Answer

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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.

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