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

Last updated date: 20th Jun 2024
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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

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