Question

prove that root3-root7 is irrational​

Answer 1

Answer:

Step-by-step explanation:

Let it be assumed that (√3-√7) is rational.

Then (√3-√7)=a where a is rational

√3=a+√7

=> (√3)² = (a+√7)²

[Squaring both sides]

=> 3 = a²+2√7a+7

=> 2√7 = 3-a²-7

=> √7 = -a²-4/2

But, in the LHS side we get an irrational number i.e. √7 whereas the RHS contains a rational number. This occurred due to the incorrect assumption of √3-√7 as rational.

Hence, √3-√7 is irrational. [Proved]