Question

Prove that \[7 + \sqrt 2 \] is irrational.

Answer 1

Hint:
We will check whether 7 and \[\sqrt 2 \] are rational numbers individually. We will try to express both of them as rational numbers. We will get a contradiction for one of them. We will use the fact that the sum of a rational and irrational number is irrational. We will prove the result using this.

Complete step by step solution:
We know that a number is a rational number if it can be expressed in the form \[\dfrac{p}{q}\] where \[p\] and \[q\] are integers with no common factor and \[q \ne 0\].
We can express 7 as \[\dfrac{7}{1}\] where 7 and 1 are integers and \[1 \ne 0\]. 7 is a rational number.
We will assume that \[\sqrt 2 \] is a rational number. We can express \[\sqrt 2 \] as \[\sqrt 2 = \dfrac{p}{q}\] where \[p\] and \[q\] are integers in their lowest form and \[q \ne 0\].
We will square both the sides:
\[\begin{array}{l} \Rightarrow {\rm{ }}2 = \dfrac{{{p^2}}}{{{q^2}}}\\{\rm{ }}2{q^2} = {p^2}{\rm{ }}\left( 1 \right)\\ \Rightarrow \sqrt 2 q = p\end{array}\]
We can see from the above equation that \[{p^2}\] is even because it is represented as a product of 2 with another number. If \[{p^2}\] is even, \[p\] must be even as well. We can express \[p\] as multiple of 2 with another number, say \[m\].
\[p = 2m\]
We will square both the sides:
\[\Rightarrow {p^2} = 4{m^2}{\rm{ }}\left( 2 \right)\]
We will compare equation (1) and equation (2):
\[\begin{array}{l} \Rightarrow {\rm{ }}2{q^2} = 4{m^2}\\ \Rightarrow {q^2} = 2{m^2}\end{array}\]
\[{q^2}\] is even because it is represented as a product of 2 with another number. If \[{q^2}\] is even, \[q\] must be even as well.
Both \[p\] and \[q\] are even. Both \[p\] and \[q\] have 2 as a common factor. This is a contradiction because we had assumed that \[p\] and \[q\] had no common factor. So, our assumption is wrong. \[\sqrt 2 \] is an irrational number.
We know that the sum of a rational and irrational number is irrational.

\[7 + \sqrt 2 \] is irrational.

Note:
We can also prove this using an alternate method.
Let us assume that \[7 + \sqrt 2 \] is rational. We can express \[7 + \sqrt 2 \] as
\[7 + \sqrt 2 = \dfrac{p}{q}\] where \[p\]and \[q\] are integers with no common factor and \[q \ne 0\].
We will subtract 7 from both sides, we get
\[\begin{array}{l} \Rightarrow 7 + \sqrt 2 - 7 = \dfrac{p}{q} - 7\\ \Rightarrow \sqrt 2 = \dfrac{p}{q} - \dfrac{7}{1}\end{array}\]
We will find the lowest common multiple (L.C.M.) of \[q\] and 1.
\[\Rightarrow \sqrt 2 = \dfrac{{p - 7q}}{q}\]
We have got an irrational number on the L.H.S. and a rational number of the R.H.S. This is not possible; we have arrived at a contradiction. So, our assumption is wrong.
 \[7 + \sqrt 2 \] is irrational.