Answer
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Hint: We will think of any 2 irrational numbers, we will find their sum and see if it is a rational number. We will think of any 2 irrational numbers, we will find their product and see if it is a rational number.
Formula used: \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
Complete step-by-step answer:
We will look at some properties of rational and irrational numbers that will help us in solving the question:
We know that a number is a rational number if it can be expressed in the form \[\dfrac{p}{q}\] where \[p\]and \[q\] are integers with no common factor and \[q \ne 0\].
We know that a number is an irrational number if it cannot be expressed in the form \[\dfrac{p}{q}\] where \[p\]and \[q\] are integers with no common factor and \[q \ne 0\].
We know that the sum of a rational and an irrational number is always an irrational number.
We know that the difference of a rational and an irrational number is always an irrational number.
We know that \[\sqrt 7 \] is an irrational number.
We can conclude from the 3rd property that \[2 + \sqrt 7 \] is an irrational number.
We can conclude from the 4th property that \[2 - \sqrt 7 \] is an irrational number.
We will take the first number as \[2 + \sqrt 7 \] and the second number as \[2 - \sqrt 7 \].
We will find the sum of the 2 numbers:
$( {2 + \sqrt 7 } + ( {2 - \sqrt 7 }$
=$ 2 + \sqrt 7 + 2 - \sqrt 7 $
=$ {2 + 2}$+${\sqrt 7 - \sqrt 7 }$
= 4
4 is a rational number.
We will find the product of the 2 numbers. We will substitute 2 for \[a\] and \[\sqrt 7 \] for \[b\]in the formula:
$(2 + \sqrt 7)$.$(2 - \sqrt 7)$
=${2^2} - ( {\sqrt 7 }$
= 4 - 7
= - 3
\[ - 3\] is a rational number.
\[\therefore 2 + \sqrt 7 \] and \[2 - \sqrt 7 \] are 2 irrational numbers whose sum as well as the product are rational numbers.
Note: We must know that the square roots of all non-negative integers except the perfect squares are irrational numbers. This will help us in choosing the correct irrational number. For example, \[\sqrt 2 ,\sqrt 3 ,\sqrt 5 ,\sqrt 7 ,\]etc are irrational numbers whereas \[\sqrt 9 ,\sqrt {16} ,\sqrt {25} ,\sqrt {49} ,\] etc are rational numbers.
Formula used: \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
Complete step-by-step answer:
We will look at some properties of rational and irrational numbers that will help us in solving the question:
We know that a number is a rational number if it can be expressed in the form \[\dfrac{p}{q}\] where \[p\]and \[q\] are integers with no common factor and \[q \ne 0\].
We know that a number is an irrational number if it cannot be expressed in the form \[\dfrac{p}{q}\] where \[p\]and \[q\] are integers with no common factor and \[q \ne 0\].
We know that the sum of a rational and an irrational number is always an irrational number.
We know that the difference of a rational and an irrational number is always an irrational number.
We know that \[\sqrt 7 \] is an irrational number.
We can conclude from the 3rd property that \[2 + \sqrt 7 \] is an irrational number.
We can conclude from the 4th property that \[2 - \sqrt 7 \] is an irrational number.
We will take the first number as \[2 + \sqrt 7 \] and the second number as \[2 - \sqrt 7 \].
We will find the sum of the 2 numbers:
$( {2 + \sqrt 7 } + ( {2 - \sqrt 7 }$
=$ 2 + \sqrt 7 + 2 - \sqrt 7 $
=$ {2 + 2}$+${\sqrt 7 - \sqrt 7 }$
= 4
4 is a rational number.
We will find the product of the 2 numbers. We will substitute 2 for \[a\] and \[\sqrt 7 \] for \[b\]in the formula:
$(2 + \sqrt 7)$.$(2 - \sqrt 7)$
=${2^2} - ( {\sqrt 7 }$
= 4 - 7
= - 3
\[ - 3\] is a rational number.
\[\therefore 2 + \sqrt 7 \] and \[2 - \sqrt 7 \] are 2 irrational numbers whose sum as well as the product are rational numbers.
Note: We must know that the square roots of all non-negative integers except the perfect squares are irrational numbers. This will help us in choosing the correct irrational number. For example, \[\sqrt 2 ,\sqrt 3 ,\sqrt 5 ,\sqrt 7 ,\]etc are irrational numbers whereas \[\sqrt 9 ,\sqrt {16} ,\sqrt {25} ,\sqrt {49} ,\] etc are rational numbers.
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