Question

Which of the following is not irrational? (a) (3+sqrt(7)) (b) (3-sqrt(7)) (c) (3+sqrt(7))(3-sqrt(7)) (d) 3sqrt(7)

Answer 1

Answer:

  Correct answer is option (c); $(3+\sqrt{7} )$ $(3-\sqrt{7} )$. It is a rational number and it is not irrational.

Explanation:

• The real numbers which can be represented in the form of ratio of two integers, as $\frac{p}{q}$, where 'q' is not equal to zero are called rational numbers.

• The real numbers which cannot be expressed in the form of ratio of two integers are called as irrational numbers.

Step 1:

Consider option (a);

• We want to check wheather the number $(3+\sqrt{7})$ is rational number or irrational number.

• If we assume that $(3+\sqrt{7})$ is a rational number. So that we can write $(3+\sqrt{7})$ = $\frac{p}{q}$, where p and q are coprime integers and q is not equal to 0.

if  $(3+\sqrt{7})$ = $\frac{p}{q}$, then $\sqrt{7}$ = $\frac{p}{q}-3$

⇒ $\sqrt{7}$ = $\frac{p-3q}{q}$

• Since p and q are integers then,  $\frac{p-3q}{q}$ will be a rational number. So $\sqrt{7}$ will be a rational number. But, this contradicts the fact that the number $\sqrt{7}$ is an irrational one. This contradiction arises because of our wrong asssemption, i.e., $(3+\sqrt{7} )$ is a rational number. Hence, our assemption is wrong and the number  $(3+\sqrt{7} )$ is an irrational number.

Similarly, consider option (b);

• We want to check wheather the number  $(3-\sqrt{7} )$ is rational number or irrational number.

• By following the same steps as earlier we can write;

$\sqrt{7}$ = $3-\frac{p}{q}$ and $\sqrt{7}$ = $\frac{3q-p}{q}$

• Since, the number $\sqrt{7}$ is an irrational one, the number $(3-\sqrt{7} )$  is an irrational number.

Step 2:

Consider option (c);

• We want to check wheather the number $(3+\sqrt{7} )$ $(3-\sqrt{7} )$ is rational number or irrational number.

• First we want to simplify the number by using the identity;

(a+b) (a-b) = $a^{2} -b^{2}$, and then want to check whether the number is rational or not.

• By using the identity;

$(3+\sqrt{7} )$ $(3-\sqrt{7} )$ = $3^{2}$ - $(\sqrt{7} )^{2}$

⇒ 9 - 7

⇒ 2

• The number 2 can be written in the form of $\frac{p}{q}$. That is $\frac{2}{1}$.

• Hence the number $(3+\sqrt{7} )$ $(3-\sqrt{7} )$ is a rational number and it is not irrational.

Step 3:

Consider option (d);

• We want to check wheather the number $(3\sqrt{7})$ is rational number or irrational number.

• Consider that the number $(3\sqrt{7})$ is rational, then

$(3\sqrt{7})$ = $\frac{p}{q}$

• Rearranging the equation, we get; $\sqrt{7}$ = $\frac{p}{3q}$.

• Since 3, p, q are integers it can be written in $\frac{p}{q}$ format and hence the number will be rational since $\sqrt{7}$ is rational.

But in actual case the number $\sqrt{7}$ is an irrational number. That is our assemption is contradiction to the real fact.

• Since $\sqrt{7}$ is irrational, also the number $(3\sqrt{7})$ is an irrational number.

• Therefore, the answer is option (c); $(3+\sqrt{7} )$ $(3-\sqrt{7} )$. It is a rational number and it is not irrational.

To know more, go through the links;

https://brainly.in/question/39249959

https://brainly.in/question/33551114

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