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Prove that \[\sqrt{3}\] is irrational.

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Last updated date: 13th Jul 2024
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Answer
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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.