Question

show that root 7 is irrational
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Answer 1

Sol: let us assume that √7 be rational. then it must in the form of p / q [q ≠ 0] [p and q are co-prime] √7 = p / q => √7 x q = p squaring on both sides => 7q2= p2 ------> (1) p2 is divisible by 7 p is divisible by 7 p = 7c [c is a positive integer] [squaring on both sides ] p2 = 49 c2 --------- > (2) subsitute p2 in equ (1) we get 7q2 = 49 c2 q2 = 7c2 => q is divisble by 7 thus q and p have a common factor 7. there is a contradiction as our assumsion p & q are co prime but it has a common factor. so that √7 is an irrational.

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Answer 2

Answer:

Explanation:

(√7)^2 = (a/b)^2

7 = a^2/b^2

Let a = 7c

7b^2= (7c)^2

7b^2 = 49

b = 49/7

b = 7

Therefore, a and b are co-prime

This is due to our wrong assumption