Question

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

Answer 1

Hint: We solve his problem by using the contradiction method.
We assume that \[\sqrt{3}\] is a rational number. Then, we use the definition of a rational number to prove that our assumption is wrong.
A number that can be represented in the form \[\dfrac{p}{q}\] where \[p,q\] are integers and co – primes that is they have only 1 common factor as 1 and \[q\ne 0\] is called a rational number.
We use one theorem that if \[a\] is a prime number and \[a\] divides \[{{k}^{2}}\] then \[a\] divides \[k\]

Complete step by step answer:
We are asked to prove that \[\sqrt{3}\] is an irrational number.
Let us use the contradiction method to solve this problem.
Let us assume that \[\sqrt{3}\] is a rational number.
We know that number that can be represented in the form \[\dfrac{p}{q}\] where \[p,q\] are integers and co – primes that is they have only 1 common factor as 1 and \[q\ne 0\] is called a rational number.
By using the above definition of rational number to \[\sqrt{3}\] then we get
\[\Rightarrow \sqrt{3}=\dfrac{p}{q}\]
Now, by squaring on both sides we get
\[\begin{align}
  & \Rightarrow \dfrac{{{p}^{2}}}{{{q}^{2}}}=3 \\
 & \Rightarrow \dfrac{{{p}^{2}}}{3}={{q}^{2}}.........equation(i) \\
\end{align}\]
Here, we can see that 3 divides \[{{p}^{2}}\]
We know that theorem that if \[a\] is a prime number and \[a\] divides \[{{k}^{2}}\] then \[a\] divides \[k\]
By using the above theorem we can say that 3 divides \[p\]
Let us assume that quotient after dividing \[p\] with q then we get
\[\Rightarrow p=3r.........equation(ii)\]
Now, let us substitute equation (ii) in equation (i) then we get
\[\begin{align}
  & \Rightarrow \dfrac{{{\left( 3r \right)}^{2}}}{3}={{q}^{2}} \\
 & \Rightarrow 3{{r}^{2}}={{q}^{2}} \\
 & \Rightarrow \dfrac{{{q}^{2}}}{3}={{r}^{2}} \\
\end{align}\]
Here, we can see that 3 divides \[{{q}^{2}}\]
Now, by using the theorem that if \[a\] is a prime number and \[a\] divides \[{{k}^{2}}\] then \[a\] divides \[k\] we can say that 3 divides \[q\]
Let us assume that the quotient after dividing \[q\] with 3 as
\[\Rightarrow q=3s........equation(iii)\]
Here, from equation (ii) and equation (iii) we can say that 3 is the common factor for \[p,q\]
But from the definition of rational number we know that \[p,q\] are co – primes that have only 1 factor as 1
So, we can say that our assumption is wrong as we get wrong result from our assumption.
Therefore, we can conclude that \[\sqrt{3}\] is an irrational number.

Note:
We can directly conclude the result of this problem.
We have a standard condition that the square root of all prime numbers is an irrational number.
Here, we can see that number 3 is a prime number.
By using the above condition we can directly conclude that \[\sqrt{3}\] is an irrational number.