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Find the number of distinct rational numbers x such that $0

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Hint: To solve this question, we must know that a rational number is expressed in the form of \[\dfrac{p}{q}\] . Also, to achieve the condition $0
Complete step-by-step answer:
Now, we start with the solution. We will be taking a fixed value of p and accordingly we will change the values of q in a particular case. So, let us start forming the cases:
Case I: - When the value of p is 1, the q can have values 2,3,4,5,6. The rational numbers formed in this case are: - \[\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6}\] . Total number of rational numbers \[=\] 5
Case II: - When the value of \[p=2\], the q can have values 3,4,5,6. The rational numbers formed in this case are: - \[\dfrac{2}{3},\dfrac{2}{4},\dfrac{2}{5},\dfrac{2}{6}\] . Total number of rational numbers obtained are equal to 4.
 Case III: - When the value of \[p=3\] , the q can have values 4,5,6. The rational numbers formed in this case are given as: - \[\dfrac{3}{4},\dfrac{3}{5},\dfrac{3}{6}\] . Total number of rational numbers obtained in this case is equal to 3.
Case IV: - When the value of \[p=4\], the q can contain values 5 and 6. The rational numbers formed in this case are: - \[\dfrac{4}{5},\dfrac{4}{6}\] . The total number of rational numbers obtained in this case equals to 2.
Case V: - When the value of \[p=5\] , the q can contain only 6. The rational number formed in the case is only \[\dfrac{5}{6}\] . Only one rational number is obtained in this case.
Now, we have to add the cases but we cannot add them because there are some similar elements. We will have to eliminate them first. The similar elements are: - ( \[\dfrac{1}{2}\] and \[\dfrac{2}{4}\] and \[\dfrac{3}{6}\] ) , ( \[\dfrac{1}{3}\] and \[\dfrac{2}{6}\] ), ( \[\dfrac{2}{3}\] and \[\dfrac{4}{6}\] ).Thus we have to eliminate four rational numbers ( \[\dfrac{2}{4}\], \[\dfrac{3}{6}\], \[\dfrac{2}{6}\] and \[\dfrac{4}{6}\] ) from total.
Thus total rational numbers formed are given by
\[=\] (Case I + Case II + Case III + Case IV + Case V) \[-4\]
\[=\left( 5+4+3+2+1 \right)-4\]
\[=15-4\]
\[=11\]

Note: We have formed cases according to the value of p. It is not necessary to form cases according to the value of p and changing the values of q in these cases. We could have done the opposite of this by forming cases according to the values of q. In both the cases, the final answer will be the same.

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