Question

Find the five rational numbers between \[1\] and \[2\]?

Answer 1

Hint: To find the rational numbers between two whole numbers, we first need to form fractions of the two whole numbers by dividing each of them by any common number with the two numbers as \[p\] and \[q\] and the denominator as \[2\left( p+q \right)\] with \[2\left( p+q \right) =1\] and \[r=2\left( p+q \right)\] , the rational numbers to be found between \[\dfrac{p}{r}\] and \[\dfrac{q}{r}\]as \[p\times \dfrac{r}{r},\dfrac{r+1}{r},\dfrac{r+2}{r},......,\dfrac{r+n}{r},q\times \dfrac{r}{r}\] with \[n\] as the total number of rational number between \[p\] and \[q\].

Complete step-by-step answer:
First let us take a random number \[r\], let us take it as \[r=2\left( p+q \right)\].
The number for the denominator should be greater than 1 but should not exceed the value of \[2\].
Now if the number between the two numbers are taken as \[p\times \dfrac{r}{r},\dfrac{r+1}{r},\dfrac{r+2}{r},......,\dfrac{r+n}{r},q\times \dfrac{r}{r}\]. Then let us put the values in the fraction as:
  \[p=1\], \[q=2\] and \[r=2\left( p+q \right)\]
Before that let us find the value of \[r\], which is \[r=2\times \left( 1+2 \right)\]
 \[\Rightarrow 2\times \left( 1+2 \right)\]
 \[\Rightarrow 6\]
Placing the values in the given number system below as:
 \[\Rightarrow 1\times \dfrac{6}{6},\dfrac{6+1}{6},\dfrac{6+2}{6},\dfrac{6+3}{6},\dfrac{6+4}{6},\dfrac{6+5}{6},2\times \dfrac{6}{6}\]
Adding numerators with \[1,2,....,5\] and dividing with \[6\], we get the middle fractions as:
 \[\Rightarrow \dfrac{7}{6},\dfrac{8}{6},\dfrac{9}{6},\dfrac{10}{6},\dfrac{11}{6}\]
Now, the number between the whole numbers \[1\] and \[2\] are \[\dfrac{7}{6},\dfrac{8}{6},\dfrac{9}{6},\dfrac{10}{6},\dfrac{11}{6}\]. Hence, these fractional numbers are the rational number between \[1\] and \[2\].
So, the correct answer is “\[\dfrac{7}{6},\dfrac{8}{6},\dfrac{9}{6},\dfrac{10}{6},\dfrac{11}{6}\].”.

Note: Another method to find the rational numbers between the two whole numbers ( \[p,q\]) is \[\dfrac{p+q}{2}\]. Now with every rational number the value of \[q\] changes by \[\dfrac{p+q}{2}\] of the previous rational number.
Taking the above formula, we find the rational numbers as \[p\] and \[\dfrac{p+q}{2}\] with \[p=1\] and \[q=2\].
Now the first rational number with \[p=1\] and \[q=2\] is \[\dfrac{1+2}{2}=\dfrac{3}{2}\].

The second rational number with \[p=1\] and new \[q\] as \[\dfrac{3}{2}\] is \[\dfrac{1+\dfrac{3}{2}}{2}=\dfrac{5}{4}\].

The third rational number with \[p=1\] and new \[q\] as \[\dfrac{5}{4}\] is \[\dfrac{1+\dfrac{5}{4}}{2}=\dfrac{9}{8}\].

The fourth rational number with \[p=1\] and new \[q\] as \[\dfrac{9}{8}\] is \[\dfrac{1+\dfrac{5}{4}}{2}=\dfrac{17}{16}\].

The fifth rational number with \[p=1\] and new \[q\] as \[\dfrac{9}{8}\] is \[\dfrac{1+\dfrac{17}{16}}{2}=\dfrac{33}{32}\].

Hence, the rational numbers are \[\dfrac{3}{2},\dfrac{5}{4},\dfrac{9}{8},\dfrac{17}{16},\dfrac{33}{32}\].
Another fact is that if there are no options given then the value of rational numbers for each student can be different as there are infinite numbers of rational numbers between any two numbers hence, all the values are valid until and unless they are between the two natural numbers given in the question.