NCERT Solutions · Class 9 Mathematics The World of Numbers

Look at the prime factors of the denominator — no division needed.First check the fraction is in lowest form. Here \(\displaystyle 18 = 2 \times 3^2\) and \(\displaystyle 125 = 5^3\); they share no prime factor, so \(\displaystyle \dfrac{18}{125}\) is already in lowest form.The test for a terminating expansion is: a rational number \(\displaystyle \dfrac{p}{q}\) in lowest form has a terminating decimal expansion exactly when \(\displaystyle q\) can be written as \(\displaystyle 2^{m} \times 5^{n}\) — only 2s and 5s, nothing else. The reason is that a terminating decimal is a fraction over a power of $\displaystyle 10$, and \(\displaystyle 10^k = 2^k \times 5^k\) contains no prime other than $\displaystyle 2$ and 5.Here \[q = 125 = 5^3 = 2^{0} \times 5^{3}, \] so \(\displaystyle m = 0\), \(\displaystyle n = 3\). Only the primes $\displaystyle 2$ and $\displaystyle 5$ appear, hence the expansion terminates.How many decimal places. To turn the denominator into a power of $\displaystyle 10$ we must balance the 2s and the 5s. There are three 5s and no 2s, so multiply top and bottom by \(\displaystyle 2^3 = 8\): \[\frac{18}{125} \;=\; \frac{18 \times 8}{125 \times 8} \;=\; \frac{144}{1000} \;=\; 0.144 . \] That is $\displaystyle 3$ decimal places, and it cannot stop sooner: the numerator $\displaystyle 144$ is not a multiple of $\displaystyle 10$, so the digit in the third place ($\displaystyle 4$) is non-zero and no trailing zero can be dropped.In general, for \(\displaystyle q = 2^{m} \times 5^{n}\) the number of decimal places is \(\displaystyle \max(m,n)\). Here \(\displaystyle \max(0,3) = 3\).Answer. Terminating, with exactly $\displaystyle 3$ decimal places \(\displaystyle \left(\dfrac{18}{125} = 0.144\right)\).

Turn the denominator into a power of $\displaystyle 10$ by balancing the 2s and 5s.Let the number be \[x = \frac{p}{2^{3} \times 5} = \frac{p}{40}, \] in lowest form, so \(\displaystyle \gcd(p, 40) = 1\).A decimal with \(\displaystyle k\) places is a fraction with denominator \(\displaystyle 10^{k} = 2^{k} \times 5^{k}\) — equal numbers of 2s and 5s. Our denominator has three 2s but only one $\displaystyle 5$, so it is short of two 5s. Multiply the numerator and denominator by \(\displaystyle 5^{2} = 25\) (this does not change the value): \[x \;=\; \frac{p \times 25}{2^{3} \times 5 \times 5^{2}} \;=\; \frac{25p}{2^{3}\times 5^{3}} \;=\; \frac{25p}{1000}. \] A whole number divided by \(\displaystyle 1000\) is a decimal with at most $\displaystyle 3$ places.Why it is exactly $\displaystyle 3$, never fewer. The expansion would stop earlier only if \(\displaystyle 25p\) ended in a zero, i.e. if \(\displaystyle 10 \mid 25p\). But \(\displaystyle \gcd(p,40)=1\) forces \(\displaystyle p\) to be odd, and for odd \(\displaystyle p\), \[25p \equiv 5p \equiv 5 \pmod{10}, \] so \(\displaystyle 25p\) always ends in the digit 5. The third decimal place therefore holds a $\displaystyle 5$, which is not zero.Check with a few such numbers: \[\frac{1}{40}=0.025,\qquad \frac{3}{40}=0.075,\qquad \frac{7}{40}=0.175,\qquad \frac{23}{40}=0.575 . \] Every one has $\displaystyle 3$ decimal places and ends in 5.The general rule this illustrates: if the lowest-form denominator is \(\displaystyle 2^{m}\times 5^{n}\), the expansion has exactly \(\displaystyle \max(m,n)\) decimal places, because you must top up whichever prime is in short supply. Here \(\displaystyle m = 3\), \(\displaystyle n = 1\), so \(\displaystyle \max(3,1) = 3\).Answer. Exactly $\displaystyle 3$ decimal places.

Put both fractions over one denominator, then stretch that denominator until there is room in between.Step $\displaystyle 1$ — a common denominator. \(\displaystyle a = \dfrac{7}{12}\) and \(\displaystyle b = \dfrac{5}{6}\). Since \(\displaystyle \operatorname{lcm}(12,6)=12\), \[a = \frac{7}{12}, \qquad b = \frac{5}{6} = \frac{10}{12}. \] Here \(\displaystyle k_1 = 7\), \(\displaystyle k_2 = 10\), \(\displaystyle m = 12\), and \(\displaystyle k_2 - k_1 = 3\). That is not more than $\displaystyle 6$, and indeed only \(\displaystyle \dfrac{8}{12}\) and \(\displaystyle \dfrac{9}{12}\) sit between them — two numbers, not five.Step $\displaystyle 2$ — make the gap bigger. Multiplying the numerator and denominator of a fraction by the same non-zero integer \(\displaystyle t\) does not change its value, but it multiplies the numerators by \(\displaystyle t\): \[a = \frac{7t}{12t}, \qquad b = \frac{10t}{12t}, \qquad k_2 - k_1 = 10t - 7t = 3t . \] We need \(\displaystyle 3t > 6\), i.e. \(\displaystyle t > 2\). The smallest whole number that works is \(\displaystyle t = 3\): \[a = \frac{21}{36}, \qquad b = \frac{30}{36} \] so \(\displaystyle k_1 = 21\), \(\displaystyle k_2 = 30\), \(\displaystyle m = 36\) and \(\displaystyle k_2 - k_1 = 9 > 6\). \(\displaystyle \checkmark\)Step $\displaystyle 3$ — five rational numbers between \(\displaystyle a\) and \(\displaystyle b\). Keep the denominator $\displaystyle 36$ and take integer numerators strictly between $\displaystyle 21$ and $\displaystyle 30$: \[\frac{22}{36},\quad \frac{23}{36},\quad \frac{24}{36},\quad \frac{25}{36},\quad \frac{26}{36} \qquad\text{i.e.}\qquad \frac{11}{18},\;\; \frac{23}{36},\;\; \frac{2}{3},\;\; \frac{25}{36},\;\; \frac{13}{18}. \] Check: \(\displaystyle \dfrac{21}{36} < \dfrac{22}{36} < \dfrac{23}{36} < \dfrac{24}{36} < \dfrac{25}{36} < \dfrac{26}{36} < \dfrac{30}{36}\), and as decimals \[0.5833\ldots < 0.6111\ldots < 0.6388\ldots < 0.6666\ldots < 0.6944\ldots < 0.7222\ldots < 0.8333\ldots \;\checkmark \] (Any five of the eight numerators \(\displaystyle 22,23,\ldots,29\) would do; the choice is not unique.)Step $\displaystyle 4$ — why the gap between \(\displaystyle k_1\) and \(\displaystyle k_2\) has to be large. Once the denominator \(\displaystyle m\) is fixed, the only numbers this method can produce are \(\displaystyle \dfrac{k}{m}\) with \(\displaystyle k\) an integer. Such a number lies strictly between \(\displaystyle \dfrac{k_1}{m}\) and \(\displaystyle \dfrac{k_2}{m}\) precisely when \[k_1 < k < k_2, \qquad\text{that is}\qquad k = k_1+1,\; k_1+2,\; \ldots,\; k_2-1 . \] Counting these numerators gives \[(k_2 - 1) - (k_1 + 1) + 1 \;=\; k_2 - k_1 - 1 \text{ choices.} \] So the method can deliver \(\displaystyle n\) numbers only if \[k_2 - k_1 - 1 \;\ge\; n, \qquad\text{i.e.}\qquad k_2 - k_1 \;\ge\; n+1 . \] If the gap is \(\displaystyle n\) or less, there are at most \(\displaystyle n-1\) integers in between and the method simply cannot produce \(\displaystyle n\) numbers — you must first enlarge \(\displaystyle m\) (multiply top and bottom by a suitable \(\displaystyle t\)) to widen the gap. This is the whole point of Step 2.The condition quoted in the question, \(\displaystyle k_2 - k_1 > n+1\), is the strict form of this requirement. It leaves \(\displaystyle k_2-k_1-1 \ge n+1\) numerators available — at least one to spare — so \(\displaystyle n\) distinct numbers can certainly be chosen. (To be exact, the borderline case \(\displaystyle k_2-k_1 = n+1\) already gives exactly \(\displaystyle n\) numerators, just enough with nothing spare; taking the gap strictly greater than \(\displaystyle n+1\) is the safe version of the rule.) For \(\displaystyle n = 5\) it reads \(\displaystyle k_2 - k_1 > 6\), which is exactly the condition asked for in the first part, and our gap of $\displaystyle 9$ satisfies it.Answer. \(\displaystyle a = \dfrac{21}{36}\), \(\displaystyle b = \dfrac{30}{36}\) (so \(\displaystyle k_1 = 21,\;k_2=30,\;m=36,\;k_2-k_1 = 9 > 6\)); five rationals between them are \(\displaystyle \dfrac{22}{36},\dfrac{23}{36},\dfrac{24}{36},\dfrac{25}{36},\dfrac{26}{36}\). The gap condition is needed because, with the denominator fixed at \(\displaystyle m\), exactly \(\displaystyle k_2-k_1-1\) integer numerators lie strictly between \(\displaystyle k_1\) and \(\displaystyle k_2\), so a large enough gap is the only way to have \(\displaystyle n\) of them.

Square the first equation and use the second.Recall the identity for the square of a sum of three terms: \[(x+y+z)^2 \;=\; x^2 + y^2 + z^2 + 2(xy + yz + zx). \] (You can see this by multiplying \(\displaystyle (x+y+z)(x+y+z)\) out: you get the three squares once each, and each of the products \(\displaystyle xy, yz, zx\) twice.)Now substitute what we are given. Since \(\displaystyle x+y+z = 0\), the left side is \(\displaystyle 0^2 = 0\); and since \(\displaystyle xy+yz+zx = 0\), the last bracket is \(\displaystyle 0\). So \[0 \;=\; x^2 + y^2 + z^2 + 2 \times 0, \] which gives \[x^2 + y^2 + z^2 \;=\; 0 . \]Finish with the sign of a square. For any rational number \(\displaystyle t\), \(\displaystyle t^2 \ge 0\) — a square is never negative. So \(\displaystyle x^2, y^2, z^2\) are three numbers, none of them negative, whose total is \(\displaystyle 0\). If even one of them were positive, the total would be positive (the other two could not pull it down, since they cannot be negative). Hence each one must be zero: \[x^2 = 0, \qquad y^2 = 0, \qquad z^2 = 0, \] and therefore \[x = 0, \qquad y = 0, \qquad z = 0 . \]Check. \(\displaystyle x=y=z=0\) does satisfy both conditions: \(\displaystyle 0+0+0=0\) and \(\displaystyle 0+0+0=0\). \(\displaystyle \checkmark\)It is worth seeing that the second condition is doing real work. The triple \(\displaystyle x = 1,\, y = -1,\, z = 0\) satisfies \(\displaystyle x+y+z = 0\), but \[xy+yz+zx = (1)(-1) + (-1)(0) + (0)(1) = -1 \neq 0, \] so it is ruled out — exactly as the proof says it must be.Answer. \(\displaystyle x^2+y^2+z^2 = 0\) follows from the two conditions, and a sum of squares vanishes only when every term does, so \(\displaystyle x = y = z = 0\) is the only possibility.

Compare the middle number with each end by subtracting.For one number to lie between two others, the two others must be different, so take \(\displaystyle a \neq b\). There are two cases, and they are mirror images; suppose \(\displaystyle a < b\) (if instead \(\displaystyle b < a\), just swap the names \(\displaystyle a\) and \(\displaystyle b\) — the expression \(\displaystyle \dfrac{a+b}{2}\) is unchanged, since \(\displaystyle a+b = b+a\)).It is bigger than \(\displaystyle a\). \[\frac{a+b}{2} - a \;=\; \frac{a+b}{2} - \frac{2a}{2} \;=\; \frac{a+b-2a}{2} \;=\; \frac{b-a}{2}. \] Because \(\displaystyle a < b\) we have \(\displaystyle b - a > 0\), and dividing a positive number by \(\displaystyle 2\) keeps it positive. So \[\frac{a+b}{2} - a > 0 \qquad\Longrightarrow\qquad \frac{a+b}{2} > a . \]It is smaller than \(\displaystyle b\). \[b - \frac{a+b}{2} \;=\; \frac{2b}{2} - \frac{a+b}{2} \;=\; \frac{2b-a-b}{2} \;=\; \frac{b-a}{2} \;>\; 0, \] by the same reasoning. So \[\frac{a+b}{2} < b . \]Putting the two together, \[a \;<\; \frac{a+b}{2} \;<\; b . \]It really is a rational number. If \(\displaystyle a\) and \(\displaystyle b\) are rational then \(\displaystyle a+b\) is rational (the sum of two rationals is rational), and dividing a rational number by the non-zero integer \(\displaystyle 2\) again gives a rational number. So \(\displaystyle \dfrac{a+b}{2}\) is a rational number lying strictly between \(\displaystyle a\) and \(\displaystyle b\).Notice also that both differences came out equal to \(\displaystyle \dfrac{b-a}{2}\): the number \(\displaystyle \dfrac{a+b}{2}\) is the exact midpoint of \(\displaystyle a\) and \(\displaystyle b\), the same distance from each.Example. With \(\displaystyle a = \dfrac{7}{12}\) and \(\displaystyle b = \dfrac{5}{6}\): \[\frac{a+b}{2} = \frac{\dfrac{7}{12}+\dfrac{10}{12}}{2} = \frac{\dfrac{17}{12}}{2} = \frac{17}{24} = 0.708\overline{3}, \] and \(\displaystyle 0.583\overline{3} < 0.708\overline{3} < 0.833\overline{3}\). \(\displaystyle \checkmark\)(If \(\displaystyle a = b\), then \(\displaystyle \dfrac{a+b}{2} = \dfrac{2a}{2} = a = b\): it equals them both, and nothing can lie strictly between two equal numbers. This is why we assumed \(\displaystyle a \neq b\).)Answer. For \(\displaystyle a \neq b\), \(\displaystyle \dfrac{a+b}{2} - a = b - \dfrac{a+b}{2} = \dfrac{b-a}{2}\), so \(\displaystyle \dfrac{a+b}{2}\) is a rational number lying exactly halfway between \(\displaystyle a\) and \(\displaystyle b\).

Every new triangle stands on the previous hypotenuse and adds one more unit leg, so each hypotenuse squared grows by exactly 1.The square root spiral is built by repeating one single move. Start at a point \(\displaystyle O\) and mark \(\displaystyle P_1\) with \(\displaystyle OP_1 = 1\) unit. Then, again and again: at the outer end of the hypotenuse you have just drawn, put up a segment of length \(\displaystyle 1\) unit perpendicular to it, and join its far end back to \(\displaystyle O\). Each step makes a new right triangle whose two legs are "the previous hypotenuse" and "$\displaystyle 1$ unit".Apply Pythagoras' theorem to each triangle in turn.First triangle \(\displaystyle OP_1P_2\), right-angled at \(\displaystyle P_1\), legs \(\displaystyle OP_1 = 1\) and \(\displaystyle P_1P_2 = 1\): \[OP_2^{\,2} = 1^2 + 1^2 = 2 \qquad\Longrightarrow\qquad OP_2 = \sqrt{2}. \]Second triangle \(\displaystyle OP_2P_3\), right-angled at \(\displaystyle P_2\), legs \(\displaystyle OP_2 = \sqrt{2}\) and \(\displaystyle P_2P_3 = 1\): \[OP_3^{\,2} = \left(\sqrt{2}\right)^2 + 1^2 = 2 + 1 = 3 \qquad\Longrightarrow\qquad OP_3 = \sqrt{3}. \]Third triangle: \[OP_4^{\,2} = \left(\sqrt{3}\right)^2 + 1^2 = 3 + 1 = 4 \qquad\Longrightarrow\qquad OP_4 = \sqrt{4} = 2. \]Fourth triangle: \[OP_5^{\,2} = 4 + 1 = 5 \qquad\Longrightarrow\qquad OP_5 = \sqrt{5}, \] and so on.The pattern. Squaring removes the root, so the new leg of length $\displaystyle 1$ always contributes exactly \(\displaystyle 1^2 = 1\): \[OP_{k+2}^{\,2} \;=\; OP_{k+1}^{\,2} + 1 . \] Starting from \(\displaystyle OP_2^{\,2} = 2\), the squares run \(\displaystyle 2, 3, 4, 5, 6, \ldots\) — the whole numbers in order. Hence the hypotenuse of the \(\displaystyle k\)-th right triangle, counting outwards from the first, is \[OP_{k+1} \;=\; \sqrt{k+1}. \]So, reading the triangles outwards along the spiral, their hypotenuses are \[\sqrt{2},\;\; \sqrt{3},\;\; \sqrt{4}=2,\;\; \sqrt{5},\;\; \sqrt{6},\;\; \sqrt{7},\;\; \sqrt{8},\;\; \sqrt{9}=3,\;\; \sqrt{10},\;\; \ldots \] Fig. $\displaystyle 3.14$ shows only finitely many of them: if the spiral drawn there has \(\displaystyle N\) triangles, their hypotenuses are \(\displaystyle \sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots, \sqrt{N+1}\), taken in that order from the list above.Why the spiral is useful. Only the perfect-square entries give whole numbers — \(\displaystyle \sqrt{4}=2\), \(\displaystyle \sqrt{9}=3\), \(\displaystyle \sqrt{16}=4\), and so on. All the others, \(\displaystyle \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}, \ldots\), are irrational. The spiral therefore constructs, with nothing but a ruler and a set square, an actual line segment of each of these irrational lengths — which is how such numbers get their place on the number line.Answer. The hypotenuses, taken outwards from the innermost triangle, are \(\displaystyle \sqrt{2},\ \sqrt{3},\ \sqrt{4}=2,\ \sqrt{5},\ \sqrt{6},\ \sqrt{7},\ \sqrt{8},\ \sqrt{9}=3,\ \ldots\); in general the \(\displaystyle k\)-th right triangle of the spiral has hypotenuse \(\displaystyle \sqrt{k+1}\).