Question

Prove that \[\sqrt 7 \]​ is an irrational number.

Answer 1

Hint: Here we will first take an assumption that \[\sqrt 7 \]​ is an rational number and then write it in the form of \[\dfrac{x}{y}\] and then square both sides and then apply the divisibility rule.
An irrational number is a number which non terminating as well as non -recurring and cannot be expressed in the form of \[\dfrac{x}{y}\]

Complete step-by-step answer:
Let us assume that \[\sqrt 7 \]​ is rational. Then, there exist coprime positive integers x and y such that
\[\sqrt 7 = \dfrac{x}{y},y \ne 0\]
\[ \Rightarrow x = \sqrt 7 y\]
Now, Squaring on both sides, we get:-
\[{\left( x \right)^2} = {\left( {\sqrt 7 y} \right)^2}\]
Simplifying it further we get:-
\[ \Rightarrow {x^2} = {\left( {\sqrt 7 } \right)^2}\left( {{y^2}} \right)\]
\[ \Rightarrow {x^2} = 7{y^2}................\left( 1 \right)\]
Therefore, this implies that \[{x^2}\] is divisible by 7 which further implies that, x is also divisible by 7
So, we can write\[x = 7p\], for some integer p.
Now we will substitute the value of x in equation 1.
Substituting the value of x we get:-
\[{\left( {7p} \right)^2} = 7{y^2}\]
\[ \Rightarrow 49{p^2} = 7{y^2}\]
\[ \Rightarrow {y^2} = 7{p^2}\]
This means, \[{y^2}\] is also divisible by 7 and therefore this implies that, y is also divisible by 7.
Therefore, x and y have at least one common factor, i.e., 7.
But, this contradicts the fact that x and y are co-prime.
Thus, our assumption is wrong.
Hence, \[\sqrt 7 \] is irrational.
Hence proved.

Note: Students should keep in mind that the only numbers which can be expressed in the form of \[\dfrac{a}{b}\] where a and b are co-prime numbers which means they have only 1 as their common factor and \[b \ne 0\] rest all are irrationals.
We can also check that \[\sqrt 7 \] is irrational by calculating its value which comes out be non- non-terminating as well as non -repeating.