefore, a and b have at least 5 ts a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that square root of 5 is rational.

So, we conclude that square root of 5 is irrational.

Some More Questions From Real Numbers Chapter

Express each number as a product of its prime factors: (iv) 5005 

Express each number as a product of its prime factors:  (v) 7429

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.  (ii) 510 and 92 

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.  (iii) 336 and 54

Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 

Find the LCM and HCF of the following integers by applying the prime factorisation method. (ii) 17, 23 and 29 

Find the LCM and HCF of the following integers by applying the prime factorisation method. (iii) 8, 9 and 25

Check whether 6n can end with the digit 0 for any natural number n

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