Prove that √n is not a rational number, if n is not a perfect square.
Answers
Answer:
☞If n is not a perfect square then is irrational
☞Let on the contrary say it is rational .
Then,
where p and q are coprime integers.
so, n =p2/q2
☞p2 =nq2
☞This shows p divides q which is a contradiction.
☞Hence it is irrational if n is not a perfect square.
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Suppose √n is a rational number.
√n = p/q ( p and q have no common factor co-primes, and q is not equal to zero)
☆ Squaring both sides ,we get
⇒ n = p²/q²
⇒ p² = nq² ...........( 1 )
⇒ n divides p²
⇒ n divides p .............( 2 )
Now,
Let p = nm
☆ Squaring both sides ,we get
⇒ p² = n²m²
Substituting the value of p² in ( 1 ) ,we get
n²m² = nq²
⇒ q² = n²m²/n
⇒ q² = nm².
⇒ n divides q²
⇒ n divides q ............( 3 )
From ( 2 ), n divides p and from ( 3 ), n divides q. It means n is a factor of both p and q. This contradicts the assumption that p and q have no common factor. So our supposition is wrong.
Hence, √5 cannot be a rational number.