
Give two irrational numbers so that their.
(i)Sum is an irrational number.
(ii)Sum is not an irrational number.
(iii)Difference is an irrational number.
(iv)Difference is not an irrational number.
(v)Product is an irrational number.
(vi)Product is not an irrational number.
(vii)Quotient is an irrational number.
(viii)Quotient is not an irrational number.
Answer
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Hint: Here we will consider two irrational numbers. We will give sum and difference and product and quotient of that number. Another solution we will give is also each one is not an irrational number. Here numbers are also taken manually.
Complete step-by-step answer:
(i)Consider the two-irrational number $2 + \sqrt 3 $ and $\sqrt 3 - 2$ .
Their sum $ = 2 + \sqrt 3 + \sqrt 3 - 2 = 2\sqrt 3 $ is an irrational number.
(ii)Consider the two irrational numbers $\sqrt 2 $ and $ - \sqrt 2 $ .
Their sum $ = \sqrt 2 + ( - \sqrt 2 ) = 0$ is a rational number.
(iii)Consider the two irrational numbers $\sqrt 3 $ and $ - \sqrt 2 $ .
Their difference $ = \sqrt 3 - \sqrt 2 $ is an irrational number.
(iv)Consider the two irrational numbers $5 + \sqrt 3 $ and $\sqrt 3 - 5$ .
Their difference $(5 + \sqrt 3 ) - (\sqrt 3 - 5) = 10$ is a rational number.
(v)Consider the two irrational numbers $\sqrt 3 $ and $\sqrt 5 $ .
Their product $ = \sqrt 3 \times \sqrt 5 = \sqrt {15} $ is an irrational number.
(vi)Consider the two irrational numbers $\sqrt {18} $ and $\sqrt 2 $ .
Their product $ = \sqrt {18} \times \sqrt 2 = \sqrt {36} = 6$ is a rational number.
(vii)Consider the two irrational numbers $\sqrt {15} $ and $\sqrt 3 $ .
Their quotient $ = \dfrac{{\sqrt {15} }}{{\sqrt 3 }} = \sqrt {\dfrac{{15}}{3} = \sqrt 5 } $ is an irrational number.
(viii)Consider the two irrational numbers $\sqrt {75} $ and $\sqrt 3 $ .
Their quotient $ = \dfrac{{\sqrt {75} }}{{\sqrt 3 }} = \sqrt {\dfrac{{75}}{3} = 5} $ is a rational number.
Additional information:
a number that can be expressed exactly by a ratio of two integers. All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.
Note: Zero is a Rational Number. As such, if the numerator is zero $(0)$ , and the denominator is any non-zero integer, the resulting quotient is itself zero.
Complete step-by-step answer:
(i)Consider the two-irrational number $2 + \sqrt 3 $ and $\sqrt 3 - 2$ .
Their sum $ = 2 + \sqrt 3 + \sqrt 3 - 2 = 2\sqrt 3 $ is an irrational number.
(ii)Consider the two irrational numbers $\sqrt 2 $ and $ - \sqrt 2 $ .
Their sum $ = \sqrt 2 + ( - \sqrt 2 ) = 0$ is a rational number.
(iii)Consider the two irrational numbers $\sqrt 3 $ and $ - \sqrt 2 $ .
Their difference $ = \sqrt 3 - \sqrt 2 $ is an irrational number.
(iv)Consider the two irrational numbers $5 + \sqrt 3 $ and $\sqrt 3 - 5$ .
Their difference $(5 + \sqrt 3 ) - (\sqrt 3 - 5) = 10$ is a rational number.
(v)Consider the two irrational numbers $\sqrt 3 $ and $\sqrt 5 $ .
Their product $ = \sqrt 3 \times \sqrt 5 = \sqrt {15} $ is an irrational number.
(vi)Consider the two irrational numbers $\sqrt {18} $ and $\sqrt 2 $ .
Their product $ = \sqrt {18} \times \sqrt 2 = \sqrt {36} = 6$ is a rational number.
(vii)Consider the two irrational numbers $\sqrt {15} $ and $\sqrt 3 $ .
Their quotient $ = \dfrac{{\sqrt {15} }}{{\sqrt 3 }} = \sqrt {\dfrac{{15}}{3} = \sqrt 5 } $ is an irrational number.
(viii)Consider the two irrational numbers $\sqrt {75} $ and $\sqrt 3 $ .
Their quotient $ = \dfrac{{\sqrt {75} }}{{\sqrt 3 }} = \sqrt {\dfrac{{75}}{3} = 5} $ is a rational number.
Additional information:
a number that can be expressed exactly by a ratio of two integers. All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.
Note: Zero is a Rational Number. As such, if the numerator is zero $(0)$ , and the denominator is any non-zero integer, the resulting quotient is itself zero.
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