NCERT Solutions · Class 9 Mathematics The World of Numbers

Put all three numbers over one common denominator, then use that as the size of the small divisions.The three numbers use different denominators, so first make them comparable. Note \(\displaystyle 1\frac{1}{2} = \frac{3}{2}\). The LCM of \(\displaystyle 3\), \(\displaystyle 4\) and \(\displaystyle 2\) is \(\displaystyle 12\), so express each as twelfths:\[\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}, \qquad -\frac{5}{4} = -\frac{5 \times 3}{4 \times 3} = -\frac{15}{12}, \qquad 1\frac{1}{2} = \frac{3}{2} = \frac{3 \times 6}{2 \times 6} = \frac{18}{12}. \]How to draw it.1. Draw a straight line, mark a point \(\displaystyle 0\) on it, and step off equal unit lengths to mark \(\displaystyle -2, -1, 0, 1, 2\). (You need to reach \(\displaystyle -2\) on the left because \(\displaystyle -\frac{5}{4}\) lies beyond \(\displaystyle -1\), and \(\displaystyle 2\) on the right because \(\displaystyle \frac{3}{2}\) lies beyond \(\displaystyle 1\).) 2. Divide every unit interval into \(\displaystyle 12\) equal parts. Each small step is now worth \(\displaystyle \frac{1}{12}\), and all three numbers land exactly on one of these marks — this is the whole point of choosing the common denominator \(\displaystyle 12\). 3. Count the small steps from \(\displaystyle 0\), to the right for a positive number and to the left for a negative one:

Number	As twelfths	Direction from \(\displaystyle 0\)	Where it lands
\(\displaystyle \frac{2}{3}\)	\(\displaystyle \frac{8}{12}\)	\(\displaystyle 8\) steps right	between \(\displaystyle 0\) and \(\displaystyle 1\), two-thirds of the way
\(\displaystyle -\frac{5}{4}\)	\(\displaystyle -\frac{15}{12}\)	\(\displaystyle 15\) steps left	past \(\displaystyle -1\), a quarter of the way on to \(\displaystyle -2\)
\(\displaystyle 1\frac{1}{2}\)	\(\displaystyle \frac{18}{12}\)	\(\displaystyle 18\) steps right	exactly halfway between \(\displaystyle 1\) and \(\displaystyle 2\)

A quicker way to place each one, if you draw them separately. Order on the line (left to right). \[-\frac{5}{4} \;<\; 0 \;<\; \frac{2}{3} \;<\; 1\frac{1}{2}, \] since as twelfths these are \(\displaystyle -15, 0, 8, 18\). In decimals: \(\displaystyle -1.25,\; 0.\overline{6},\; 1.5\) — which confirms the order and the positions above.So on a single line marked in twelfths, \(\displaystyle -\frac{5}{4}\) sits \(\displaystyle 15\) small steps to the left of \(\displaystyle 0\), \(\displaystyle \frac{2}{3}\) sits \(\displaystyle 8\) small steps to the right, and \(\displaystyle 1\frac{1}{2}\) sits \(\displaystyle 18\) small steps to the right (exactly midway between \(\displaystyle 1\) and \(\displaystyle 2\)).

Rewrite both numbers with the same denominator, then the numbers in between become easy to read off.The LCM of \(\displaystyle 2\) and \(\displaystyle 4\) is \(\displaystyle 4\), but a denominator of \(\displaystyle 4\) leaves only \(\displaystyle -\frac{1}{4}\) and \(\displaystyle 0\) strictly in between — enough here, but it is safer to make room. Multiply up to eighths instead:\[-\frac{1}{2} = -\frac{4}{8}, \qquad \frac{1}{4} = \frac{2}{8}. \]Now the question reads: which eighths lie strictly between \(\displaystyle -\frac{4}{8}\) and \(\displaystyle \frac{2}{8}\)? The numerators strictly between \(\displaystyle -4\) and \(\displaystyle 2\) are \(\displaystyle -3, -2, -1, 0, 1\), giving\[-\frac{3}{8},\quad -\frac{2}{8} = -\frac{1}{4},\quad -\frac{1}{8},\quad 0,\quad \frac{1}{8}. \]Choose any three of these, for instance\[-\frac{3}{8}, \qquad -\frac{1}{8}, \qquad \frac{1}{8}. \]Check (convert to decimals and confirm the order). \[-\frac{1}{2} = -0.5 \;<\; -\frac{3}{8} = -0.375 \;<\; -\frac{1}{8} = -0.125 \;<\; \frac{1}{8} = 0.125 \;<\; \frac{1}{4} = 0.25. \] All three lie strictly inside, and all three are distinct.Other answers are equally valid. There are in fact infinitely many rational numbers between any two distinct rational numbers: given any two of them you can always take their average (the midpoint), which lies strictly between, and repeat forever. For example the midpoint of \(\displaystyle -\frac{1}{2}\) and \(\displaystyle \frac{1}{4}\) is \[\frac{1}{2}\left(-\frac{1}{2} + \frac{1}{4}\right) = \frac{1}{2} \times \left(-\frac{1}{4}\right) = -\frac{1}{8}, \] and repeating the trick generates as many as you like. So \(\displaystyle -\frac{1}{4},\; 0,\; \frac{1}{5}\) would be just as correct an answer as the one above.Three such numbers: \(\displaystyle -\dfrac{3}{8},\; -\dfrac{1}{8},\; \dfrac{1}{8}\) (any three rationals strictly between \(\displaystyle -0.5\) and \(\displaystyle 0.25\) are acceptable).

Making the denominators the same. Two fractions can only be added when their pieces are the same size, so the first job is to rewrite both over a common denominator. The LCM of \(\displaystyle 4\) and \(\displaystyle 12\) is \(\displaystyle 12\).\[-\frac{1}{4} \;=\; -\frac{1\times 3}{4\times 3} \;=\; -\frac{3}{12}\](Multiplying top and bottom by the same number does not change a fraction's value — it only cuts each piece into $\displaystyle 3$ smaller ones.)Now the two fractions can be added by adding numerators:\[-\frac{3}{12}+\frac{5}{12} \;=\; \frac{-3+5}{12} \;=\; \frac{2}{12}\]Reduce to lowest terms by dividing numerator and denominator by their HCF, which is \(\displaystyle 2\):\[\frac{2}{12} \;=\; \frac{2\div 2}{12\div 2} \;=\; \frac{1}{6}\]Check with decimals. \(\displaystyle -\tfrac14=-0.25\) and \(\displaystyle \tfrac{5}{12}=0.41\overline{6}\); their sum is \(\displaystyle 0.1\overline{6}=\tfrac16\). \(\displaystyle \checkmark\)Answer: \(\displaystyle \dfrac{1}{6}\).

Division of mixed fractions. "How many lengths of \(\displaystyle 2\frac14\) m fit inside \(\displaystyle 15\frac34\) m?" is a division question: total length \(\displaystyle \div\) length per kurta.First turn both mixed fractions into improper fractions:\[15\frac{3}{4}=\frac{15\times 4+3}{4}=\frac{63}{4}\,\text{m},\qquad 2\frac{1}{4}=\frac{2\times 4+1}{4}=\frac{9}{4}\,\text{m}\]Divide by multiplying by the reciprocal:\[\frac{63}{4}\div\frac{9}{4}=\frac{63}{4}\times\frac{4}{9}=\frac{63}{9}=7\]Why the answer is exact. Both lengths have denominator \(\displaystyle 4\), i.e. both are a whole number of quarter-metres: the tailor has \(\displaystyle 63\) quarter-metres and each kurta uses \(\displaystyle 9\) of them. Since \(\displaystyle 9\times 7=63\) exactly, there is no silk left over — so "exactly how many" has a clean answer with no remainder to worry about.Check. \(\displaystyle 7\times 2\frac14 = 7\times\frac94=\frac{63}{4}=15\frac34\) m. \(\displaystyle \checkmark\)Answer: $\displaystyle 7$ kurtas, using up all the silk exactly (nothing left over).

Insert one more decimal place. The two numbers agree up to the fourth decimal place and differ only there. Write both with one extra decimal place, which does not change their values:\[3.1415 = 3.14150,\qquad 3.1416 = 3.14160\]Now the fifth decimal digit is free to be anything from \(\displaystyle 1\) to \(\displaystyle 9\), and every such choice gives a number squeezed between the two. Pick three of them:\[3.14151,\qquad 3.14152,\qquad 3.14153\]Why these are rational. Each one terminates, so each is a whole number over a power of ten:\[3.14151=\frac{314151}{100000},\quad 3.14152=\frac{314152}{100000},\quad 3.14153=\frac{314153}{100000}\]Check the order. \[3.14150 < 3.14151 < 3.14152 < 3.14153 < 3.14160\ \checkmark\]The choice is not unique — \(\displaystyle 3.14155\), \(\displaystyle 3.141592\), and infinitely many others also work. (Note that \(\displaystyle 3.141592\) is rational even though it looks like \(\displaystyle \pi\): it stops, and \(\displaystyle \pi\) does not.)Answer: for example \(\displaystyle 3.14151,\ 3.14152,\ 3.14153\), i.e. \(\displaystyle \dfrac{314151}{100000},\ \dfrac{314152}{100000},\ \dfrac{314153}{100000}\). Other answers are equally correct.

Yes — several. This question is open: the methods below are all correct, and other correct methods exist. Take two rational numbers \(\displaystyle a<b\).Way $\displaystyle 1$ — the average. The midpoint \[\frac{a+b}{2}\] lies exactly halfway between them, so \(\displaystyle a<\frac{a+b}{2}<b\). It is rational because adding two rationals and dividing by \(\displaystyle 2\) can never leave the rationals.Way $\displaystyle 2$ — weighted points. Instead of halving, cut the gap into three (or any number of) equal parts: \[\frac{2a+b}{3},\qquad \frac{a+2b}{3}\] More generally \(\displaystyle \dfrac{ma+nb}{m+n}\) is rational and lies strictly between \(\displaystyle a\) and \(\displaystyle b\) for any positive integers \(\displaystyle m,n\). This one method produces as many in-between numbers as you like in a single step.Way $\displaystyle 3$ — scale up the common denominator. Write \(\displaystyle a=\frac{p}{q}\) and \(\displaystyle b=\frac{r}{q}\) over the same denominator. If \(\displaystyle p\) and \(\displaystyle r\) are consecutive integers there is no room yet, so multiply top and bottom of both by \(\displaystyle 10\): \[a=\frac{10p}{10q},\qquad b=\frac{10r}{10q}\] Now nine integers sit strictly between \(\displaystyle 10p\) and \(\displaystyle 10r\), giving nine fractions between \(\displaystyle a\) and \(\displaystyle b\).Way $\displaystyle 4$ — decimal insertion. Write both as decimals and change a digit far enough to the right that the number stays inside the gap — the method used in Question 5.Way $\displaystyle 5$ — the mediant. If \(\displaystyle a=\frac{p}{q}\) and \(\displaystyle b=\frac{r}{s}\) with \(\displaystyle q,s>0\), then the "wrong-looking" fraction \[\frac{p+r}{q+s}\] always lies strictly between them. For example between \(\displaystyle \frac14\) and \(\displaystyle \frac{5}{12}\) the mediant is \(\displaystyle \frac{1+5}{4+12}=\frac{6}{16}=\frac38\), and indeed \(\displaystyle 0.25<0.375<0.41\overline{6}\). \(\displaystyle \checkmark\) (Warning: this is a special trick for the mediant only — you may never add fractions this way.)One consequence. Whichever method you choose, it can be applied again to the new, smaller gap, and again, without ever running out. So between any two distinct rational numbers there are infinitely many rational numbers.Answer: yes — the average \(\displaystyle \dfrac{a+b}{2}\), weighted means \(\displaystyle \dfrac{ma+nb}{m+n}\), scaling to a larger common denominator, inserting decimal digits, or the mediant \(\displaystyle \dfrac{p+r}{q+s}\). All are valid, so this answer is not unique.