Give an example of each, of two irrational numbers whose:
Question:

Give an example of each, of two irrational numbers whose:
(i) difference is a rational number.
(ii) difference is an irrational number.
(iii) sum is a rational number.
(iv) sum is an irrational number.
(v) product is an rational number.
(vi) product is an irrational number.
(vii) quotient is a rational number.
(viii) quotient is an irrational number.

Solution:

(i) Let $\sqrt{2}, 1+\sqrt{2}$

And, so $1+\sqrt{2}-\sqrt{2}=1$

Therefore, $\sqrt{2}$ and $1+\sqrt{2}$ are two irrational numbers and their difference is a rational number

(ii) Let $4 \sqrt{3}, 3 \sqrt{3}$ are two irrational numbers and their difference is an irrational number

Because $4 \sqrt{3}-3 \sqrt{3}=\sqrt{3}$ is an irrational number

(iii) Let $\sqrt{5},-\sqrt{5}$ are two irrational numbers and their sum is a rational number

That is $\sqrt{5}+(-\sqrt{5})=0$

(iv) Let $2 \sqrt{5}, 3 \sqrt{5}$ are two irrational numbers and their sum is an irrational number

That is $2 \sqrt{5}+3 \sqrt{5}=5 \sqrt{5}$

(v) Let $\sqrt{8}, \sqrt{2}$ are two irrational numbers and their product is a rational number

That is $\sqrt{8} \times \sqrt{2}=\sqrt{16}=4$

(vi) Let $\sqrt{2}, \sqrt{3}$ are two irrational numbers and their product is an irrational number

That is $\sqrt{2} \times \sqrt{3}=\sqrt{6}$

(vii) Let $\sqrt{8}, \sqrt{2}$ are two irrational numbers and their quotient is a rational number

That is $\frac{\sqrt{8}}{\sqrt{2}}=\frac{2 \sqrt{2}}{\sqrt{2}}=2$

(viii) Let $\sqrt{2}, \sqrt{3}$ are two irrational numbers and their quotient is an irrational number

That is $\sqrt{2} \div \sqrt{3}=\frac{\sqrt{2}}{\sqrt{3}}$

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