Question

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

Answer 1

Hint: To prove this we need to know the definition of rational and irrational numbers. We know that rational numbers can be written as a ratio of two integers, but some numbers cannot be expressed as a ratio of two integers, that is we call them irrational numbers. First we consider \[2 + \sqrt 3 \] is a rational number then we prove that it is false. Then we can say that \[2 + \sqrt 3 \] is an irrational number.

Complete step by step answer:
Let's consider \[2 + \sqrt 3 \]. We know that \[\sqrt 3 \] is an irrational number because it cannot be expressed as a ratio of two integers. Now let us assume that \[2 + \sqrt 3 \] is a rational number. Then by the definition of rational number there must be two integer’s $p$ and $q$ such that
\[ \Rightarrow 2 + \sqrt 3 = \dfrac{p}{q}\] , where \[q \ne 0\].
Now subtracting two on both sides we have
\[ \Rightarrow \sqrt 3 = \dfrac{p}{q} - 2\]
Taking LCM and simplifying we have
\[ \Rightarrow \sqrt 3 = \dfrac{{\left( {p - 2q} \right)}}{q}\]
That is \[\sqrt 3 \] is expressed as a ratio of two integers. Then \[\sqrt 3 \] is a rational number. Which contradicts the fact that \[\sqrt 3 \] is an irrational number. So, our assumption \[2 + \sqrt 3 \] is a rational number is false.

Hence \[2 + \sqrt 3 \] is an irrational number and it is proved.

Note: Rational number includes numbers, which are finite or are recurring in nature. For example \[\dfrac{1}{2}\], 3.7676. Irrational numbers consist of numbers, which are non-terminating and non-repeating in nature. For example \[\sqrt 2 ,\sqrt 3 \] etc… Also \[\pi \] is a famous irrational number. We can not write down a simple fraction that is equal to \[\pi \].