Answer
Verified
430.5k+ 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.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE