Question

Prove that the following numbers are irrational:
1/ (3 - √5)

no faltu answers​

Answer 1

Answer

To prove

• 1/ (3 - √5) is irrational number

Proof

Let us assume 1/ (3 - √5) to be a rational number. Thus it could be expressed in the form of p/q where p & q are integers and q ≠ 0

Thus ,

=> $\sf \dfrac {1}{ (3 - \sqrt5) }= \dfrac{p}{q}$

=> $\sf 1 = \dfrac{p}{q} \times (3 - \sqrt5)$

=> $\sf \dfrac{q}{p} = 3 - \sqrt5$

=> $\sf \dfrac{q}{p} - 3 = - \sqrt5$

=> $\sf -\dfrac{q}{p} + 3 = \sqrt5$

=> $\sf 3 - \dfrac{q}{p}= \sqrt5$

=> $\sf \dfrac{3p - q}{p}= \sqrt5$

=> $\sf \sqrt5 = \dfrac{3p - q}{p}$

Here in RHS 3 , p , q are integers i.e the RHS is rational thus LHS need to be rational too i.e √5 is rational but this contradicts the fact that √5 is an irrational number , this contradiction has been arised due to our wrong assumption that 1/ (3 - √5) is a rational number thus 1/ (3 - √5) is an irrational number

Answer 2

To prove

• 1/ (3 - √5) is irrational number

Proof

Let us assume 1/ (3 - √5) to be a rational number. Thus it could be expressed in the form of p/q where p & q are integers and q ≠ 0

Thus ,

=> $\sf \dfrac {1}{ (3 - \sqrt5) }= \dfrac{p}{q}$

=> $\sf 1 = \dfrac{p}{q} \times (3 - \sqrt5)$

=> $\sf \dfrac{q}{p} = 3 - \sqrt5$

=> $\sf \dfrac{q}{p} - 3 = - \sqrt5$

=> $\sf -\dfrac{q}{p} + 3 = \sqrt5$

=> $\sf 3 - \dfrac{q}{p}= \sqrt5$

=> $\sf \dfrac{3p - q}{p}= \sqrt5$

=> $\sf \sqrt5 = \dfrac{3p - q}{p}$

Here in RHS 3 , p , q are integers i.e the RHS is rational thus LHS need to be rational too i.e √5 is rational but this contradicts the fact that √5 is an irrational number , this contradiction has been arised due to our wrong assumption that 1/ (3 - √5) is a rational number thus 1/ (3 - √5) is an irrational number