Question

Prove that \[3+\sqrt{5}\] is an irrational number.

Answer 1

Hint: To prove that \[3+\sqrt{5}\] is an irrational number, first assume it to be a rational number. Then we contradict this statement by proving it to be an irrational number.

Complete step by step answer:
Let us assume \[3+\sqrt{5}\] is a rational number.
If it is a rational number then it can be expressed in the form of \[\dfrac{p}{q}\] where p and q are integers and \[q\ne 0\].
Therefore:
$\Rightarrow 3+\sqrt{5}=\dfrac{p}{q}$
$\Rightarrow \sqrt{5}=\dfrac{p}{q}-3$
Taking q is the common denominator, numerator becomes \[p-3q\]:
\[\Rightarrow \sqrt{5}=\dfrac{p-3q}{q}\]
\[\Rightarrow \sqrt{5}=\dfrac{p-3q}{q}\]
As both p and q are integers, so \[p-3q\] is also an integer.
As q is not equal to 0, \[\dfrac{p-3q}{q}\] is a rational number.
\[\Rightarrow \sqrt{5}\text{ }=\text{ }\dfrac{p-3q}{q}\text{ }\] is also a rational number.
But this contradicts the fact that \[\sqrt{5}\] is irrational because we can’t write \[\sqrt{5}\] in standard form of a rational number which is $\dfrac{p}{q}$ form. This contradiction arises because of our false assumption that \[3+\sqrt{5}\] is a rational number.
Therefore our assumption was wrong and \[3+\sqrt{5}\] is an irrational number.
Hence proved.

Note: While solving this type of question we have to be careful about perfect squares. If there is a square root of any perfect square number we can write it in form of integer and integer is rational because we can write an integer in standard form of a rational number.
For example if we have $\sqrt{25}$.
As we know 25 is a square of 5 because $5\times 5=25$.
So we can write $\sqrt{25}$ as below
$\Rightarrow \sqrt{25}=5$
$\Rightarrow \sqrt{25}=\dfrac{5}{1}$
Hence $\sqrt{25}$ is a rational number.