NCERT Solutions · Class 9 Mathematics The World of Numbers

Long division, watching the remainder. The division stops when a remainder of \(\displaystyle 0\) appears, and repeats when an earlier remainder comes back.(i) \(\displaystyle \dfrac{3}{50}\)Before dividing, look at the denominator: \(\displaystyle 50=2\times 5^2\) is built only from \(\displaystyle 2\)s and \(\displaystyle 5\)s, so we should expect a terminating decimal.\(\displaystyle 50\) into \(\displaystyle 30\) goes \(\displaystyle 0\) times, remainder \(\displaystyle 30\) \(\displaystyle \Rightarrow\) first decimal digit \(\displaystyle 0\). \(\displaystyle 50\) into \(\displaystyle 300\) goes \(\displaystyle 6\) times (\(\displaystyle 6\times 50=300\)), remainder \(\displaystyle 0\) \(\displaystyle \Rightarrow\) next digit \(\displaystyle 6\), and the division stops.\[\frac{3}{50}=0.06\]Check the other way: \(\displaystyle \frac{3}{50}=\frac{3\times 2}{50\times 2}=\frac{6}{100}=0.06\). \(\displaystyle \checkmark\) Terminating.(ii) \(\displaystyle \dfrac{2}{9}\)Here \(\displaystyle 9=3^2\) contains the prime \(\displaystyle 3\), so expect a repeating decimal.\(\displaystyle 9\) into \(\displaystyle 20\) goes \(\displaystyle 2\) times (\(\displaystyle 18\)), remainder \(\displaystyle 2\) \(\displaystyle \Rightarrow\) digit \(\displaystyle 2\). But the remainder \(\displaystyle 2\) is exactly the number we began with, so the very same step happens again, and again, for ever.\[\frac{2}{9}=0.2222\ldots=0.\overline{2}\]Check: \(\displaystyle 0.\overline{2}\) means \(\displaystyle x\) with \(\displaystyle 10x-x=2\), so \(\displaystyle 9x=2\) and \(\displaystyle x=\frac29\). \(\displaystyle \checkmark\) Non-terminating and repeating.Answer: (i) \(\displaystyle \dfrac{3}{50}=0.06\), a terminating decimal. (ii) \(\displaystyle \dfrac{2}{9}=0.\overline{2}\), a non-terminating repeating decimal.

Proof by contradiction. We assume the opposite of what we want, and show it forces something impossible.The fact we shall use. If a prime \(\displaystyle p\) divides \(\displaystyle a^2\), then \(\displaystyle p\) divides \(\displaystyle a\). (Reason: the prime factorisation of \(\displaystyle a^2\) is just the factorisation of \(\displaystyle a\) with every exponent doubled, so no new prime can appear in \(\displaystyle a^2\) that was not already in \(\displaystyle a\).) We use it with \(\displaystyle p=5\).The proof.Suppose, for the sake of argument, that \(\displaystyle \sqrt5\) is rational. Then it can be written as a fraction \[\sqrt5=\frac{p}{q}\] where \(\displaystyle p\) and \(\displaystyle q\) are integers, \(\displaystyle q\neq 0\), and the fraction is in lowest terms — that is, \(\displaystyle p\) and \(\displaystyle q\) have no common factor other than \(\displaystyle 1\). (Assuming lowest terms costs nothing, since every fraction can be reduced to lowest terms.)Square both sides: \[5=\frac{p^{2}}{q^{2}}\qquad\Longrightarrow\qquad p^{2}=5q^{2}\tag{1}\]From \(\displaystyle (1)\), \(\displaystyle p^{2}\) is \(\displaystyle 5\) times an integer, so \(\displaystyle 5\) divides \(\displaystyle p^{2}\). By the fact above, \(\displaystyle 5\) divides \(\displaystyle p\). So we may write \[p=5m\quad\text{for some integer } m\]Substitute this into \(\displaystyle (1)\): \[(5m)^{2}=5q^{2}\;\Longrightarrow\;25m^{2}=5q^{2}\;\Longrightarrow\;q^{2}=5m^{2}\tag{2}\]From \(\displaystyle (2)\), \(\displaystyle 5\) divides \(\displaystyle q^{2}\), and by the same fact, \(\displaystyle 5\) divides \(\displaystyle q\).The contradiction. We have now shown that \(\displaystyle 5\) divides \(\displaystyle p\) and \(\displaystyle 5\) divides \(\displaystyle q\). So \(\displaystyle p\) and \(\displaystyle q\) share the common factor \(\displaystyle 5\) — which flatly contradicts our statement that \(\displaystyle \frac pq\) was in lowest terms.Every step after the assumption was valid, so the fault must lie in the assumption itself. Therefore \(\displaystyle \sqrt5\) cannot be written as \(\displaystyle \frac pq\).Answer: \(\displaystyle \sqrt5\) is irrational.(Its decimal begins \(\displaystyle 2.2360679\ldots\) and never terminates or repeats — but no amount of computing decimals could ever prove this, since you can only ever check finitely many digits. Only the argument above settles it.)

Two methods, depending on the decimal.Terminating decimal: write the digits over the matching power of \(\displaystyle 10\), then reduce.Repeating decimal: call the number \(\displaystyle x\), then multiply by two powers of \(\displaystyle 10\) chosen so that the endless tails line up exactly — one power to push the repeating block up to the decimal point, and a second power one whole block further. Subtracting cancels the tails and leaves whole numbers.(i) \(\displaystyle 12.6\) — terminating, one decimal place. \[12.6=\frac{126}{10}=\frac{126\div 2}{10\div 2}=\frac{63}{5}\](ii) \(\displaystyle 0.0120\) — terminating; the final \(\displaystyle 0\) adds nothing, so this is \(\displaystyle 0.012\), three decimal places. \[0.012=\frac{12}{1000}=\frac{12\div 4}{1000\div 4}=\frac{3}{250}\] Check: \(\displaystyle \frac{3}{250}=\frac{12}{1000}=0.012\). \(\displaystyle \checkmark\)(iii) \(\displaystyle 3.0\overline{52}=3.0525252\ldots\) — one digit before the block, block length \(\displaystyle 2\). Let \(\displaystyle x=3.0525252\ldots\) \[10x=30.525252\ldots\qquad(\text{block now starts right after the point})\] \[1000x=3052.525252\ldots\qquad(\text{a further } 2 \text{ places, i.e. one whole block})\] Subtract, and the identical tails vanish: \[1000x-10x=3052-30=3022\;\Longrightarrow\;990x=3022\;\Longrightarrow\;x=\frac{3022}{990}=\frac{1511}{495}\] Check: \(\displaystyle 1511\div 495=3.0525252\ldots\) \(\displaystyle \checkmark\)(iv) \(\displaystyle 1.2\overline{35}=1.2353535\ldots\) — one digit before the block, block length \(\displaystyle 2\). \[10x=12.353535\ldots,\qquad 1000x=1235.353535\ldots\] \[990x=1235-12=1223\;\Longrightarrow\;x=\frac{1223}{990}\] Already lowest terms: \(\displaystyle 990=2\times 3^2\times 5\times 11\), while \(\displaystyle 1223\) is odd, has digit sum \(\displaystyle 8\) (not a multiple of \(\displaystyle 3\)), does not end in \(\displaystyle 0\) or \(\displaystyle 5\), and its alternating digit sum \(\displaystyle 3-2+2-1=2\) is not a multiple of \(\displaystyle 11\).(v) \(\displaystyle 0.\overline{23}=0.232323\ldots\) — block length \(\displaystyle 2\), nothing before it. \[100x=23.232323\ldots\;\Longrightarrow\;100x-x=23\;\Longrightarrow\;99x=23\;\Longrightarrow\;x=\frac{23}{99}\] (\(\displaystyle 23\) is prime and does not divide \(\displaystyle 99=9\times 11\), so this is lowest terms.)(vi) \(\displaystyle 2.0\overline{5}=2.05555\ldots\) — one digit before the block, block length \(\displaystyle 1\). \[10x=20.5555\ldots,\qquad 100x=205.5555\ldots\] \[90x=205-20=185\;\Longrightarrow\;x=\frac{185}{90}=\frac{37}{18}\] Check: \(\displaystyle 37\div 18=2.0555\ldots\) \(\displaystyle \checkmark\)(vii) \(\displaystyle 2.12\overline{5}=2.125555\ldots\) — two digits before the block, block length \(\displaystyle 1\). \[100x=212.5555\ldots,\qquad 1000x=2125.5555\ldots\] \[900x=2125-212=1913\;\Longrightarrow\;x=\frac{1913}{900}\] Lowest terms: \(\displaystyle 900=2^2\times 3^2\times 5^2\); \(\displaystyle 1913\) is odd, digit sum \(\displaystyle 14\) (not a multiple of \(\displaystyle 3\)), and does not end in \(\displaystyle 0\) or \(\displaystyle 5\).(viii) \(\displaystyle 3.12\overline{5}=3.125555\ldots\) — same shape as (vii). \[100x=312.5555\ldots,\qquad 1000x=3125.5555\ldots\] \[900x=3125-312=2813\;\Longrightarrow\;x=\frac{2813}{900}\] Neat check: (viii) is exactly \(\displaystyle 1\) more than (vii), and indeed \(\displaystyle \frac{2813}{900}-\frac{1913}{900}=\frac{900}{900}=1\). \(\displaystyle \checkmark\)(ix) \(\displaystyle 2.\overline{1625}=2.16251625\ldots\) — block length \(\displaystyle 4\), nothing before it. \[10000x=21625.16251625\ldots\;\Longrightarrow\;10000x-x=21625-2=21623\] \[9999x=21623\;\Longrightarrow\;x=\frac{21623}{9999}\] Lowest terms: \(\displaystyle 9999=3^2\times 11\times 101\); \(\displaystyle 21623\) has digit sum \(\displaystyle 14\) (not a multiple of \(\displaystyle 3\)), alternating sum \(\displaystyle 3-2+6-1+2=8\) (not a multiple of \(\displaystyle 11\)), and \(\displaystyle 101\times 214=21614\) leaves remainder \(\displaystyle 9\).Answer.

	Decimal	\(\displaystyle \dfrac{p}{q}\)
(i)	\(\displaystyle 12.6\)	\(\displaystyle \dfrac{63}{5}\)
(ii)	\(\displaystyle 0.0120\)	\(\displaystyle \dfrac{3}{250}\)
(iii)	\(\displaystyle 3.0\overline{52}\)	\(\displaystyle \dfrac{1511}{495}\)
(iv)	\(\displaystyle 1.2\overline{35}\)	\(\displaystyle \dfrac{1223}{990}\)
(v)	\(\displaystyle 0.\overline{23}\)	\(\displaystyle \dfrac{23}{99}\)
(vi)	\(\displaystyle 2.0\overline{5}\)	\(\displaystyle \dfrac{37}{18}\)
(vii)	\(\displaystyle 2.12\overline{5}\)	\(\displaystyle \dfrac{1913}{900}\)
(viii)	\(\displaystyle 3.12\overline{5}\)	\(\displaystyle \dfrac{2813}{900}\)
(ix)	\(\displaystyle 2.\overline{1625}\)	\(\displaystyle \dfrac{21623}{9999}\)

Successive magnification. A decimal digit tells you which tenth of the current strip to zoom into, so read the digits one at a time and magnify once per digit. Draw each strip on your own page as you follow the steps below.(i) \(\displaystyle 0.532\)Step 1. Draw the segment from \(\displaystyle 0\) to \(\displaystyle 1\) and divide it into \(\displaystyle 10\) equal parts: \(\displaystyle 0.1,0.2,\dots,0.9\). The first decimal digit is \(\displaystyle 5\), so the point lies in the strip between \(\displaystyle \mathbf{0.5}\) and \(\displaystyle \mathbf{0.6}\).Step 2. Magnify that strip — redraw \(\displaystyle 0.5\) to \(\displaystyle 0.6\) as a full-length segment — and divide it into \(\displaystyle 10\) equal parts marked \(\displaystyle 0.50,\ 0.51,\ \dots,\ 0.60\). The second digit is \(\displaystyle 3\), so the point lies between \(\displaystyle \mathbf{0.53}\) and \(\displaystyle \mathbf{0.54}\).Step 3. Magnify \(\displaystyle 0.53\) to \(\displaystyle 0.54\) and divide into \(\displaystyle 10\) parts marked \(\displaystyle 0.530,\ 0.531,\ \dots,\ 0.540\). The third digit is \(\displaystyle 2\), so the point is the 2nd mark, i.e. exactly \(\displaystyle 0.532\).Because the decimal terminates, the process finishes: three magnifications land on the point precisely.(ii) \(\displaystyle 1.1\overline{5}=1.15555\ldots\)First pin down the exact value, so the position is certain rather than guessed. Let \(\displaystyle x=1.15555\ldots\) \[10x=11.5555\ldots,\qquad 100x=115.5555\ldots\] \[90x=115-11=104\;\Longrightarrow\;x=\frac{104}{90}=\frac{52}{45}\] Check: \(\displaystyle 52\div 45=1.15555\ldots\) \(\displaystyle \checkmark\) So the point is exactly \(\displaystyle \dfrac{52}{45}\).Step 1. On the segment from \(\displaystyle 1\) to \(\displaystyle 2\) marked in tenths, the first decimal digit \(\displaystyle 1\) puts the point between \(\displaystyle \mathbf{1.1}\) and \(\displaystyle \mathbf{1.2}\).Step 2. Magnify \(\displaystyle 1.1\) to \(\displaystyle 1.2\) into ten parts. The next digit \(\displaystyle 5\) puts it between \(\displaystyle \mathbf{1.15}\) and \(\displaystyle \mathbf{1.16}\).Step 3. Magnify \(\displaystyle 1.15\) to \(\displaystyle 1.16\) into ten parts. The next digit \(\displaystyle 5\) puts it between \(\displaystyle \mathbf{1.155}\) and \(\displaystyle \mathbf{1.156}\).Step 4. Magnify again: between \(\displaystyle \mathbf{1.1555}\) and \(\displaystyle \mathbf{1.1556}\) — and so on for ever, since the \(\displaystyle 5\)s never stop.Why this still locates a single point. Each new strip is \(\displaystyle 10\) times shorter than the one before, so the strips shrink towards nothing while always containing the number. There is exactly one point common to all of them, and that point is \(\displaystyle \frac{52}{45}\). In a practical drawing you mark it just past the middle of the \(\displaystyle 1.15\)-to-\(\displaystyle 1.16\) strip (it sits \(\displaystyle \tfrac59\) of the way along it).Answer: (i) \(\displaystyle 0.532\) is reached exactly after three magnifications, as the 2nd of ten marks inside the strip \(\displaystyle 0.53\) to \(\displaystyle 0.54\). (ii) \(\displaystyle 1.1\overline{5}=\dfrac{52}{45}\), located as the single point common to the endlessly shrinking strips \(\displaystyle 1.1\)–\(\displaystyle 1.2\), \(\displaystyle 1.15\)–\(\displaystyle 1.16\), \(\displaystyle 1.155\)–\(\displaystyle 1.156\), \(\displaystyle \dots\)

Give both numbers a common denominator big enough to leave room. Between the integers \(\displaystyle 3\) and \(\displaystyle 4\) there is no integer at all, so writing them as whole numbers gives us nothing to pick from. Rewriting them as fractions creates the room.We want \(\displaystyle 6\) numbers, so choose a denominator of \(\displaystyle 7\) (one more than \(\displaystyle 6\)) — that leaves exactly \(\displaystyle 6\) gaps:\[3=\frac{3\times 7}{7}=\frac{21}{7},\qquad 4=\frac{4\times 7}{7}=\frac{28}{7}\]The fractions with denominator \(\displaystyle 7\) lying strictly between \(\displaystyle \frac{21}{7}\) and \(\displaystyle \frac{28}{7}\) are the ones whose numerators are the integers strictly between \(\displaystyle 21\) and \(\displaystyle 28\):\[\frac{22}{7},\quad \frac{23}{7},\quad \frac{24}{7},\quad \frac{25}{7},\quad \frac{26}{7},\quad \frac{27}{7}\]That is exactly \(\displaystyle 6\) numbers, and each is rational (an integer over an integer).Check the two ends. \(\displaystyle \frac{22}{7}\approx 3.143>3\) \(\displaystyle \checkmark\) and \(\displaystyle \frac{27}{7}\approx 3.857<4\) \(\displaystyle \checkmark\), and since the numerators increase the whole list is in order.Other answers work just as well — for instance \(\displaystyle 3.1,\,3.2,\,3.3,\,3.4,\,3.5,\,3.6\), or six copies of the "take the average" step. In fact there are infinitely many rationals between \(\displaystyle 3\) and \(\displaystyle 4\), so the answer is not unique.Answer: \(\displaystyle \dfrac{22}{7},\ \dfrac{23}{7},\ \dfrac{24}{7},\ \dfrac{25}{7},\ \dfrac{26}{7},\ \dfrac{27}{7}\).

Scale up the common denominator. The two fractions already share the denominator \(\displaystyle 5\), but their numerators \(\displaystyle 2\) and \(\displaystyle 3\) are consecutive integers, so there is no fraction with denominator \(\displaystyle 5\) between them. Multiplying top and bottom of both by the same number does not change their values but does create room.We need \(\displaystyle 5\) numbers in between, so multiply by \(\displaystyle 6\) (leaving \(\displaystyle 6\) gaps, which is more than enough):\[\frac{2}{5}=\frac{2\times 6}{5\times 6}=\frac{12}{30},\qquad \frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}\]Now every numerator strictly between \(\displaystyle 12\) and \(\displaystyle 18\) gives a fraction strictly between the two:\[\frac{13}{30},\quad \frac{14}{30}=\frac{7}{15},\quad \frac{15}{30}=\frac{1}{2},\quad \frac{16}{30}=\frac{8}{15},\quad \frac{17}{30}\]Check in decimals. \[0.4\;<\;0.4\overline{3}\;<\;0.4\overline{6}\;<\;0.5\;<\;0.5\overline{3}\;<\;0.5\overline{6}\;<\;0.6\ \checkmark\] (the outer two being \(\displaystyle \frac25=0.4\) and \(\displaystyle \frac35=0.6\)).Other answers are equally correct — multiplying by \(\displaystyle 10\) instead would give \(\displaystyle \frac{21}{50},\frac{22}{50},\dots\), and there are infinitely many possibilities.Answer: \(\displaystyle \dfrac{13}{30},\ \dfrac{7}{15},\ \dfrac{1}{2},\ \dfrac{8}{15},\ \dfrac{17}{30}\).

Put both fractions over a common denominator. Once the denominators match, comparing and filling the gap is just a matter of looking at the numerators.The LCM of \(\displaystyle 6\) and \(\displaystyle 5\) is \(\displaystyle 30\): \[\frac{1}{6}=\frac{1\times 5}{6\times 5}=\frac{5}{30},\qquad \frac{2}{5}=\frac{2\times 6}{5\times 6}=\frac{12}{30}\]The integers strictly between \(\displaystyle 5\) and \(\displaystyle 12\) are \(\displaystyle 6,7,8,9,10,11\) — six of them, and we need five. Taking the first five and reducing where possible:\[\frac{6}{30}=\frac{1}{5},\quad \frac{7}{30},\quad \frac{8}{30}=\frac{4}{15},\quad \frac{9}{30}=\frac{3}{10},\quad \frac{10}{30}=\frac{1}{3}\]Check in decimals. \[\frac16\approx 0.167\;<\;0.2\;<\;0.2\overline{3}\;<\;0.2\overline{6}\;<\;0.3\;<\;0.\overline{3}\;<\;0.4=\frac25\ \checkmark\]The answer is not unique — \(\displaystyle \frac{11}{30}\) would also qualify, and multiplying by a larger factor (say using denominator \(\displaystyle 60\) or \(\displaystyle 300\)) produces many more.Answer: \(\displaystyle \dfrac{1}{5},\ \dfrac{7}{30},\ \dfrac{4}{15},\ \dfrac{3}{10},\ \dfrac{1}{3}\).

Combine the left-hand side over one denominator. The two terms on the left cannot be added while their denominators differ, so rewrite both over the LCM of \(\displaystyle 3\) and \(\displaystyle 5\), which is \(\displaystyle 15\).\[\frac{x}{3}=\frac{x\times 5}{3\times 5}=\frac{5x}{15},\qquad \frac{x}{5}=\frac{x\times 3}{5\times 3}=\frac{3x}{15}\]Add them: \[\frac{5x}{15}+\frac{3x}{15}=\frac{5x+3x}{15}=\frac{8x}{15}\]So the equation becomes \[\frac{8x}{15}=\frac{16}{15}\]Both sides now have the same denominator, \(\displaystyle 15\). Two fractions with equal denominators are equal only when their numerators are equal, so \[8x=16\qquad\Longrightarrow\qquad x=\frac{16}{8}=2\](The same thing, done in one line: multiply both sides of the original equation by \(\displaystyle 15\) to clear all denominators, giving \(\displaystyle 5x+3x=16\).)Check by substituting \(\displaystyle x=2\) back into the original equation. \[\frac{2}{3}+\frac{2}{5}=\frac{10}{15}+\frac{6}{15}=\frac{16}{15}\ \checkmark\]Answer: \(\displaystyle x=2\).

Rearrange until \(\displaystyle ab\) appears as a single quantity. We are told not to try numbers, so the argument must come from the equation itself.We are given that \(\displaystyle a\) and \(\displaystyle b\) are non-zero rationals with \[a+\frac{1}{b}=0\] (\(\displaystyle b\neq0\) is exactly what makes \(\displaystyle \frac1b\) meaningful.)Move \(\displaystyle \frac1b\) to the other side: \[a=-\frac{1}{b}\]Now multiply both sides by \(\displaystyle b\) — allowed, since \(\displaystyle b\neq 0\): \[ab=-\frac{1}{b}\times b=-1\]So \(\displaystyle ab\) is not merely negative; it is exactly \(\displaystyle -1\), for every pair \(\displaystyle (a,b)\) satisfying the condition. And since \(\displaystyle -1<0\), \(\displaystyle ab\) is negative.A second way to see the sign, without the algebra. The equation \(\displaystyle a=-\frac1b\) says that \(\displaystyle a\) is the opposite of \(\displaystyle \frac1b\). A number and its reciprocal always carry the same sign (a positive times a positive is positive, a negative times a negative is positive, so \(\displaystyle b\) and \(\displaystyle \frac1b\) agree in sign). Hence \(\displaystyle a\) has the opposite sign to \(\displaystyle b\) — one is positive and the other negative — and the product of a positive and a negative number is negative.Confirmation with numbers (only as a check, not as the reasoning): \(\displaystyle b=4\Rightarrow a=-\frac14\Rightarrow ab=-1\); \(\displaystyle b=-\frac23\Rightarrow a=\frac32\Rightarrow ab=\frac32\times\left(-\frac23\right)=-1\). \(\displaystyle \checkmark\)Answer: \(\displaystyle ab\) is negative — in fact \(\displaystyle ab=-1\) always.

Multiplying by a power of $\displaystyle 10$ clears a terminating decimal.Let the rational number be \(\displaystyle x\). Its decimal expansion stops, and the last non-zero digit sits in the 4th decimal place. So nothing at all is written after the 4th place, and \(\displaystyle x\) looks like \[x \;=\; \pm\, N.d_1d_2d_3d_4 , \] where \(\displaystyle N\) is the whole-number part and \(\displaystyle d_1,d_2,d_3,d_4\) are digits with \(\displaystyle d_4 \neq 0\).Multiplying by \(\displaystyle 10\) moves the decimal point one place to the right, so multiplying by \(\displaystyle 10^4\) moves it four places to the right — exactly enough to swallow all four decimal digits: \[10^4 x \;=\; \pm\, N d_1 d_2 d_3 d_4 \quad(\text{now read as a whole number}). \] Call this integer \(\displaystyle p\). Dividing both sides by \(\displaystyle 10^4\), \[x \;=\; \frac{p}{10^{4}} . \]Why \(\displaystyle p\) is not divisible by 10. The units digit of \(\displaystyle p\) is \(\displaystyle d_4\), and \(\displaystyle d_4 \neq 0\) because the 4th decimal place holds the last non-zero digit. An integer is a multiple of $\displaystyle 10$ exactly when its units digit is \(\displaystyle 0\). Since the units digit of \(\displaystyle p\) is not \(\displaystyle 0\), \(\displaystyle 10 \nmid p\).For example \(\displaystyle 0.0625 = \dfrac{625}{10^4}\), \(\displaystyle 12.3456 = \dfrac{123456}{10^4}\), \(\displaystyle 0.0001 = \dfrac{1}{10^4}\) — in each case the numerator ends in a non-zero digit.Second part: yes, it is necessary.Write \(\displaystyle 10^4 = 2^4 \times 5^4\), so \(\displaystyle x = \dfrac{p}{2^4 \times 5^4}\). To reduce this to lowest form we cancel \(\displaystyle g = \gcd(p, 10^4)\); the lowest-form denominator is \(\displaystyle \dfrac{10^{4}}{g}\). Since \(\displaystyle 10^4\) has only the primes $\displaystyle 2$ and $\displaystyle 5$, \(\displaystyle g = 2^a \times 5^b\) with \(\displaystyle 0 \le a \le 4\) and \(\displaystyle 0 \le b \le 4\).Now use \(\displaystyle 10 \nmid p\). If \(\displaystyle p\) had both a factor $\displaystyle 2$ and a factor $\displaystyle 5$, it would have the factor \(\displaystyle 2\times 5 = 10\) — which it does not. So at least one of "\(\displaystyle 2 \mid p\)", "\(\displaystyle 5 \mid p\)" fails, giving two cases.Case $\displaystyle 1$: \(\displaystyle p\) is odd. Then no $\displaystyle 2$ can be cancelled, so \(\displaystyle a = 0\) and \(\displaystyle g = 5^{b}\). The lowest-form denominator is \[\frac{2^4 \times 5^4}{5^{b}} = 2^{4} \times 5^{\,4-b}, \] which is divisible by \(\displaystyle 2^4 = 16\).Case $\displaystyle 2$: \(\displaystyle p\) is not a multiple of 5. Then \(\displaystyle b = 0\) and \(\displaystyle g = 2^{a}\). The lowest-form denominator is \[\frac{2^4 \times 5^4}{2^{a}} = 2^{\,4-a} \times 5^{4}, \] which is divisible by \(\displaystyle 5^4 = 625\).Every \(\displaystyle p\) with \(\displaystyle 10 \nmid p\) falls into at least one of these two cases, so the conclusion always holds. The reason in one line: cancelling can strip the 2s or the 5s, but never both, because \(\displaystyle p\) does not carry both factors.Some checks: \[0.0625 = \frac{625}{10^4} = \frac{1}{16},\qquad 16 = 2^4 \;\checkmark \] \[0.0016 = \frac{16}{10^4} = \frac{1}{625},\qquad 625 = 5^4 \;\checkmark \] \[0.1234 = \frac{1234}{10^4} = \frac{617}{5000},\qquad 5000 = 2^3 \times 5^4 \;\checkmark \] \[0.0001 = \frac{1}{10^4},\qquad 10^4 = 2^4 \times 5^4 \;\checkmark \text{ (both)} \]Answer. Such a number is \(\displaystyle x = \dfrac{p}{10^{4}}\) with \(\displaystyle p\) an integer whose units digit is the non-zero 4th decimal digit, so \(\displaystyle 10 \nmid p\). And yes — in lowest form its denominator must be divisible by \(\displaystyle 2^4\) or by \(\displaystyle 5^4\) (by both when \(\displaystyle p\) is coprime to $\displaystyle 10$), because \(\displaystyle p\) cannot supply factors of $\displaystyle 2$ and of $\displaystyle 5$ at the same time.