Question

Prove that \[2 - 3\sqrt 5 \] is an irrational number.

Answer 1

Hint:First of all, consider the given statement as the opposite. Then show that this statement is contradictory and therefore prove that our assumption is wrong and the given statement is correct. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
We have to prove that \[2 - 3\sqrt 5 \] is irrational and we know that \[\sqrt 5 \] is an irrational number.
Let us assume the opposite, i.e., \[2 - 3\sqrt 5 \] is rational
Hence, \[2 - 3\sqrt 5 \] can be written in the form \[\dfrac{a}{b}\] where \[a{\text{ and }}b\] are co-prime and \[b \ne 0\]
Thus, we have
\[
  2 - 3\sqrt 5 = \dfrac{a}{b} \\
  3\sqrt 5 = \dfrac{a}{b} + 2 = \dfrac{{a + 2b}}{b} \\
  \sqrt 5 = \dfrac{{a + 2b}}{{3b}} \\
\]
Here clearly \[\sqrt 5 \] is an irrational number and \[\dfrac{{a + 2b}}{{3b}}\] is a rational number.
Since rational \[ \ne \] irrational
This is a contradiction.
Therefore, our assumption is incorrect.
Hence, \[2 - 3\sqrt 5 \] is an irrational number.

Note: An irrational number is a number that cannot be expressed as a fraction for any integers and irrational numbers have decimal expansion that neither terminate nor become periodic. Remember that \[\sqrt 5 \] is an irrational number.