Answer
Verified
424.2k+ views
Hint: In this question we have to comment upon whether the number is a rational number or not. A rational number is one which can be represented in the form of $\dfrac{a}{b}$ where $b \ne 0$. So use the concept of contradiction by proof by initially taking the given number to be a rational number. This will help you get the right solution.
Complete step-by-step answer:
The following proof is a proof by contradiction.
Let us assume $\left( {\sqrt 6 + \sqrt 9 } \right)$ is a rational number.
Where a rational number is a number which is represented in the form of $\dfrac{a}{b}$ , where $b \ne 0$ and a and b has not any common factors except 1.
Then it can be represented as a fraction of two integers.
Let the lowest terms representation be:
$\left( {\sqrt 6 + \sqrt 9 } \right) = \dfrac{a}{b}$ , where$b \ne 0$.
Now as we know$\sqrt 9 = 3$.
So substitute this value in above equation we have,
$\left( {\sqrt 6 + 3} \right) = \dfrac{a}{b}$
$ \Rightarrow \sqrt 6 = \dfrac{a}{b} - 3 = \dfrac{{a - 3b}}{b}$
Now let $a - 3b = p$
$ \Rightarrow \sqrt 6 = \dfrac{p}{b}$
Now squaring on both sides we have,
$ \Rightarrow 6 = \dfrac{{{p^2}}}{{{b^2}}}$
$ \Rightarrow {p^2} = 6{b^2}$………………………………. (1)
From above ${p^2}$ is even therefore p should also be even.
Let $p = 2c$ (c is constant and 2c is an even number)
Squaring both sides we have,
$ \Rightarrow {p^2} = 4{c^2}$………………… (2)
From equation (1) and (2) we can say that
$4{c^2} = 6{b^2}$
Divide by 2 in above equation we have,
$ \Rightarrow 2{c^2} = 3{b^2}$
Therefore from above $3{b^2}$ is even therefore ${b^2}$ should also be even and again b is even.
Therefore p and b both are even.
Therefore p and b have some common factors.
But p and b were in lowest form and both cannot be even.
Hence, the assumption was wrong and hence, $\left( {\sqrt 6 + \sqrt 9 } \right)$ is an irrational number.
So, $\left( {\sqrt 6 + \sqrt 9 } \right)$ is an irrational number.
Note: Whenever we face such types of problems the key concept involved is the method of proof by contradiction. We have to assume that the given number is a rational number then carry forward with the simplification part by using the basic definition of rational number. In the end if the assumption is wrong then we can say that the number was irrational or else rational only.
Complete step-by-step answer:
The following proof is a proof by contradiction.
Let us assume $\left( {\sqrt 6 + \sqrt 9 } \right)$ is a rational number.
Where a rational number is a number which is represented in the form of $\dfrac{a}{b}$ , where $b \ne 0$ and a and b has not any common factors except 1.
Then it can be represented as a fraction of two integers.
Let the lowest terms representation be:
$\left( {\sqrt 6 + \sqrt 9 } \right) = \dfrac{a}{b}$ , where$b \ne 0$.
Now as we know$\sqrt 9 = 3$.
So substitute this value in above equation we have,
$\left( {\sqrt 6 + 3} \right) = \dfrac{a}{b}$
$ \Rightarrow \sqrt 6 = \dfrac{a}{b} - 3 = \dfrac{{a - 3b}}{b}$
Now let $a - 3b = p$
$ \Rightarrow \sqrt 6 = \dfrac{p}{b}$
Now squaring on both sides we have,
$ \Rightarrow 6 = \dfrac{{{p^2}}}{{{b^2}}}$
$ \Rightarrow {p^2} = 6{b^2}$………………………………. (1)
From above ${p^2}$ is even therefore p should also be even.
Let $p = 2c$ (c is constant and 2c is an even number)
Squaring both sides we have,
$ \Rightarrow {p^2} = 4{c^2}$………………… (2)
From equation (1) and (2) we can say that
$4{c^2} = 6{b^2}$
Divide by 2 in above equation we have,
$ \Rightarrow 2{c^2} = 3{b^2}$
Therefore from above $3{b^2}$ is even therefore ${b^2}$ should also be even and again b is even.
Therefore p and b both are even.
Therefore p and b have some common factors.
But p and b were in lowest form and both cannot be even.
Hence, the assumption was wrong and hence, $\left( {\sqrt 6 + \sqrt 9 } \right)$ is an irrational number.
So, $\left( {\sqrt 6 + \sqrt 9 } \right)$ is an irrational number.
Note: Whenever we face such types of problems the key concept involved is the method of proof by contradiction. We have to assume that the given number is a rational number then carry forward with the simplification part by using the basic definition of rational number. In the end if the assumption is wrong then we can say that the number was irrational or else rational only.
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE