Question

Write down ten positive rational numbers such that the numerator–denominator for each of them is -2. Write them in increasing order.

Answer 1

Hint: To find the type of rational numbers given in question, we will write a rational number of the form such that \(P{\rm{ - q = - 2}}\) where \(\dfrac{P}{q}\) is any rational number.

Complete step-by-step answer:

Before solving the question, we must first know what a rational number is. A rational number is a number that can be expressed as the quotient or fraction \(\dfrac{P}{q}\) of two numbers, a numerator P and a non-zero denominator q. In the question, it is given that numerator–denominator = -2 i.e. \(P{\rm{ - q = - 2}}\) . Thus, we can also write this as: \(q{\rm{ = P + 2}}\). So the rational number in our case becomes of the form\[ = {\rm{ }}\dfrac{P}{q}{\rm{ = }}\dfrac{P}{{P{\rm{ + 2}}}}\]. Now, we will put different values of P in the above form and find the ten different rational numbers. So first let us put \(P{\rm{ = 1}}{\rm{.}}\)Thus the rational number we get \(\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{1}{{1{\rm{ + 2}}}}{\rm{ = }}\dfrac{1}{3}{\rm{ = 0}}{\rm{.333}}\).
Now, we will put \(P{\rm{ = 2}}{\rm{.}}\)In this case, the rational number we will obtain
\(\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{2}{{2{\rm{ + 2}}}}{\rm{ = }}\dfrac{2}{4}{\rm{ = }}\dfrac{1}{2}{\rm{ = 0}}{\rm{.5}}\). Now, we will put \(P{\rm{ = 3,}}\)the rational number in this case will be\(\dfrac{3}{{3{\rm{ + 2}}}}{\rm{ = }}\dfrac{3}{5}{\rm{ = 0}}{\rm{.6}}\). Similarly:
When we put \(P{\rm{ = 4,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{4}{{{\rm{4 + 2}}}}{\rm{ = }}\dfrac{4}{6}{\rm{ = }}\dfrac{2}{3}{\rm{ = 0}}{\rm{.66}}\)
When we put \(P{\rm{ = 5,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{5}{{{\rm{5 + 2}}}}{\rm{ = }}\dfrac{5}{7}{\rm{ = 0}}{\rm{.71}}\)
When we put \(P{\rm{ = 6,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{6}{{{\rm{6 + 2}}}}{\rm{ = }}\dfrac{6}{8}{\rm{ = }}\dfrac{3}{4}{\rm{ = 0}}{\rm{.75}}\)
When we put \(P{\rm{ = 7,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{7}{{{\rm{7 + 2}}}}{\rm{ = }}\dfrac{7}{9}{\rm{ = 0}}{\rm{.78}}\)
When we put \(P{\rm{ = 8,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{8}{{{\rm{8 + 2}}}}{\rm{ = }}\dfrac{8}{{10}}{\rm{ = }}\dfrac{4}{5}{\rm{ = 0}}{\rm{.80}}\)
When we put \(P{\rm{ = 9,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{9}{{{\rm{9 + 2}}}}{\rm{ = }}\dfrac{9}{{11}}{\rm{ = 0}}{\rm{.81}}\)
When we put \(P{\rm{ = 10,}}\)we will get \( = {\rm{ }}\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{{10}}{{{\rm{10 + 2}}}}{\rm{ = }}\dfrac{{10}}{{12}}{\rm{ = }}\dfrac{5}{6}{\rm{ = 0}}{\rm{.83}}\)
Now we have to write them in increasing order. This means that the smallest number should be written first and the largest at last. Thus, we get:
\(\dfrac{1}{3}{\rm{, }}\dfrac{2}{4},{\rm{ }}\dfrac{3}{5},{\rm{ }}\dfrac{4}{6},{\rm{ }}\dfrac{5}{7},{\rm{ }}\dfrac{6}{8},{\rm{ }}\dfrac{7}{9},{\rm{ }}\dfrac{8}{{10}},{\rm{ }}\dfrac{9}{{11}},{\rm{ }}\dfrac{{10}}{{12}}\)

Note: We can put any integer in place of P except -2. When\(P{\rm{ = - 2}}\), the rational number becomes:
\(\dfrac{P}{{P{\rm{ + 2}}}}{\rm{ = }}\dfrac{{ - 2}}{{ - 2{\rm{ + 2}}}}{\rm{ = }}\dfrac{{ - 2}}{0}\)
Thus, we got \( - \infty \)as the number.