NCERT Solutions · Class 9 Mathematics The World of Numbers

Look at the prime factors of the denominator. Reduce the fraction to lowest terms first, then factorise the denominator.A decimal is just a fraction whose denominator is a power of ten, and \(\displaystyle 10=2\times 5\). So a fraction in lowest terms can be rewritten with a denominator \(\displaystyle 10^n\) — and therefore terminates — exactly when its denominator is built only from the primes \(\displaystyle 2\) and \(\displaystyle 5\). If any other prime (\(\displaystyle 3,7,11,13,\dots\)) survives in the denominator, no amount of multiplying can turn it into a power of ten, the long division never gives remainder \(\displaystyle 0\), and the decimal repeats forever.Predictions.

Fraction	Denominator in primes	Prediction
\(\displaystyle \dfrac{7}{20}\)	\(\displaystyle 20=2^2\times 5\)	terminating
\(\displaystyle \dfrac{4}{15}\)	\(\displaystyle 15=3\times 5\) (a $\displaystyle 3$ is present)	repeating
\(\displaystyle \dfrac{13}{250}\)	\(\displaystyle 250=2\times 5^3\)	terminating

(All three are already in lowest terms: \(\displaystyle \gcd(7,20)=1\), \(\displaystyle \gcd(4,15)=1\), and \(\displaystyle 13\) is prime and does not divide \(\displaystyle 250\).)Now the long divisions.\(\displaystyle \dfrac{7}{20}\): \(\displaystyle 20\) into \(\displaystyle 70\) goes \(\displaystyle 3\) times (\(\displaystyle 60\)), remainder \(\displaystyle 10\); \(\displaystyle 20\) into \(\displaystyle 100\) goes \(\displaystyle 5\) times (\(\displaystyle 100\)), remainder \(\displaystyle 0\) — stop. \[\frac{7}{20}=0.35\] Shortcut check: \(\displaystyle \frac{7}{20}=\frac{7\times 5}{20\times 5}=\frac{35}{100}=0.35\). \(\displaystyle \checkmark\)\(\displaystyle \dfrac{4}{15}\): \(\displaystyle 15\) into \(\displaystyle 40\) goes \(\displaystyle 2\) (\(\displaystyle 30\)), remainder \(\displaystyle 10\); \(\displaystyle 15\) into \(\displaystyle 100\) goes \(\displaystyle 6\) (\(\displaystyle 90\)), remainder \(\displaystyle 10\). The remainder \(\displaystyle 10\) has come back, so the digit \(\displaystyle 6\) must repeat for ever. \[\frac{4}{15}=0.2666\ldots=0.2\overline{6}\]\(\displaystyle \dfrac{13}{250}\): \(\displaystyle 250\) into \(\displaystyle 130\) goes \(\displaystyle 0\), remainder \(\displaystyle 130\); \(\displaystyle 250\) into \(\displaystyle 1300\) goes \(\displaystyle 5\) (\(\displaystyle 1250\)), remainder \(\displaystyle 50\); \(\displaystyle 250\) into \(\displaystyle 500\) goes \(\displaystyle 2\), remainder \(\displaystyle 0\) — stop. \[\frac{13}{250}=0.052\] Shortcut check: \(\displaystyle \frac{13}{250}=\frac{13\times 4}{250\times 4}=\frac{52}{1000}=0.052\). \(\displaystyle \checkmark\)Every prediction matched the division.Answer: \(\displaystyle \dfrac{7}{20}=0.35\) (terminating), \(\displaystyle \dfrac{4}{15}=0.2\overline{6}\) (non-terminating repeating), \(\displaystyle \dfrac{13}{250}=0.052\) (terminating).

Long division, watching the remainders. The remainders are what tell you when the block closes.Dividing \(\displaystyle 1\) by \(\displaystyle 13\):

Bring down	Digit of quotient	New remainder
\(\displaystyle 10\)	\(\displaystyle 0\)	\(\displaystyle 10\)
\(\displaystyle 100\)	\(\displaystyle 7\), since \(\displaystyle 7\times 13=91\)	\(\displaystyle 9\)
\(\displaystyle 90\)	\(\displaystyle 6\), since \(\displaystyle 6\times 13=78\)	\(\displaystyle 12\)
\(\displaystyle 120\)	\(\displaystyle 9\), since \(\displaystyle 9\times 13=117\)	\(\displaystyle 3\)
\(\displaystyle 30\)	\(\displaystyle 2\), since \(\displaystyle 2\times 13=26\)	\(\displaystyle 4\)
\(\displaystyle 40\)	\(\displaystyle 3\), since \(\displaystyle 3\times 13=39\)	\(\displaystyle 1\)

The remainder is \(\displaystyle 1\) again — the number we started with — so from here every step repeats exactly.\[\frac{1}{13}=0.\overline{076923}\qquad\text{repeating block } \mathbf{076923},\ \text{length } 6\]Why the length is 6. Dividing by \(\displaystyle 13\) can only ever leave remainders \(\displaystyle 1,2,\dots,12\), and never \(\displaystyle 0\) (no power of \(\displaystyle 10\) is a multiple of \(\displaystyle 13\)). So the block can be at most \(\displaystyle 12\) digits long. Here the remainders that actually appeared were \(\displaystyle 1,10,9,12,3,4\) — six of the twelve — and then the cycle closed.Now \(\displaystyle \dfrac{2}{13}\): \(\displaystyle 20\div13\to 1\) r \(\displaystyle 7\); \(\displaystyle 70\div13\to 5\) r \(\displaystyle 5\); \(\displaystyle 50\div13\to 3\) r \(\displaystyle 11\); \(\displaystyle 110\div13\to 8\) r \(\displaystyle 6\); \(\displaystyle 60\div13\to 4\) r \(\displaystyle 8\); \(\displaystyle 80\div13\to 6\) r \(\displaystyle 2\) — back to the start. \[\frac{2}{13}=0.\overline{153846}\]Is \(\displaystyle 153846\) a rotation of \(\displaystyle 076923\)? The rotations of \(\displaystyle 076923\) are \[076923,\ 769230,\ 692307,\ 923076,\ 230769,\ 307692\] and \(\displaystyle 153846\) is not among them. So \(\displaystyle 13\) does not behave quite like \(\displaystyle 7\).All twelve:

\(\displaystyle k\)	\(\displaystyle k/13\)	\(\displaystyle k\)	\(\displaystyle k/13\)
\(\displaystyle 1\)	\(\displaystyle 0.\overline{076923}\)	\(\displaystyle 7\)	\(\displaystyle 0.\overline{538461}\)
\(\displaystyle 2\)	\(\displaystyle 0.\overline{153846}\)	\(\displaystyle 8\)	\(\displaystyle 0.\overline{615384}\)
\(\displaystyle 3\)	\(\displaystyle 0.\overline{230769}\)	\(\displaystyle 9\)	\(\displaystyle 0.\overline{692307}\)
\(\displaystyle 4\)	\(\displaystyle 0.\overline{307692}\)	\(\displaystyle 10\)	\(\displaystyle 0.\overline{769230}\)
\(\displaystyle 5\)	\(\displaystyle 0.\overline{384615}\)	\(\displaystyle 11\)	\(\displaystyle 0.\overline{846153}\)
\(\displaystyle 6\)	\(\displaystyle 0.\overline{461538}\)	\(\displaystyle 12\)	\(\displaystyle 0.\overline{923076}\)

What you notice. The behaviour is cyclic, but it splits into two families rather than one: No block from one family is a rotation of a block in the other, and the two families together use all twelve values of \(\displaystyle k\).Compared with \(\displaystyle \dfrac17\). There \(\displaystyle 1/7=0.\overline{142857}\) has block length \(\displaystyle 6=7-1\), so the single cycle uses up every possible remainder, and all six of \(\displaystyle 1/7,\dots,6/7\) are rotations of one block. For \(\displaystyle 13\) the block is only \(\displaystyle 6\) long while there are \(\displaystyle 12\) possible remainders, so the remainders have to split into two separate cycles of six.Answer: \(\displaystyle \dfrac{1}{13}=0.\overline{076923}\), repeating block \(\displaystyle 076923\) of length \(\displaystyle 6\); \(\displaystyle \dfrac{2}{13}=0.\overline{153846}\). The fractions \(\displaystyle k/13\) are cyclic, but in two families of six rotations each — \(\displaystyle \{1,3,4,9,10,12\}\) rotate \(\displaystyle 076923\) and \(\displaystyle \{2,5,6,7,8,11\}\) rotate \(\displaystyle 153846\) — unlike \(\displaystyle 1/7\), where all six form a single cycle.

Test: a number is rational exactly when its decimal either stops or eventually repeats a fixed block. Anything non-terminating and non-repeating is irrational.(i) \(\displaystyle \sqrt{81}\). Since \(\displaystyle 9\times 9=81\), \(\displaystyle \sqrt{81}=9\). An integer is a rational number. Rational, \(\displaystyle \sqrt{81}=9=\dfrac{9}{1}\).(ii) \(\displaystyle \sqrt{12}\). First, \(\displaystyle \sqrt{12}=\sqrt{4\times 3}=2\sqrt3\), and \(\displaystyle 9<12<16\) gives \(\displaystyle 3<\sqrt{12}<4\), so it is not a whole number. Suppose it were rational, say \(\displaystyle \sqrt{12}=\frac pq\) in lowest terms. Then \(\displaystyle p^2=12q^2\), so \(\displaystyle 3\mid p^2\), and since \(\displaystyle 3\) is prime, \(\displaystyle 3\mid p\). Write \(\displaystyle p=3m\): then \(\displaystyle 9m^2=12q^2\), i.e. \(\displaystyle 3m^2=4q^2\). So \(\displaystyle 3\mid 4q^2\), and as \(\displaystyle 3\nmid 4\) we get \(\displaystyle 3\mid q^2\), hence \(\displaystyle 3\mid q\). Now \(\displaystyle 3\) divides both \(\displaystyle p\) and \(\displaystyle q\), contradicting "lowest terms". Irrational (\(\displaystyle \approx 3.4641016\ldots\)). No fraction exists.(iii) \(\displaystyle 0.33333\ldots\) — the block \(\displaystyle 3\) repeats. Let \(\displaystyle x=0.\overline{3}\). Then \(\displaystyle 10x=3.\overline{3}\), and subtracting, \(\displaystyle 9x=3\), so \(\displaystyle x=\frac39\). Rational, \(\displaystyle =\dfrac{1}{3}\).(iv) \(\displaystyle 0.123451234512345\ldots\) — the block \(\displaystyle 12345\) repeats, length \(\displaystyle 5\). Let \(\displaystyle x=0.\overline{12345}\). Then \(\displaystyle 100000x=12345.\overline{12345}\), so \(\displaystyle 99999x=12345\) and \[x=\frac{12345}{99999}=\frac{12345\div 3}{99999\div 3}=\frac{4115}{33333}\] Rational, \(\displaystyle =\dfrac{4115}{33333}\). (Check: \(\displaystyle 4115\div 33333=0.123451234512345\ldots\) \(\displaystyle \checkmark\))(v) \(\displaystyle 1.01001000100001\ldots\) — look at the runs of zeros: \(\displaystyle 1\) zero, then \(\displaystyle 2\), then \(\displaystyle 3\), then \(\displaystyle 4\), and so on. Is a single block repeating? No. If some block of length \(\displaystyle L\) repeated, the pattern would be periodic from some point on; but the runs of zeros keep growing and eventually exceed \(\displaystyle L\), so no such block can exist. The decimal is non-terminating and non-repeating. Irrational. No fraction exists.(vi) \(\displaystyle 23.560185612239874790120\). As printed this decimal stops after \(\displaystyle 21\) places (there is no "\(\displaystyle \ldots\)"), and every terminating decimal is rational: put the digits over the matching power of ten. \[23.560185612239874790120=\frac{23560185612239874790120}{10^{21}}\] Both parts are divisible by \(\displaystyle 40\) (the numerator ends in \(\displaystyle 120\)), giving \[=\frac{589004640305996869753}{25000000000000000000}\] which is in lowest terms because the numerator is odd and does not end in \(\displaystyle 0\) or \(\displaystyle 5\). Rational. (One caution: this is only true because the decimal is written as terminating. If the digits were meant to run on for ever with no repeating block, the number would be irrational.)Answer: (i) rational, \(\displaystyle 9=\frac{9}{1}\); (ii) irrational; (iii) rational, \(\displaystyle \frac13\); (iv) rational, \(\displaystyle \frac{4115}{33333}\); (v) irrational; (vi) rational (it terminates), \(\displaystyle \frac{589004640305996869753}{25000000000000000000}\).

Algebra on a repeating decimal. The trick is to multiply by a power of ten that makes the endless tail line up with itself, so that subtraction wipes it out.Let \[x=0.\overline{9}=0.9999\ldots\]The repeating block is a single digit, so shift by one place — that is, multiply by \(\displaystyle 10\): \[10x=9.9999\ldots\]Look carefully at what each side is. After the decimal point, \(\displaystyle 10x\) has \(\displaystyle .9999\ldots\) and \(\displaystyle x\) has \(\displaystyle .9999\ldots\) — the same endless tail, because pushing an endless string of 9s one place along still leaves an endless string of 9s. So when we subtract, the tails cancel completely:\[\begin{array}{r} 10x = 9.9999\ldots\\ -\quad x = 0.9999\ldots\\ \hline 9x = 9.0000\ldots \end{array} \]\[9x=9\qquad\Longrightarrow\qquad x=1\]A second way, to be sure. You already know \(\displaystyle \dfrac13=0.333\ldots\). Multiply both sides by \(\displaystyle 3\): \[3\times\frac13=1\qquad\text{and}\qquad 3\times 0.333\ldots=0.999\ldots\] Two names for the result of the same multiplication, so \(\displaystyle 0.999\ldots=1\).A third way — how big is the gap? If \(\displaystyle 0.999\ldots\) were less than \(\displaystyle 1\), the difference \(\displaystyle 1-0.999\ldots\) would be a positive number. But it is smaller than \(\displaystyle 1-0.9=0.1\), smaller than \(\displaystyle 1-0.99=0.01\), smaller than \(\displaystyle 0.001\), and so on — smaller than every positive number of the form \(\displaystyle 10^{-n}\). The only number that small is \(\displaystyle 0\). So the gap is \(\displaystyle 0\) and the two are the same number.What this does and does not say. It does not say \(\displaystyle 0.999\ldots\) "creeps up to" \(\displaystyle 1\) or "is almost" \(\displaystyle 1\). \(\displaystyle 0.999\ldots\) is a name for one fixed number, and that number is \(\displaystyle 1\) — exactly as \(\displaystyle \frac24\) and \(\displaystyle \frac12\) are two names for one number.Answer: \(\displaystyle 0.\overline{9}=1\) exactly.

What makes \(\displaystyle 1/7\) special. \(\displaystyle \frac17=0.\overline{142857}\) has a block of length \(\displaystyle 6\), and \(\displaystyle 6=7-1\). That is the whole story: dividing by \(\displaystyle 7\) can leave remainders \(\displaystyle 1,2,3,4,5,6\), and this single cycle uses every one of them before returning to the start. So \(\displaystyle 2/7,3/7,\dots,6/7\) simply enter the same cycle at a different point, and their blocks are rotations of \(\displaystyle 142857\): \[\tfrac17=0.\overline{142857},\ \tfrac27=0.\overline{285714},\ \tfrac37=0.\overline{428571},\ \tfrac47=0.\overline{571428},\ \tfrac57=0.\overline{714285},\ \tfrac67=0.\overline{857142}\]So the search is: find \(\displaystyle n\) whose block length for \(\displaystyle 1/n\) is exactly \(\displaystyle n-1\). (\(\displaystyle n\) must be prime, and cannot be \(\displaystyle 2\) or \(\displaystyle 5\), since those give terminating decimals.)How to test one by hand. Divide \(\displaystyle 1\) by \(\displaystyle n\) and count the steps until the remainder \(\displaystyle 1\) reappears. If that count equals \(\displaystyle n-1\), the number is cyclic.Results.

\(\displaystyle n\)	Repeating block of \(\displaystyle 1/n\)	Length	\(\displaystyle =n-1\)?
\(\displaystyle 7\)	\(\displaystyle 142857\)	\(\displaystyle 6\)	yes
\(\displaystyle 11\)	\(\displaystyle 09\)	\(\displaystyle 2\)	no
\(\displaystyle 13\)	\(\displaystyle 076923\)	\(\displaystyle 6\)	no (\(\displaystyle 12\) needed)
\(\displaystyle 17\)	\(\displaystyle 0588235294117647\)	\(\displaystyle 16\)	yes
\(\displaystyle 19\)	\(\displaystyle 052631578947368421\)	\(\displaystyle 18\)	yes
\(\displaystyle 23\)	\(\displaystyle 0434782608695652173913\)	\(\displaystyle 22\)	yes
\(\displaystyle 29\)	\(\displaystyle 0344827586206896551724137931\)	\(\displaystyle 28\)	yes
\(\displaystyle 31\)	\(\displaystyle 032258064516129\)	\(\displaystyle 15\)	no
\(\displaystyle 37\)	\(\displaystyle 027\)	\(\displaystyle 3\)	no
\(\displaystyle 41\)	\(\displaystyle 02439\)	\(\displaystyle 5\)	no

Below \(\displaystyle 100\) the complete list is \[n=7,\ 17,\ 19,\ 23,\ 29,\ 47,\ 59,\ 61,\ 97\]Spot-check that \(\displaystyle 17\) really is cyclic. \(\displaystyle \frac{2}{17}=0.\overline{1176470588235294}\) — start reading the block of \(\displaystyle \frac1{17}\), namely \(\displaystyle 0588235294117647\), from the "\(\displaystyle 11\)" near its end and wrap around: \(\displaystyle 1176470588235294\). Same digits, rotated. \(\displaystyle \checkmark\)Contrast with \(\displaystyle 13\) (Question $\displaystyle 2$): its block is \(\displaystyle 6\) long but there are \(\displaystyle 12\) remainders, so the remainders split into two cycles and the twelve fractions \(\displaystyle k/13\) fall into two families of rotations, not one.---The note at the end: non-uniqueness of decimals.Yes — every terminating decimal has a second name ending in repeating 9s, and the algebra of Question $\displaystyle 4$ proves each case. For \(\displaystyle 2.47\): \[\text{let } x=2.46\overline{9};\quad 100x=246.\overline{9}=246+0.\overline{9}=246+1=247,\ \text{so } x=2.47\ \checkmark\]Is it surprising? It is, the first time. The reason most people guess \(\displaystyle 0.999\ldots<1\) is that they picture the decimal being built digit by digit, and every finite stage — \(\displaystyle 0.9,\ 0.99,\ 0.999\) — genuinely is less than \(\displaystyle 1\). But \(\displaystyle 0.999\ldots\) is not any one of those stages; it is the single number the whole endless list closes in on, and that number is \(\displaystyle 1\). Note also that only terminating decimals get two names — \(\displaystyle \frac13=0.333\ldots\) has just one.Answer: the required \(\displaystyle n\) are the primes whose \(\displaystyle 1/n\) has a full-length block \(\displaystyle n-1\): below \(\displaystyle 100\) these are \(\displaystyle 7,\,17,\,19,\,23,\,29,\,47,\,59,\,61,\,97\). And yes, \(\displaystyle 0.999\ldots=1\); every terminating decimal has an alternative repeating-$\displaystyle 9$ form.