Question

Prove that following statement is incorrect:
$6+\sqrt{2}$ is a rational number.

Answer 1

Hint: To solve this question we need to have a knowledge of numbers like rational and irrational. The question is solved by a contradiction method. The first step will be to assume $6+\sqrt{2}$ as a rational number by equating it to a different rational number.

Complete step by step answer:
The question asks us whether $6+\sqrt{2}$ is a rational number or not . So, in the first step we will take an assumption. The number in this question will be solved using the contradiction method which means we will firstly consider the number to be a rational number and then prove it to be wrong. So, the assumption that will be to consider is $6+\sqrt{2}$as a rational number.
We will write the number $6+\sqrt{2}$ as the fraction which is $\dfrac{a}{b}$. We will equate the number with the fraction. On considering this we will get:
$\Rightarrow 6+\sqrt{2}=\dfrac{a}{b}$
On rearranging the above equation in such a way that the expression in Left Hand Side and Right Hand Side contradict each other, making the assumption to be incorrect. On rearranging with the help of cross multiplication we get:
$\Rightarrow b\left( 6+\sqrt{2} \right)=a$
We will be applying the commutative rule in the above expression which says $p\left( x+y \right)=px+py$. On applying the rule we get:
$\Rightarrow 6b+b\sqrt{2}=a$
On further rearrangement we get:
$\Rightarrow b\sqrt{2}=a-6b$
$\Rightarrow \sqrt{2}=\dfrac{\left( a-6b \right)}{b}$
Now on considering the expression given above we see that in the Left Hand Side we have $\sqrt{2}$ which is an irrational number while in the Right Hand Side we have $\dfrac{\left( a-6b \right)}{b}$, which is a rational number, as it consist of all rational numbers like $a,b$ and $6$.
The rational number cannot be equal to the irrational number which is $\sqrt{2}$, so this means the assumption we have made is incorrect. So the number $6+\sqrt{2}$ is an irrational number.
$\therefore $ The statement, $6+\sqrt{2}$ is a rational number is incorrect.

Note: Do remember that the sum or product of a rational and irrational number is always an irrational number, for example $6+\sqrt{2}$ and $5\sqrt{3}$. We can use the same method to prove the other irrational numbers like $\sqrt{7},\sqrt[3]{3},\sqrt{5}$ etc.