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Hint: Consider \[\sqrt{3}\] as rational. Prove that they are not coprime i.e. they have common factors other than 1. Take, \[\sqrt{3}=\dfrac{a}{b}\], where a and b are co-prime. Finally, you will get 3 divides \[{{a}^{2}}\]and 3 divides \[{{b}^{2}}\].

“Complete step-by-step answer:”

We have to prove that \[\sqrt{3}\] is irrational.

Let us assume the opposite that \[\sqrt{3}\] is rational.

Hence, \[\sqrt{3}\]can be written in the form \[\dfrac{a}{b}\].

Where a and b\[\left( b\ne 0 \right)\] are co-prime, they have no common factor other than 1.

Hence, \[\sqrt{3}=\dfrac{a}{b}\]

\[\Rightarrow a=\sqrt{3}b\]

Now, squaring on both sides

\[\begin{align}

& {{\left( \sqrt{3}b \right)}^{2}}={{a}^{2}} \\

& 3{{b}^{2}}={{a}^{2}} \\

& \therefore \dfrac{{{a}^{2}}}{3}={{b}^{2}} \\

\end{align}\]

Hence, 3 divides \[{{a}^{2}}\].

By theorem: If p is a prime number, and p divides \[{{a}^{2}}\], then p divides a, where a is a positive number.

\[\therefore \]3 also divides a …………………………….(1)

Hence, we can say,

\[\dfrac{a}{3}=c\], where c is some integer.

\[\therefore a=3c\]

We know, a=3c and \[3{{b}^{2}}={{a}^{2}}\]

Substituting the value of a

\[\begin{align}

& 3{{b}^{2}}={{\left( 3c \right)}^{2}} \\

& \Rightarrow 3{{b}^{2}}=9{{c}^{2}} \\

& \therefore {{b}^{2}}=3{{c}^{2}} \\

& \dfrac{{{b}^{2}}}{3}={{c}^{2}} \\

\end{align}\]

Hence, 3 divides \[{{b}^{2}}\].

By theorem: If p is a prime number and p divides \[{{b}^{2}}\], then p divides b, where b is a positive number.

\[\therefore \]3 also divides b …………………………...(2)

By (1) and (2) \[\Rightarrow 3\] divides both a & b.

Hence, 3 is a factor of 3.

\[\therefore \] a and b are not coprime as they have a factor 3 other than 1.

\[\therefore \] Hence, our assumptions are wrong.

\[\therefore \]\[\sqrt{3}\] is irrational.

Note: By proving that 3 divides both \[{{a}^{2}}\]and \[{{b}^{2}}\], we come to the conclusion that they are divisible by 3 other than 1. So, they are co-prime.

“Complete step-by-step answer:”

We have to prove that \[\sqrt{3}\] is irrational.

Let us assume the opposite that \[\sqrt{3}\] is rational.

Hence, \[\sqrt{3}\]can be written in the form \[\dfrac{a}{b}\].

Where a and b\[\left( b\ne 0 \right)\] are co-prime, they have no common factor other than 1.

Hence, \[\sqrt{3}=\dfrac{a}{b}\]

\[\Rightarrow a=\sqrt{3}b\]

Now, squaring on both sides

\[\begin{align}

& {{\left( \sqrt{3}b \right)}^{2}}={{a}^{2}} \\

& 3{{b}^{2}}={{a}^{2}} \\

& \therefore \dfrac{{{a}^{2}}}{3}={{b}^{2}} \\

\end{align}\]

Hence, 3 divides \[{{a}^{2}}\].

By theorem: If p is a prime number, and p divides \[{{a}^{2}}\], then p divides a, where a is a positive number.

\[\therefore \]3 also divides a …………………………….(1)

Hence, we can say,

\[\dfrac{a}{3}=c\], where c is some integer.

\[\therefore a=3c\]

We know, a=3c and \[3{{b}^{2}}={{a}^{2}}\]

Substituting the value of a

\[\begin{align}

& 3{{b}^{2}}={{\left( 3c \right)}^{2}} \\

& \Rightarrow 3{{b}^{2}}=9{{c}^{2}} \\

& \therefore {{b}^{2}}=3{{c}^{2}} \\

& \dfrac{{{b}^{2}}}{3}={{c}^{2}} \\

\end{align}\]

Hence, 3 divides \[{{b}^{2}}\].

By theorem: If p is a prime number and p divides \[{{b}^{2}}\], then p divides b, where b is a positive number.

\[\therefore \]3 also divides b …………………………...(2)

By (1) and (2) \[\Rightarrow 3\] divides both a & b.

Hence, 3 is a factor of 3.

\[\therefore \] a and b are not coprime as they have a factor 3 other than 1.

\[\therefore \] Hence, our assumptions are wrong.

\[\therefore \]\[\sqrt{3}\] is irrational.

Note: By proving that 3 divides both \[{{a}^{2}}\]and \[{{b}^{2}}\], we come to the conclusion that they are divisible by 3 other than 1. So, they are co-prime.

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