Question

Prove that $5 - \sqrt 3 $ is irrational.

Answer 1

Hint: In this question we have to prove that the given number is irrational. Irrational numbers are those which cannot be represented into the form $\dfrac{a}{b}$ where $b \ne 0$. Firstly assume the given number to be a rational number then using the definition of irrational number prove by contradiction that the given number is irrational.

Complete step-by-step answer:

The following proof is a proof by contradiction.

Let us assume $\left( {5 - \sqrt 3 } \right)$ is a rational number.

Where a rational number is a number which is represented in the form of $\dfrac{a}{b}$ ,

where $b \ne 0$ and a and b has not any common factors except 1.

Then it can be represented as a fraction of two integers.

Let the lowest terms representation be:

$\left( {5 - \sqrt 3 } \right) = \dfrac{a}{b}$ , where$b \ne 0$.

$ \Rightarrow \sqrt 3 = - \dfrac{a}{b} + 5 = \dfrac{{ - a + 5b}}{b}$

Now let $ - a + 5b = p$

$ \Rightarrow \sqrt 3 = \dfrac{p}{b}$

Now squaring on both sides we have,

$ \Rightarrow 3 = \dfrac{{{p^2}}}{{{b^2}}}$

$ \Rightarrow {p^2} = 3{b^2}$………………………………. (1)

From above let ${p^2}$ is even therefore p should also be even.

Let $p = 2c$ (c is constant and 2c is an even number)

Squaring both sides we have,

$ \Rightarrow {p^2} = 4{c^2}$………………… (2)

From equation (1) and (2) we can say that

$4{c^2} = 3{b^2}$

Therefore from above $3{b^2}$ is even therefore ${b^2}$ should also be even and again b is even.

Therefore p and b both are even.

Therefore p and b have some common factors.

But p and b were in lowest form and both cannot be even.

Hence, the assumption was wrong and hence, $\left( {5 - \sqrt 3 } \right)$ is an irrational number.

So, $\left( {5 - \sqrt 3 } \right)$ is an irrational number.

Note: Whenever we face such types of problems the key concept is to have the knowledge of the basic definition of rational and irrational numbers, these questions are always proved by the method of contradiction, this concept will help you get on the right track to solve such types of problems.