Question

Prove that \[\sqrt 2 + \sqrt 3 \] is an irrational number?

Answer 1

Hint: We are given a number and we have to prove that it is an irrational number. We will try to do this by considering it a rational number and use the definition of rational number and then come to a conclusion where we are proved wrong. In this way we will try to prove it is an irrational number.
Formula used: We have used following formula here,
A rational number can be expressed as the ratio of two integers such that denominator is not zero as \[\dfrac{p}{q}(q \ne 0),p,q \in \mathbb{Z}\], here \[p,q\] are integers.
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\](whole square formula of sum of two numbers).

Complete step-by-step solution:
We are given the number \[\sqrt 2 + \sqrt 3 \], and we have to prove it is an irrational number.
Let us consider the given number as a rational number. We know that a rational number is of the form \[\dfrac{p}{q}(q \ne 0),p,q \in \mathbb{Z}\]. So, we let \[\sqrt 2 + \sqrt 3 \] equal to \[\dfrac{p}{q}\] where \[\dfrac{p}{q}\] follows all the condition of rationality and \[p,q\] are co-prime as well, so that they cannot be divided further. So,
\[\sqrt 2 + \sqrt 3 = \dfrac{p}{q}\]
Now on squaring both sides,
\[ \Rightarrow {\left( {\sqrt 2 + \sqrt 3 } \right)^2} = {\left( {\dfrac{p}{q}} \right)^2}\]
We know that \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], using this above we get,
\[ \Rightarrow {\sqrt 2 ^2} + {\sqrt 3 ^2} + 2 \times \sqrt 3 \times \sqrt 2 = \left( {\dfrac{{{p^2}}}{{{q^2}}}} \right) \\
   \Rightarrow 2 + 3 + 2\sqrt 6 = \dfrac{{{p^2}}}{{{q^2}}} \\
   \Rightarrow 6 + 2\sqrt 6 = \dfrac{{{p^2}}}{{{q^2}}} \\
   \Rightarrow 2\sqrt 6 = \dfrac{{{p^2}}}{{{q^2}}} - 6 \]
Since \[p,q\] are integers, \[\dfrac{{{p^2}}}{{{q^2}}}\] is rational and also \[\dfrac{{{p^2}}}{{{q^2}}} - 6\] is rational but we know that \[2\sqrt 6 \] is irrational. LHS is irrational while RHS is rational. But a rational number cannot be equal to an irrational number. Hence this contradicts our assumption that \[\sqrt 2 + \sqrt 3 \] is rational. So, we say that \[\sqrt 2 + \sqrt 3 \] is an irrational number. Hence proved.

Note: We can also prove this by proving the given terms \[\sqrt 2 \] and \[\sqrt 3 \] as irrational. After doing this we know that the sum of two irrational numbers is an irrational number. Using this we can say that the sum of two irrational numbers \[\sqrt 2 \] and \[\sqrt 3 \] i.e. \[\sqrt 2 + \sqrt 3 \] is an irrational number.