Q. 13.9( 541 Votes )

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Answer :

Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]


√5 = a/b


5 b = a


Squaring both sides,


(5 b)2 = a2


5b2 = a2


a2/5 = b2


Hence, 5 divides a2


By theorem, if p is a prime number and p divides a2, then p divides a, where a is a positive number


So, 5 divides a too


Hence, we can say a/5 = c where, c is some integer


So, a = 5c


Now we know that,


5b2 = a2


Putting a = 5c,


5b2 = (5c)2


5b2 = 25c2


b2 = 5c2


b2/5 = c2


Hence, 5 divides b2


By theorem, if p is a prime number and p divides a2, then p divides a, where a is a positive number


So, 5 divides b too


By earlier deductions, 5 divides both a and b


Hence, 5 is a factor of a and b


a and b are not co-prime.


Hence, the assumption is wrong.


By contradiction,


√5 is irrational

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