
Prove that $3 + \sqrt 7 $ is an irrational number.
Answer
486.9k+ 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.
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.
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