Question

Prove that the number \[\sqrt{5}-\sqrt{3}\] is not a rational number.

Answer 1

Hint: First of all consider \[\sqrt{5}-\sqrt{3}\] as rational number and equate it to \[\dfrac{p}{q}\text{ where }q\ne 0\]. Now, square both sides of this equation and finally find the value of \[\sqrt{15}\] in terms of p and q. As LHS is irrational, RHS must be irrational which won’t be the case. From this, prove the desired result.

Complete step-by-step solution -
In this question, we have to prove that \[\sqrt{5}-\sqrt{3}\] is not a rational number. Let us assume that \[\sqrt{5}-\sqrt{3}\] is a rational number of the form \[\dfrac{p}{q}\] where p and q are integers and \[q\ne 0\]. So, we get,
\[\sqrt{5}-\sqrt{3}=\dfrac{p}{q}\]
By squaring both the sides of the above equation, we get,
\[{{\left( \sqrt{5}-\sqrt{3} \right)}^{2}}=\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. By using this, we get,
\[{{\left( \sqrt{5} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}-2\left( \sqrt{5} \right)\left( \sqrt{3} \right)=\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
\[\Rightarrow 5+3-2\sqrt{15}=\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
\[\Rightarrow 8-2\sqrt{15}=\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
\[\Rightarrow 2\sqrt{15}=8-\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
By dividing both the equations by 2, we get,
\[\Rightarrow \sqrt{15}=\dfrac{8}{2}-\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
\[\Rightarrow \sqrt{15}=4-\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
\[\Rightarrow \sqrt{15}=\dfrac{4{{q}^{2}}-{{p}^{2}}}{{{q}^{2}}}....\left( i \right)\]
We know that any number of the form \[\sqrt{a}\] is an irrational number if ‘a’ is not a perfect square. So, we get, \[\sqrt{15}\] as an irrational number as 15 is not a perfect square.
Now, in equation (i), since we get LHS that is \[\sqrt{15}\] is an irrational number. So, RHS must also be an irrational number. But as we know that p and q are integers and \[q\ne 0\]. So, \[\dfrac{4{{q}^{2}}-{{p}^{2}}}{{{q}^{2}}}\] would be a rational number and we know that irrational is not equal to rational. So, our assumption was wrong and \[\sqrt{5}-\sqrt{3}\] is not a rational number. In fact, it is an irrational number.
Hence Proved.

Note: In this type of question, first of all, always assume the number to be rational and then prove that the assumption is wrong. We can also solve this question as we know that \[\sqrt{5}\text{ and }\sqrt{3}\] are irrational numbers and we know that the addition and subtraction of irrational numbers give irrational numbers. So, we can say that \[\sqrt{5}-\sqrt{3}\] is an irrational number.