# Prove that $3 + \sqrt 7 $ is an irrational number.

Verified

147k+ views

**Hint:**Here in this question we must know rational and irrational numbers. Below is a brief mentioning some of the properties which will get used in this question:-

Rational numbers $ \ne $ Irrational numbers

Rational numbers are in the form of $\dfrac{a}{b}$ where a and b are co-prime and b$ \ne $0

**Complete step-by-step answer:**

We have to prove that $3 + \sqrt 7 $ is an irrational number so let us assume that $3 + \sqrt 7 $ is a rational number. Therefore if $3 + \sqrt 7 $ is a rational number then it can be written in the form of $\dfrac{a}{b}$ where a and b are co-prime and b $ \ne $0

Hence it can be written as $3 + \sqrt 7 = \dfrac{a}{b}$

$ \Rightarrow \sqrt 7 = \dfrac{a}{b} - 3$

$ \Rightarrow \sqrt 7 = \dfrac{{a - 3b}}{b}$

As we can see that $\dfrac{{a - 3b}}{b}$ is a rational term and $\sqrt 7 $ is an irrational number which cannot be equal so this contradicts the assumption we have taken in the starting which proves that $3 + \sqrt 7 $ is an irrational number.

**Note:**Some students may find difficulty in understanding some terms being used in solution part like coprime, rational and irrational numbers so here below is a brief explanation of these terms are given: -

Co-prime: -Two integers are said to be coprime if the common number that divides both is 1 only. Some of the examples are 3 and 7, 8 and 13.

Rational number: -A rational number is a number that can be in the form p/q where p and q are integers and q is not equal to zero.

Irrational number: -A real number that cannot be made by dividing two integers is an irrational number also in irrational number decimal value goes on forever without repeating. For example: -

$\pi $ is an irrational number which goes on forever without repeating.