
How do you prove that square root $ 15 $ is irrational?
Answer
446.4k+ views
Hint: Here, using contradiction method we will assume contradictory to the requirement and then will prove accordingly as per requirement. So, here we will assume that square root $ 15 $ is rational.
Complete step-by-step solution:
A rational number is the number which can be expressed as the ratio of two numbers or which can be expressed as the p/q form or as the quotient or the fraction with non-zero denominator whereas, the numbers which are not represented as the rational are known as the irrational number.
Convert the given word statement in the form of mathematical expression. Also, we will assume that square root $ 15 $ is rational.
Let $ \sqrt {15} $ is rational.
$ \therefore \sqrt {15} = \dfrac{a}{b} $ [Here “a” and “b” are not the common factor.]
Take square on both the side of the equation.
\[\therefore {\left( {\sqrt {15} } \right)^2} = {\left( {\dfrac{a}{b}} \right)^2}\]
Square and square root cancel each other on the left hand side of the equation.
\[\therefore 15 = \left( {\dfrac{{{a^2}}}{{{b^2}}}} \right)\]
Do cross multiplication, where the denominator of one side is multiplied with the numerator of the opposite side.
$ \Rightarrow 15{b^2} = {a^2} $
The left hand side of the equation has factors of $ 3 $ and $ 5. $ So $ {a^2} $ must be divisible by $ 3 $ and $ 5. $ Also, by the unique prime factorization theorem, “a” must be also divisible by $ 3 $ and $ 5. $
So, assume $ a = 3.5.k $ where $ k \in N $
$ \Rightarrow 15{b^2} = {a^2} = {(15k)^2} = 15.(15{k^2}) $
Divide both the ends of the equation with $ 15 $ to find:
$ \Rightarrow {b^2} = (15{k^2}) $
Make constant term the subject-
$ \Rightarrow \dfrac{{{b^2}}}{{{k^2}}} = 15 $
Take the square root on both the sides.
$ \Rightarrow \sqrt {\dfrac{{{b^2}}}{{{k^2}}}} = \sqrt {15} $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow \dfrac{b}{k} = \sqrt {15} $
The above expression is in the form of an irrational number.
So, our assumption is not correct and hence $ \sqrt {15} $ is the irrational number.
Note: In irrational numbers are in the decimal form and are the non-repeating and non-terminating numbers. Remember zero is the rational number. Also, refer to other terminologies for natural numbers, whole numbers and integers, fractions and know the difference between them.
Complete step-by-step solution:
A rational number is the number which can be expressed as the ratio of two numbers or which can be expressed as the p/q form or as the quotient or the fraction with non-zero denominator whereas, the numbers which are not represented as the rational are known as the irrational number.
Convert the given word statement in the form of mathematical expression. Also, we will assume that square root $ 15 $ is rational.
Let $ \sqrt {15} $ is rational.
$ \therefore \sqrt {15} = \dfrac{a}{b} $ [Here “a” and “b” are not the common factor.]
Take square on both the side of the equation.
\[\therefore {\left( {\sqrt {15} } \right)^2} = {\left( {\dfrac{a}{b}} \right)^2}\]
Square and square root cancel each other on the left hand side of the equation.
\[\therefore 15 = \left( {\dfrac{{{a^2}}}{{{b^2}}}} \right)\]
Do cross multiplication, where the denominator of one side is multiplied with the numerator of the opposite side.
$ \Rightarrow 15{b^2} = {a^2} $
The left hand side of the equation has factors of $ 3 $ and $ 5. $ So $ {a^2} $ must be divisible by $ 3 $ and $ 5. $ Also, by the unique prime factorization theorem, “a” must be also divisible by $ 3 $ and $ 5. $
So, assume $ a = 3.5.k $ where $ k \in N $
$ \Rightarrow 15{b^2} = {a^2} = {(15k)^2} = 15.(15{k^2}) $
Divide both the ends of the equation with $ 15 $ to find:
$ \Rightarrow {b^2} = (15{k^2}) $
Make constant term the subject-
$ \Rightarrow \dfrac{{{b^2}}}{{{k^2}}} = 15 $
Take the square root on both the sides.
$ \Rightarrow \sqrt {\dfrac{{{b^2}}}{{{k^2}}}} = \sqrt {15} $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow \dfrac{b}{k} = \sqrt {15} $
The above expression is in the form of an irrational number.
So, our assumption is not correct and hence $ \sqrt {15} $ is the irrational number.
Note: In irrational numbers are in the decimal form and are the non-repeating and non-terminating numbers. Remember zero is the rational number. Also, refer to other terminologies for natural numbers, whole numbers and integers, fractions and know the difference between them.
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