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Hint: Suppose two numbers, one of the rational form and other as an irrational. Use the contradiction method to prove the statement i.e. suppose sum of both rational and irrational as a rational number that it can be represented in form of $\dfrac{p}{q}$ , where $q\ne 0$ and p, q are the integers.

Complete step-by-step answer:

Let us suppose two numbers ‘a’ and ‘b’ which are rational (a) and irrational (b) respectively.

Now, we know the fundamental definition of a rational number that a rational number can be written in the form of fraction $\dfrac{p}{q}$ , where q will never be 0. And we know that an irrational number cannot be written in the form of $\dfrac{p}{q}$ (Ex: 0.1102345………..)

So, let us use the contradiction method for providing the statement that the sum of a rational number and an irrational number is always irrational.

Hence, the sum of a rational number i.e. ‘a’ (as supposed above) and an irrational number i.e. ‘b’ (as supposed above) will give a rational number. It means we can represent sum of ‘a’ and ‘b’ in terms of fraction $\dfrac{p}{q}$, where q will never be zero and p and q are the integers i.e. both are rational and fraction is in simplified form.

Hence, we can equate $\dfrac{p}{q}$ to a + b and get relation as

$\dfrac{p}{q}=a+b$

Now, subtract ‘a’ from both sides of the equations and hence, we get

$\dfrac{p}{q}-a=a+b-a$

Or

$\dfrac{p}{q}-a=b$………………. (i)

Now, we observe the relation given in equation (i) and get that the left hand side of the equation is representing a rational number and the right hand side of the equation i.e. ‘b’ is an irrational term as supposed in the starting of the solution. So it cannot be possible to equate any rational number to an irrational term. So, we get

$\dfrac{p}{q}-a\ne b$

$\text{rational }\ne \text{ irrational}$

So, it contradicts our assumption that the sum of rational and irrational numbers will be a rational number. Hence, it is proved that the sum of rational and irrational numbers is an irrational number i.e. opposite to our assumption.

Note: One can prove the given statement by taking the examples of rational numbers and irrational numbers as well.

Example: As a sum of $\sqrt{3}=1.7325.........$ and 1 cannot give a rational number, it will give $2.7325..........$ , which is an irrational number. Contradiction method plays an important role with these types of questions. It is applied with a combination of statements, ideas or features which are opposed to one another.

Complete step-by-step answer:

Let us suppose two numbers ‘a’ and ‘b’ which are rational (a) and irrational (b) respectively.

Now, we know the fundamental definition of a rational number that a rational number can be written in the form of fraction $\dfrac{p}{q}$ , where q will never be 0. And we know that an irrational number cannot be written in the form of $\dfrac{p}{q}$ (Ex: 0.1102345………..)

So, let us use the contradiction method for providing the statement that the sum of a rational number and an irrational number is always irrational.

Hence, the sum of a rational number i.e. ‘a’ (as supposed above) and an irrational number i.e. ‘b’ (as supposed above) will give a rational number. It means we can represent sum of ‘a’ and ‘b’ in terms of fraction $\dfrac{p}{q}$, where q will never be zero and p and q are the integers i.e. both are rational and fraction is in simplified form.

Hence, we can equate $\dfrac{p}{q}$ to a + b and get relation as

$\dfrac{p}{q}=a+b$

Now, subtract ‘a’ from both sides of the equations and hence, we get

$\dfrac{p}{q}-a=a+b-a$

Or

$\dfrac{p}{q}-a=b$………………. (i)

Now, we observe the relation given in equation (i) and get that the left hand side of the equation is representing a rational number and the right hand side of the equation i.e. ‘b’ is an irrational term as supposed in the starting of the solution. So it cannot be possible to equate any rational number to an irrational term. So, we get

$\dfrac{p}{q}-a\ne b$

$\text{rational }\ne \text{ irrational}$

So, it contradicts our assumption that the sum of rational and irrational numbers will be a rational number. Hence, it is proved that the sum of rational and irrational numbers is an irrational number i.e. opposite to our assumption.

Note: One can prove the given statement by taking the examples of rational numbers and irrational numbers as well.

Example: As a sum of $\sqrt{3}=1.7325.........$ and 1 cannot give a rational number, it will give $2.7325..........$ , which is an irrational number. Contradiction method plays an important role with these types of questions. It is applied with a combination of statements, ideas or features which are opposed to one another.

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