Question

Prove that: \[4-5\sqrt{3}\] is irrational.

Answer 1

Hint: Assume that \[4-5\sqrt{3}\] is a rational number and apply the method of contradiction to prove that the assumption is wrong. Use the information that difference, multiplication, addition, and division of two rational numbers is always a rational number.

Complete step by step answer:
Here, we have been provided with a number \[4-5\sqrt{3}\] and we have to prove it an irrational number. So, let us apply the method of contradiction for the proof
Now, let us assume that \[4-5\sqrt{3}\] is a rational number. So, it can be written in the form \[\dfrac{a}{b}\], where a and b are co – primes and \[b\ne 0\]. Here, the term co – primes means consecutive primes that says that a and b have no common factor other than 1. In other words, \[4-5\sqrt{3}\] can be written as \[\dfrac{a}{b}\] in simplest fractional form. So, we have,
\[\begin{align}
  & \Rightarrow 4-5\sqrt{3}=\dfrac{a}{b} \\
 & \Rightarrow -5\sqrt{3}=\dfrac{a}{b}-4 \\
\end{align}\]
Taking L.C.M in the R.H.S, we get,
\[\Rightarrow -5\sqrt{3}=\dfrac{a-4b}{b}\]
Dividing both sides with 5 we get,
\[\Rightarrow \sqrt{3}=\dfrac{a-4b}{5b}\]
Multiplying both sides with (-1) we get,
\[\begin{align}
  & \Rightarrow \left( -1 \right)\times \left( -\sqrt{3} \right)=\left( -1 \right)\times \dfrac{a-4b}{5b} \\
 & \Rightarrow \sqrt{3}=\dfrac{4b-a}{5b} \\
\end{align}\]
Now, in the above relation we can clearly see that in the L.H.S. we have \[\sqrt{3}\] which we already know is an irrational number. In the R.H.S we have the expression \[\left( \dfrac{4b-a}{5b} \right)\] in which we can see that in the numerator we have a difference of two rational ‘4b’ and ‘a’ and in the denominator we have a product of two rational ‘5’ and ‘b’. We already know that the sum, difference, product and division of two or more rationals is a rational number. So, the expression in the R.H.S is a rational number. So, according to our assumption have,
\[\Rightarrow \] Irrational = Rational
The above relation is not possible because we know that an irrational number and a rational number can never be equal. Therefore, we have,
\[\Rightarrow \sqrt{3}\ne \dfrac{4b-a}{5b}\]
Hence, our assumption that \[4-5\sqrt{3}\] is rational is wrong. So, a contradiction arises and we can conclude that \[4-5\sqrt{3}\] is an irrational number.
Hence, proved

Note:
One may note that there are other methods also to solve the above question. Here, we have used the result that \[\sqrt{3}\] is irrational but we haven’t proved it because it was not necessary. You may also use the contradiction method to prove \[\sqrt{3}\] an irrational number first then use the information that the product and difference of an irrational and a rational is always an irrational number to solve the question.