Question

By using the method of contradiction prove that √7 is irrational.

Answer 1

Let us assume √7 is a rational number
√7=p/q
squaring both sides we get
7= p^2/q^2
p²=2q². 1
7 is factor of p²
7 is factor of p also
p=7m where m is an integer
p²=49m²
7q²=49m²
q²=7m²
7 is factor of q² also q
so both have 7 as a common factor which contradict the fact that p and q have no common factor.√7 is irrational

Answer 2

Hi Friend ✋

Your Answer is :-


Let us Assume that √7 is a rational number.

Then , √7 = p/q .....................[ Where 'p' and 'q' are coprimes]

Squaring both sides,

7 = p²/q²

p² = 7 × q² ............(1)

q² = p²/7 [ therefore 7 divides p² and 7 will also divide 7 ]


p = 7 × C

From (1)

7 × q² = 49 × C²

C² = q²/7 [ therefore 7 divides q² and 7 will also divide q ]

therefore p and q have more than 2 factors [ i.e 1, 7 and itself]

but it contradicts our assumption that p and q are coprimes.

so √7 ≠ p/q

therefore √7 is rational.


hope it helps