Question

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

Answer 1

Hint: Try to recall the definition of rational and irrational numbers. You need to use the method of contradiction to prove that $\sqrt{3}+\sqrt{5}$ is an irrational number.

Complete step-by-step solution -
Before moving to the options, let us talk about the definitions of rational numbers followed by irrational numbers.
So, rational numbers are those real numbers that can be written in the form of $\dfrac{p}{q}$ such that both p and q are integers and $q0$ . In other words, we can say that the numbers which are either terminating or recurring when converted to decimal form are called rational numbers. All the integers fall under this category.
Now, moving to irrational numbers.
Those real numbers which are non-terminating and non-recurring are termed as irrational numbers.
The roots of the numbers which are not perfect squares fall under the category of irrational numbers. $\pi \text{ and }e$ are also the standard examples of irrational numbers.
Now let us move to the solution to the above question.
We take $\sqrt{3}+\sqrt{5}$ to be a rational number. Therefore, using the definition of rational number we can say that:
$\sqrt{3}+\sqrt{5}=\dfrac{p}{q}$
Where p and q both are integers. Now if we square both sides of the equation, we get
${{\left( \sqrt{3}+\sqrt{5} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}}$
Now if we use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we get
$3+5+2\sqrt{15}={{\left( \dfrac{p}{q} \right)}^{2}}$
$\Rightarrow 2\sqrt{15}={{\left( \dfrac{p}{q} \right)}^{2}}-8$
$\Rightarrow \sqrt{15}=\dfrac{{{\left( \dfrac{p}{q} \right)}^{2}}-8}{2}$
Now if we see, the right-hand side has all rational terms, and until and unless the denominator is zero, the operation between rational numbers yields a rational number. Therefore, the right-hand side of our equation is rational. But as $\sqrt{15}$ is irrational, so the left-hand side of the equation is irrational. So, the equation is not true.
So, it contradicts our assumption that $\sqrt{3}+\sqrt{5}$ is a rational number. So, $\sqrt{3}+\sqrt{5}$ is irrational.

Note: Remember the property that any operation between two or more rational numbers yields a rational number, provided the denominator is not becoming equal to zero.