Question

Prove that \[\sqrt{5}\] is irrational by the method of contradiction.

Answer 1

Hint: To solve this question, we will first of all assume \[\sqrt{5}\] to be a rational number. Using the definition of rational number \[\sqrt{5}\] can be written as the form of \[\dfrac{p}{q}\] where \[q\ne 0\] and p and q are integers which are coprime. Using that we will arrive at a contradiction of getting a common factor of p and q part from 1.

Complete step-by-step answer:
We have to prove that the number \[\sqrt{5}\] is an irrational number. And that is to be done using the contradiction method. So on the contrary, let us assume that \[\sqrt{5}\] be a rational number. Any rational number can be written in the form of \[\dfrac{p}{q}\] where \[q\ne 0\] and p and q are integers and both are coprime. \[\sqrt{5}\] being rational can be written as in the form of \[\dfrac{p}{q}\] where \[q\ne 0\] and p and q are coprime.
\[\Rightarrow \sqrt{5}=\dfrac{p}{q}\]
\[\Rightarrow \sqrt{5}\times q=p\]
Squaring both the sides, we have,
\[\Rightarrow {{\left( \sqrt{5}\times q \right)}^{2}}={{p}^{2}}\]
\[\Rightarrow 5{{q}^{2}}={{p}^{2}}.....\left( i \right)\]
\[{{p}^{2}}\] is divisible by 5 as \[{{q}^{2}}=\dfrac{{{p}^{2}}}{5}\] and q is an integer of the form, \[{{q}^{2}}=\dfrac{{{p}^{2}}}{5}=\text{integer}\text{.}\]
p is also divisible by 5. p can be written as 5 times some number. Let some number be ‘c’.
\[\Rightarrow p=5c\]
Squaring both the sides, we have,
\[\Rightarrow {{p}^{2}}={{\left( 5c \right)}^{2}}\]
\[\Rightarrow {{p}^{2}}=25{{c}^{2}}.....\left( ii \right)\]
Comparing equation (i) and (ii), we get,
\[\Rightarrow {{p}^{2}}=25{{c}^{2}}=5{{q}^{2}}\]
\[\Rightarrow 25{{c}^{2}}=5{{q}^{2}}\]
\[\Rightarrow \dfrac{25}{5}{{c}^{2}}={{q}^{2}}\]
\[\Rightarrow 5{{c}^{2}}={{q}^{2}}\]
q is divisible by 5.
Hence, we see that p and q are both divisible by 5. But this is not possible as p and q are assumed to be of the form \[\sqrt{5}=\dfrac{p}{q}\] where \[\sqrt{5}\] is rational and p and q are coprime.
Hence, this is a contradiction. Therefore, we have assumed is wrong and \[\sqrt{5}\] is an irrational number.

Note: Let us define the coprime number. Coprime numbers are the numbers which have their gcd (greatest common divisor) as 1 means if p and q are coprime. \[\Rightarrow \gcd \left( p,q \right)=1.\] So, they cannot have a common factor apart from 1. The procedure for solving this question must be followed as such and no steps must be missed in between.