Class 9 Mathematics Chapter 3: The World of Numbers

A rational number is one that can be written as $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. Since $-6 = \dfrac{-6}{1}$, with both parts integers and the denominator non-zero, $-6$ is a rational number.

Taking the average, $\dfrac{\tfrac{2}{5} + \tfrac{4}{5}}{2} = \dfrac{\tfrac{6}{5}}{2} = \tfrac{3}{5}$. Since $\tfrac{2}{5} < \tfrac{3}{5} < \tfrac{4}{5}$, it does lie between them.

Their sum is $-\tfrac{2}{6} + \tfrac{1}{6} = -\tfrac{1}{6}$, so the average is $-\tfrac{1}{12}$. Checking, $-\tfrac{1}{3} < -\tfrac{1}{12} < \tfrac{1}{6}$.

Dividing, 3 goes into 23 seven times with 2 left over, so $\tfrac{23}{3} = 7\tfrac{2}{3}$. The whole part is 7 and the fractional part is less than 1, so the number lies between 7 and 8.

A fraction $\tfrac{p}{0}$ would be asking which number multiplied by zero gives $p$. Every number multiplied by zero gives zero, so for $p \neq 0$ no number works at all, and for $p = 0$ every number would work. Neither case gives one definite value, so the expression is left undefined.

Let $m = \dfrac{a + b}{2}$. Since $a < b$, adding $a$ to both sides gives $2a < a + b$, and halving gives $a < m$. Again from $a < b$, adding $b$ to both sides gives $a + b < 2b$, and halving gives $m < b$. So $a < m < b$, which is what was required.

Write the two numbers as $\tfrac{p}{q}$ and $\tfrac{r}{s}$ with all four letters integers and $q$ and $s$ non-zero. Their sum is $\dfrac{ps + qr}{qs}$, whose top and bottom are both integers and whose bottom is non-zero, so the sum is rational. Halving multiplies the denominator by 2, which is still an integer and still non-zero. So the average is a ratio of integers with a non-zero denominator, and is therefore rational.

Over a denominator of 5 the numerators are 1 and 2, which leave no room. Multiplying the top and bottom of each by 8 gives $\tfrac{8}{40}$ and $\tfrac{16}{40}$. The numerators 9, 10, 11 and 12 all sit between 8 and 16, so $\tfrac{9}{40}$, $\tfrac{10}{40}$, $\tfrac{11}{40}$ and $\tfrac{12}{40}$ are four such numbers.

Dividing, $\tfrac{51}{4} \div \tfrac{3}{4} = \tfrac{51}{4} \times \tfrac{4}{3} = \tfrac{51}{3} = 17$ pieces exactly. Working in fractions keeps every value exact, so the answer comes out as a whole number with no remainder. Converting to 12.75 and 0.75 would have given the same result here but invites rounding in general, and a rounded division can turn an exact answer into an approximate one.

First convert to a mixed number: $\tfrac{7}{4} = 1\tfrac{3}{4}$. The denominator 4 says to divide each unit on the line into four equal parts. The minus sign means counting to the left of zero, so move one full unit to $-1$ and then three more quarter-parts in the same direction. The point therefore falls between $-1$ and $-2$.

(i) A number is rational when it can be written as $\dfrac{p}{q}$ with $p$ and $q$ integers and $q \neq 0$.

(ii) $12 = \tfrac{12}{1}$, $-4 = \tfrac{-4}{1}$ and $0 = \tfrac{0}{1}$. Each is a ratio of integers with a non-zero denominator, so all three are rational.

(iii) The average is $\dfrac{\tfrac{3}{7} + \tfrac{5}{7}}{2} = \dfrac{\tfrac{8}{7}}{2} = \tfrac{4}{7}$, and $\tfrac{3}{7} < \tfrac{4}{7} < \tfrac{5}{7}$.

(iv) Over 30 the two numbers are $\tfrac{5}{30}$ and $\tfrac{10}{30}$. The numerators 6, 7, 8 and 9 lie between, giving $\tfrac{6}{30}$, $\tfrac{7}{30}$, $\tfrac{8}{30}$ and $\tfrac{9}{30}$.

(v) Whatever number were proposed as the next one, the average of it and $\tfrac{1}{6}$ would be rational and would lie strictly between the two, so the proposed number was not immediately next after all. Since this works against any candidate, no such number exists.

(i) $\tfrac{77}{8} = 9\tfrac{5}{8}$ metres and $\tfrac{11}{8} = 1\tfrac{3}{8}$ metres.

(ii) $\tfrac{77}{8} \div \tfrac{11}{8} = \tfrac{77}{8} \times \tfrac{8}{11} = \tfrac{77}{11} = 7$ shelves.

(iii) Seven shelves use $7 \times \tfrac{11}{8} = \tfrac{77}{8}$ metres, which is the whole plank, so nothing is left over.

(iv) The second plank is $\tfrac{41}{4} = \tfrac{82}{8}$ metres. Dividing, $\tfrac{82}{8} \div \tfrac{11}{8} = \tfrac{82}{11}$, which is $7\tfrac{5}{11}$. So 7 whole shelves can be cut, using $\tfrac{77}{8}$ metres, and $\tfrac{82}{8} - \tfrac{77}{8} = \tfrac{5}{8}$ of a metre is left.

(v) Every length here has a denominator of 8 or 4, so the fractions stay exact and the division cancels neatly. As decimals the second plank would be 10.25 and the shelf 1.375, and the division gives 7.4545…, a value that must be interpreted rather than read off.

(i) The natural numbers are $1, 2, 3, \ldots$, the counting numbers, first used to record how many of something there were.

(ii) The whole numbers add zero to that list. Brahmagupta, in 628 CE, set out the rules for calculating with zero, making it a number in its own right rather than merely an absence.

(iii) The integers add the negatives, giving $\ldots, -2, -1, 0, 1, 2, \ldots$ Brahmagupta described the negative numbers as debts and the positive numbers as fortunes.

(iv) A rational number is one expressible as $\dfrac{p}{q}$ with $p$ and $q$ integers and $q \neq 0$. Every integer $n$ can be written $\dfrac{n}{1}$, which fits that description, so the integers all lie inside the rationals.

(v) Between any two different rational numbers there is always another rational number, and hence infinitely many. This is the density property, and it is why no rational number has a next one.

(i) The average of $\tfrac{1}{4}$ and $\tfrac{1}{2}$ is $\dfrac{\tfrac{3}{4}}{2} = \tfrac{3}{8}$. Averaging $\tfrac{1}{4}$ and $\tfrac{3}{8}$ next gives $\dfrac{\tfrac{5}{8}}{2} = \tfrac{5}{16}$.

(ii) Over 4 the fractions are $\tfrac{1}{4}$ and $\tfrac{2}{4}$, with no room between the numerators. Multiplying top and bottom of each by 6 gives $\tfrac{6}{24}$ and $\tfrac{12}{24}$, so $\tfrac{7}{24}$, $\tfrac{8}{24}$ and $\tfrac{9}{24}$ are three answers.

(iii) Averaging produces one new number per step, so three numbers take three separate calculations, each with a more awkward denominator than the last. Scaling produces as many as are wanted from a single conversion.

(iv) To obtain $n$ numbers the two numerators must differ by more than $n$, so that at least $n$ whole numbers lie strictly between them.

(v) Take $\tfrac{7}{24}$. Since $\tfrac{1}{4} = \tfrac{6}{24}$ and $\tfrac{1}{2} = \tfrac{12}{24}$, and $6 < 7 < 12$, the number $\tfrac{7}{24}$ does lie between the two given fractions.

The fraction is already in lowest terms, and $16 = 2^4$. Since the denominator holds no prime other than 2, the expansion terminates.

The fraction is in lowest terms, and $12 = 2^2 \times 3$. The prime 3 is neither a 2 nor a 5, so the expansion repeats.

There are three decimal places, so $0.048 = \tfrac{48}{1000}$. Dividing top and bottom by 8 gives $\tfrac{6}{125}$.

The first stops after two digits and equals $\tfrac{41}{100}$. The second has a bar over the block, meaning 41 repeats for ever as $0.414141\ldots$, and it equals $\tfrac{41}{99}$. They are different numbers.

In lowest terms the denominator is $160 = 2^5 \times 5$. There are five 2s and one 5, and the number of places is the larger of these counts, so the expansion runs to five places.

Let $x = 0.888\ldots$ The repeating block is one digit long, so multiply by 10 to shift the point past exactly one block: $10x = 8.888\ldots$ Subtracting the first line from the second, the endless tails are identical and cancel, leaving $9x = 8$. So $x = \tfrac{8}{9}$.

Let $x = 0.414141\ldots$ The block is two digits long, so the multiplier must be $10^2 = 100$, which shifts the point past exactly one whole block. Then $100x = 41.4141\ldots$, and subtracting gives $99x = 41$, so $x = \tfrac{41}{99}$. Choosing 10 instead would have left a mismatched tail and the subtraction would not have cancelled.

When $\tfrac{p}{q}$ is divided out, each step leaves a remainder that must be smaller than $q$, so it is one of $0, 1, 2, \ldots, q-1$. If a remainder of 0 ever appears, the division stops and the decimal terminates. If not, only $q - 1$ non-zero values are available, so within at most $q - 1$ steps some remainder must occur a second time. Each step depends only on the current remainder, so from that point everything repeats, and the decimal loops for ever.

The denominator test must be applied only after the fraction has been reduced to lowest terms, because cancelling can remove the prime in question. Here $\tfrac{24}{60}$ cancels by 12 to give $\tfrac{2}{5}$, and the 3 disappears with it. The reduced denominator 5 holds only the prime 5, so the expansion in fact terminates, at $0.4$.

A rational number $\tfrac{p}{q}$ can have a repeating block of up to $q - 1$ digits, because that is how many distinct non-zero remainders the division can pass through before one recurs. For a denominator such as 17 the block could be sixteen digits long, so ten digits are not nearly enough to have seen it through even once. A number is irrational only if its expansion never repeats at all, which no finite stretch of digits can establish.

(i) $20 = 2^2 \times 5$, which holds only 2s and 5s, so $\tfrac{7}{20}$ terminates.

(ii) $15 = 3 \times 5$. The prime 3 survives, so $\tfrac{4}{15}$ repeats.

(iii) $250 = 2 \times 5^3$, again only 2s and 5s, so $\tfrac{13}{250}$ terminates.

(iv) For $\tfrac{7}{20}$ there are two 2s and one 5, so the larger count is 2 and it runs to two places, giving $0.35$. For $\tfrac{13}{250}$ there is one 2 and three 5s, so the larger count is 3 and it runs to three places, giving $0.052$.

(v) After cancelling to lowest terms, the expansion terminates exactly when the denominator has no prime factor other than 2 and 5; the number of places is then the larger of the counts of 2s and 5s.

(i) The block is six digits long. Dividing by 7, each remainder must be one of 1, 2, 3, 4, 5 or 6, since 0 would end the division. There are only six such values, so a remainder must recur within six steps and the block cannot exceed six digits.

(ii) $142857 \times 2 = 285714$ and $142857 \times 3 = 428571$.

(iii) Both products use exactly the same six digits as the original, in the same cyclic order, but started at a different place.

(iv) The six remainders of sevenths form a single loop. Multiplying by 2 or 3 amounts to entering that same loop at a different remainder, so the same cycle of digits comes out, beginning elsewhere. Numbers behaving like this are called cyclic numbers.

(v) $\tfrac{2}{7} = 0.\overline{285714}$.

(i) Two decimal places, so $0.35 = \tfrac{35}{100}$, which cancels by 5 to $\tfrac{7}{20}$.

(ii) Let $x = 0.777\ldots$ Then $10x = 7.777\ldots$, and subtracting gives $9x = 7$, so $x = \tfrac{7}{9}$.

(iii) Let $x = 0.181818\ldots$ Then $100x = 18.1818\ldots$, and subtracting gives $99x = 18$, so $x = \tfrac{18}{99}$, which cancels by 9 to $\tfrac{2}{11}$.

(iv) The multiplier is the power of 10 matching the length of the repeating block: 10 for one digit, 100 for two, 1000 for three. That shifts the decimal point past exactly one whole block.

(v) Because the multiplier moves the point past exactly one block, the two lines have identical infinite tails after the point. Subtracting removes those tails completely and leaves a whole number, which is what makes the equation solvable.

(i) $\tfrac{35}{50}$ cancels by 5 to $\tfrac{7}{10}$. The fraction $\tfrac{7}{12}$ is already in lowest terms.

(ii) $10 = 2 \times 5$, holding only 2s and 5s, so $\tfrac{7}{10}$ terminates. $12 = 2^2 \times 3$, and the surviving 3 means $\tfrac{7}{12}$ repeats.

(iii) $\tfrac{7}{10} = 0.7$.

(iv) The uncancelled denominator 50 factorises as $2 \times 5^2$, which also holds only 2s and 5s, so the test happens to give the same verdict. That is luck: the factor cancelled away was a 5, which was harmless. Had it been any other prime the two tests would have disagreed.

(v) Take $\tfrac{14}{21}$. Uncancelled, the denominator 21 shows the primes 3 and 7 and the test says it repeats. Cancelled, the fraction is $\tfrac{2}{3}$, whose denominator still shows a 3, so it does repeat — a better example is $\tfrac{21}{35}$, whose uncancelled denominator $35 = 5 \times 7$ suggests repetition, while the reduced form $\tfrac{3}{5}$ terminates at $0.6$.

An irrational number is a number on the number line that cannot be written as a ratio of two integers. One example is $\sqrt2$: the assumption that it equals a fraction in simplest form leads to a contradiction, so no such fraction exists.

Since $14 \times 14 = 196$, the number $\sqrt{196}$ equals the integer 14 and so is rational. The number 20 is not a perfect square, so $\sqrt{20}$ cannot be written as a ratio of integers and is irrational.

The number $\pi$ was proved irrational by Lambert in 1761, so no ratio of integers can equal it. The value $\tfrac{22}{7}$ is only a convenient approximation, and writing an equals sign between them states something false.

Since $98 = 49 \times 2$, we have $\sqrt{98} = 7\sqrt2$. Because 98 is not a perfect square the value is irrational, and the factor $\sqrt2$ that remains confirms it.

An irrational number has a decimal expansion that never ends and never settles into a repeating block. A rational number always does one of those two things: either the division reaches a zero remainder and stops, or a remainder recurs and the digits loop for ever.

Mark $OA = 1$ unit along the line from the origin, and draw a line through $A$ perpendicular to it. On that perpendicular mark $B$ with $AB = 1$ unit, and join $OB$. By the Baudhāyana–Pythagoras theorem $OB^2 = 1^2 + 1^2 = 2$, so $OB = \sqrt2$ exactly. Finally, with centre $O$ and radius $OB$, draw an arc cutting the number line at $P$. Then $OP = \sqrt2$, and $P$ is the required point, placed exactly rather than approximately.

The proof ends by showing that $p$ and $q$ are both even, so they share a common factor of 2. That is only a contradiction if the fraction was assumed to have no common factor in the first place. Without the simplest-form condition, two even numbers would be entirely unremarkable and nothing would have been proved. The clash with that condition is the whole force of the argument.

The spiral is built from a chain of right triangles. The first has two legs of 1 unit, so its hypotenuse squares to 2. Each new triangle then uses the previous hypotenuse as one leg and a fresh unit length as the other, standing at right angles to it. So each new hypotenuse squares to the previous squared length plus 1. Starting at $1$ and adding 1 each time gives squared lengths $1, 2, 3, 4, \ldots$, so the $n$th spoke measures $\sqrt{n}$.

The word āsanna means an approximation, so he was signalling that the value was close but not exact, and he indicated that an exact fraction was unlikely to be found. Later mathematics confirmed this: Lambert proved in 1761 that $\pi$ is irrational, which means no ratio of integers whatsoever can equal it. Āryabhaṭa's caution was therefore not modesty but a correct judgement about the kind of number $\pi$ is.

Each term of the series is a fraction, and adding finitely many fractions always produces another fraction, so every partial sum is rational. But $\pi$ is irrational, so no partial sum can ever equal it exactly. That is why the series must be infinite: the totals get closer and closer to $\pi$ without any finite stage arriving at it, which is precisely what it means for an irrational number to need an endless process.

(i) Assume, for contradiction, that $\sqrt2$ is rational. Then $\sqrt2 = \dfrac{p}{q}$ where $p$ and $q$ are integers, $q \neq 0$, and the fraction is in its simplest form, so $p$ and $q$ share no common factor other than 1.

(ii) Squaring gives $2 = \dfrac{p^2}{q^2}$, and multiplying by $q^2$ gives $2q^2 = p^2$.

(iii) The equation shows $p^2$ is twice an integer, so $p^2$ is even. An odd number multiplied by an odd number is always odd, so an odd $p$ could never produce an even $p^2$. Hence $p$ is even, and we may write $p = 2k$ for some integer $k$.

(iv) Substituting gives $2q^2 = (2k)^2 = 4k^2$, and dividing by 2 gives $q^2 = 2k^2$. So $q^2$ is even, and by the same reasoning as before $q$ is even.

(v) Both $p$ and $q$ are even, so they share the common factor 2. This contradicts the assumption in (i) that the fraction was in its simplest form. Since every step after the assumption was forced, the assumption itself must be false, and therefore $\sqrt2$ is irrational.

(i) Since $16 \times 16 = 256$, we have $\sqrt{256} = 16$, an integer, so it is rational.

(ii) $27 = 9 \times 3$, so $\sqrt{27} = 3\sqrt3$. Because 27 is not a perfect square the value is irrational, and the surviving $\sqrt3$ shows it.

(iii) $\sqrt{n}$ is irrational exactly when $n$ is not a perfect square; when $n$ is a perfect square the root is an integer and therefore rational.

(iv) The squared hypotenuse is $9 + 16 = 25$, so the hypotenuse is 5. That is rational.

(v) The squared hypotenuse is $1 + 3 = 4$, so the hypotenuse is exactly 2, which is also rational. This is worth noting: an irrational leg does not force an irrational hypotenuse.

(i) $\pi$ is the ratio of a circle's circumference to its diameter, the same for every circle.

(ii) Āryabhaṭa, in 499 CE, gave $\pi \approx \dfrac{3927}{1250} = 3.1416$, and called it āsanna, meaning an approximation. He indicated that an exact fraction was unlikely to be found.

(iii) Lambert proved in 1761 that $\pi$ is irrational. That rules out every fraction, including $\tfrac{22}{7}$ and Āryabhaṭa's own value — no ratio of integers can equal $\pi$ exactly, however large its numbers.

(iv) Mādhava of Sangamagrama, who founded the Kerala School of Mathematics in the 14th century, gave $\pi = 4\left(1 - \tfrac{1}{3} + \tfrac{1}{5} - \tfrac{1}{7} + \cdots\right)$.

(v) It means taking the value that the running total gets closer and closer to as more and more terms are included. No finite stage equals $\pi$ — each is a sum of finitely many fractions and so is rational — but the totals approach it without limit.

(i) Between any two different rational numbers there is always another rational number, and hence infinitely many.

(ii) Density says only that no two rationals sit side by side with a gap between them. It says nothing about every point of the line being rational, and in fact some points are not.

(iii) The point at distance $\sqrt2$ from the origin is one. It can be constructed exactly as the diagonal of a unit square swung onto the line, so it is a genuine point; yet assuming $\sqrt2$ is a fraction in simplest form forces its numerator and denominator both to be even, which is a contradiction. So no rational number sits there.

(iv) The real numbers are the rational and irrational numbers taken together, and they fill the number line completely, leaving no gaps.

(v) A rational number's decimal expansion either terminates or repeats; an irrational number's never ends and never repeats. That test separates the two families exactly.

A rational number is one that can be written as $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. Since $-6 = \dfrac{-6}{1}$, with both parts integers and the denominator non-zero, $-6$ is a rational number.

Taking the average, $\dfrac{\tfrac{2}{5} + \tfrac{4}{5}}{2} = \dfrac{\tfrac{6}{5}}{2} = \tfrac{3}{5}$. Since $\tfrac{2}{5} < \tfrac{3}{5} < \tfrac{4}{5}$, it does lie between them.

Their sum is $-\tfrac{2}{6} + \tfrac{1}{6} = -\tfrac{1}{6}$, so the average is $-\tfrac{1}{12}$. Checking, $-\tfrac{1}{3} < -\tfrac{1}{12} < \tfrac{1}{6}$.

The fraction is already in lowest terms, and $16 = 2^4$. Since the denominator holds no prime other than 2, the expansion terminates.

The fraction is in lowest terms, and $12 = 2^2 \times 3$. The prime 3 is neither a 2 nor a 5, so the expansion repeats.

There are three decimal places, so $0.048 = \tfrac{48}{1000}$. Dividing top and bottom by 8 gives $\tfrac{6}{125}$.

An irrational number is a number on the number line that cannot be written as a ratio of two integers. One example is $\sqrt2$: the assumption that it equals a fraction in simplest form leads to a contradiction, so no such fraction exists.

Since $14 \times 14 = 196$, the number $\sqrt{196}$ equals the integer 14 and so is rational. The number 20 is not a perfect square, so $\sqrt{20}$ cannot be written as a ratio of integers and is irrational.

The number $\pi$ was proved irrational by Lambert in 1761, so no ratio of integers can equal it. The value $\tfrac{22}{7}$ is only a convenient approximation, and writing an equals sign between them states something false.