Key Concepts & Formulae at a Glance
- Natural numbers: \(\mathbb{N} = \{1, 2, 3, \ldots\}\), born from one-to-one correspondence (tally marks).
- Brahmagupta's rules for zero: \(a+0=a\), \(a-0=a\), \(a \times 0 = 0\).
- Integers: \(\mathbb{Z} = \{\ldots,-2,-1,0,1,2,\ldots\}\). Fortune (+) and debt (−) rules: fortune + fortune = fortune; debt + debt = debt; debt × fortune = debt; debt × debt = fortune.
- Rational numbers: \(\mathbb{Q} = \left\{\dfrac{p}{q} : p, q \in \mathbb{Z},\ q \neq 0\right\}\). Equality: \(\dfrac{a}{b}=\dfrac{c}{d}\) iff \(ad=bc\).
- Rational numbers are dense: the average \(\dfrac{a+b}{2}\) always lies strictly between any two distinct rationals \(a\) and \(b\).
- Irrational numbers cannot be written as \(\dfrac{p}{q}\); e.g. \(\sqrt{2}, \sqrt{3}, \pi\). First proved for \(\sqrt{2}\) by Hippasus using proof by contradiction.
- Real numbers: \(\mathbb{R} = \mathbb{Q} \cup \mathbb{I}\) (rationals united with irrationals) form the complete, unbroken number line.
- Decimal test: \(\dfrac{p}{q}\) (lowest terms) has a terminating decimal iff the only prime factors of \(q\) are 2 and/or 5; otherwise it is non-terminating repeating.
- Cyclic numbers: the repeating block of \(\frac{1}{7}\) (142857) cyclically permutes when multiplied by 1–6 — a hallmark of full-reptend primes.
Exercise Set 3.1
1A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?
The rate is 15 ingots for every 2 bags, so the number of ingots per bag is \(\dfrac{15}{2}\).
For 12 bags, the total number of ingots is:
\(\dfrac{15}{2} \times 12 = 15 \times 6 = 90\)
2Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.
Checking each number: 11, 13, 17 and 19 have no factors other than 1 and themselves — they are all prime numbers.
Continuing the sequence of primes after 19: the next three primes are 23, 29 and 31 (20, 21, 22, 24, ... , 30 are all composite, e.g. \(20=2\times10\), \(21=3\times7\), \(22=2\times11\), and so on).
3We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.
No, natural numbers are not closed under subtraction.
Example 1: \(3 - 5 = -2\), and \(-2 \notin \mathbb{N}\) since \(\mathbb{N} = \{1,2,3,\ldots\}\) contains no negative numbers.
Example 2: \(4 - 4 = 0\), and \(0 \notin \mathbb{N}\) either, since our natural numbers start from 1.
4*Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?
Each finger has 3 countable joints.
One hand has 4 fingers (excluding the thumb), and each finger has 3 joints:
Total joints \(= 4 \times 3 = 12\)
The thumb is used to touch each of these 12 joints in turn, so a person can count all the way up to 12 using just one hand.
This directly explains why several ancient civilisations adopted base-12 (duodecimal) counting systems: since one full hand naturally counts out a "dozen," units based on 12 became convenient — this is the same reason we still have 12 inches in a foot, 12 months in a year, and 12 hours on a clock face.
3.3.1 The Arithmetic of Integers
Think and Reflect
TRWhy does a negative times a negative equal a positive? Think of it in terms of action and debt.
Think of a negative number as representing a debt, and multiplication by a negative number as the removal (taking away) of that debt, rather than the addition of it.
Suppose you have four debts, each worth ₹3, i.e. each debt is \(-3\). Having these four debts means your total is \(4 \times (-3) = -12\) (you owe ₹12).
Now suppose someone takes away (removes) these four debts of ₹3 each. Removing a debt is the opposite of having it, so this action is represented by multiplying by \(-4\) (i.e. "negative four debts," or the removal of four debts):
\((-3) \times (-4)\)
Since each debt of ₹3 that is removed makes you ₹3 richer, removing 4 such debts makes you \(4 \times 3 = 12\) rupees richer, i.e. the result is \(+12\).
Exercise Set 3.2
1The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight temperature?
Temperature drops from 4°C at noon to −11°C by midnight.
Noon temperature = 4 °C. A drop of 15 °C means we subtract 15 from the noon temperature:
Midnight temperature \(= 4 - 15 = -11\) °C
2A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.
Running balance: start at 0 → −850 (loan) → +350 (after profit) → −100 (final, after loss).
Representing the loan as a debt (negative), the profit as a fortune (positive), and the loss as a debt (negative):
\((-850) + 1200 + (-450)\)
\(= -850 + 1200 - 450\)
\(= 350 - 450\)
\(= -100\)
3Calculate the following using Brahmagupta's laws: (i) (–12) × 5 (ii) (–8) × (–7) (iii) 0 – (–14) (iv) (–20) ÷ 4
(i) The product of a debt and a fortune is a debt: \((-12) \times 5 = -60\)
(ii) The product of two debts is a fortune: \((-8) \times (-7) = 56\)
(iii) Subtracting a debt is the same as adding a fortune: \(0 - (-14) = 0 + 14 = 14\)
(iv) A debt divided by a fortune is a debt: \((-20) \div 4 = -5\)
4Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g., 10 – (–5) = 15).
Suppose you have ₹10, but you also owe someone a debt of ₹5 (this debt is represented as \(-5\)).
Now imagine that debt of ₹5 is forgiven — it is taken away from your situation. Removing a debt of ₹5 has the same effect as gaining ₹5, because you no longer have to pay it back.
So starting with ₹10 and having a debt of ₹5 removed is written as \(10 - (-5)\), and the result is the same as if you had simply been given ₹5 more: \(10 + 5 = 15\).
Think and Reflect
TRCan you explain why we need q ≠ 0 in the definition of a rational number?
A rational number is defined as \(\dfrac{p}{q}\), which represents dividing \(p\) into \(q\) equal parts.
Division by zero is undefined — it is not mathematically possible to split a quantity into "zero parts," since there is no consistent value that \(\dfrac{p}{0}\) could equal (for any \(p \neq 0\), no number times 0 gives back \(p\)).
TR1. While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal? 2. Verify the distributive law for rational numbers.
1. Find the LCM (Least Common Multiple) of the two denominators. Then convert each fraction into an equivalent fraction with this LCM as the common denominator, by multiplying its numerator and denominator by the same suitable factor. Once both fractions share the same denominator, simply add or subtract their numerators, keeping the common denominator unchanged.
2. The distributive law states \(p(q+r) = pq+pr\). Let us verify it using \(p=\dfrac{1}{2}\), \(q=\dfrac{1}{3}\), \(r=\dfrac{1}{4}\).
LHS: \(q+r = \dfrac{1}{3}+\dfrac{1}{4} = \dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\), so \(p(q+r) = \dfrac{1}{2} \times \dfrac{7}{12} = \dfrac{7}{24}\)
RHS: \(pq = \dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}=\dfrac{4}{24}\); \(pr=\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1}{8}=\dfrac{3}{24}\); so \(pq+pr = \dfrac{4}{24}+\dfrac{3}{24}=\dfrac{7}{24}\)
Exercise Set 3.3
1Prove that the following rational numbers are equal: (i) 2/3 and 4/6 (ii) 5/4 and 10/8 (iii) −3/5 and −6/10 (iv) 9/3 and 3
Two rationals \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\) are equal if \(ad = bc\).
(i) \(\dfrac{2}{3}\) and \(\dfrac{4}{6}\): \(2 \times 6 = 12\) and \(3 \times 4 = 12\). Since \(12=12\), they are equal.
(ii) \(\dfrac{5}{4}\) and \(\dfrac{10}{8}\): \(5 \times 8 = 40\) and \(4 \times 10 = 40\). Since \(40=40\), they are equal.
(iii) \(-\dfrac{3}{5}\) and \(-\dfrac{6}{10}\): \((-3)\times 10 = -30\) and \(5 \times (-6) = -30\). Since \(-30=-30\), they are equal.
(iv) \(\dfrac{9}{3}\) and \(3\): writing \(3\) as \(\dfrac{3}{1}\), we check \(9 \times 1 = 9\) and \(3 \times 3 = 9\). Since \(9=9\), they are equal.
2Find the sum: (i) 2/5 + 3/10 (ii) 7/12 + 5/8 (iii) −4/7 + 3/14
(i) LCM of 5 and 10 is 10: \(\dfrac{2}{5}+\dfrac{3}{10} = \dfrac{4}{10}+\dfrac{3}{10}=\dfrac{7}{10}\)
(ii) LCM of 12 and 8 is 24: \(\dfrac{7}{12}+\dfrac{5}{8} = \dfrac{14}{24}+\dfrac{15}{24}=\dfrac{29}{24}\)
(iii) LCM of 7 and 14 is 14: \(-\dfrac{4}{7}+\dfrac{3}{14} = -\dfrac{8}{14}+\dfrac{3}{14}=-\dfrac{5}{14}\)
3Find the difference: (i) 5/6 − 1/4 (ii) 11/8 − 3/4 (iii) −7/9 − (−2/3)
(i) LCM of 6 and 4 is 12: \(\dfrac{5}{6}-\dfrac{1}{4} = \dfrac{10}{12}-\dfrac{3}{12}=\dfrac{7}{12}\)
(ii) LCM of 8 and 4 is 8: \(\dfrac{11}{8}-\dfrac{3}{4} = \dfrac{11}{8}-\dfrac{6}{8}=\dfrac{5}{8}\)
(iii) \(-\dfrac{7}{9}-\left(-\dfrac{2}{3}\right) = -\dfrac{7}{9}+\dfrac{2}{3}\). LCM of 9 and 3 is 9: \(-\dfrac{7}{9}+\dfrac{6}{9}=-\dfrac{1}{9}\)
4Find the product: (i) 2/3 × 3/10 (ii) 7/11 × 5/8 (iii) −4/7 × 5/14
(i) \(\dfrac{2}{3}\times\dfrac{3}{10} = \dfrac{6}{30}=\dfrac{1}{5}\)
(ii) \(\dfrac{7}{11}\times\dfrac{5}{8} = \dfrac{35}{88}\)
(iii) \(-\dfrac{4}{7}\times\dfrac{5}{14} = -\dfrac{20}{98}=-\dfrac{10}{49}\)
5Find the quotient: (i) 2/3 ÷ 3/10 (ii) 7/11 ÷ 5/8 (iii) −4/7 ÷ 5/14
(i) \(\dfrac{2}{3}\div\dfrac{3}{10} = \dfrac{2}{3}\times\dfrac{10}{3}=\dfrac{20}{9}\)
(ii) \(\dfrac{7}{11}\div\dfrac{5}{8} = \dfrac{7}{11}\times\dfrac{8}{5}=\dfrac{56}{55}\)
(iii) \(-\dfrac{4}{7}\div\dfrac{5}{14} = -\dfrac{4}{7}\times\dfrac{14}{5}=-\dfrac{56}{35}=-\dfrac{8}{5}\)
6Show that: (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3.
LHS: \(\dfrac{1}{2}+\dfrac{3}{4} = \dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\), so \(\left(\dfrac{5}{4}\right)\times\dfrac{8}{3} = \dfrac{40}{12}=\dfrac{10}{3}\)
RHS: \(\dfrac{1}{2}\times\dfrac{8}{3}=\dfrac{8}{6}=\dfrac{4}{3}\); \(\dfrac{3}{4}\times\dfrac{8}{3}=\dfrac{24}{12}=2\); sum \(=\dfrac{4}{3}+2=\dfrac{4}{3}+\dfrac{6}{3}=\dfrac{10}{3}\)
7Simplify the following using the distributive property: 7/9 (6/7 − 3/4).
\(\dfrac{7}{9}\left(\dfrac{6}{7}-\dfrac{3}{4}\right) = \dfrac{7}{9}\times\dfrac{6}{7} - \dfrac{7}{9}\times\dfrac{3}{4}\)
\(= \dfrac{6}{9} - \dfrac{21}{36} = \dfrac{2}{3} - \dfrac{7}{12}\)
LCM of 3 and 12 is 12: \(\dfrac{8}{12}-\dfrac{7}{12}=\dfrac{1}{12}\)
8Find the rational number x such that: 5/6 (x + 3/5) = 5/6 x + 1/2.
Expanding the LHS using the distributive law:
\(\dfrac{5}{6}\left(x+\dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{5}{6}\times\dfrac{3}{5} = \dfrac{5}{6}x + \dfrac{15}{30} = \dfrac{5}{6}x+\dfrac{1}{2}\)
This is exactly equal to the RHS, \(\dfrac{5}{6}x+\dfrac{1}{2}\), for every value of x.
3.4.1 Representation of Rational Numbers on the Number Line
Think and Reflect
TRTry and represent 8/5 and −7/4 on a number line.
\(\dfrac{8}{5} = 1\dfrac{3}{5} = 1.6\), which lies between 1 and 2. Divide the unit interval between 1 and 2 into 5 equal parts, and mark the point 3 parts to the right of 1.
\(-\dfrac{7}{4} = -1\dfrac{3}{4} = -1.75\), which lies between −2 and −1. Divide the unit interval between −2 and −1 into 4 equal parts, and mark the point 3 parts to the right of −2 (equivalently, 1 part to the left of −1).
8/5 = 1.6 sits 3/5 of the way from 1 to 2; −7/4 = −1.75 sits 3/4 of the way from −2 to −1.
Exercise Set 3.4
1Represent the rational numbers 2/3, −5/4 and 1½ on a single number line.
\(\dfrac{2}{3} \approx 0.67\), lying between 0 and 1 — divide that unit into 3 equal parts and mark 2 parts from 0.
\(-\dfrac{5}{4} = -1\dfrac{1}{4} = -1.25\), lying between −2 and −1 — divide that unit into 4 equal parts and mark 1 part to the right of −2 (i.e. 3 parts left of −1).
\(1\dfrac{1}{2} = \dfrac{3}{2} = 1.5\), lying exactly halfway between 1 and 2.
2/3, −5/4 and 1½ marked on the same number line.
2Find three distinct rational numbers that lie strictly between −1/2 and 1/4.
Three rational numbers between −1/2 and 1/4 on a number line.
Converting to a common denominator of 4: \(-\dfrac{1}{2}=-\dfrac{2}{4}\) and \(\dfrac{1}{4}\) stays as is.
We need numbers strictly between \(-\dfrac{2}{4}=-0.5\) and \(\dfrac{1}{4}=0.25\). Three such numbers are: \(-\dfrac{1}{4}=-0.25\), \(0\), and \(\dfrac{1}{8}=0.125\) — all lie strictly between −0.5 and 0.25.
3Simplify the expression: (−1/4) + (5/12).
LCM of 4 and 12 is 12: \(-\dfrac{1}{4}+\dfrac{5}{12} = -\dfrac{3}{12}+\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
4A tailor has 15¾ metres of fine silk. If making one kurta requires 2¼ metres of silk, exactly how many kurtas can he make?
\(15\dfrac{3}{4} = \dfrac{63}{4}\) metres available; \(2\dfrac{1}{4} = \dfrac{9}{4}\) metres needed per kurta.
Number of kurtas \(= \dfrac{63}{4} \div \dfrac{9}{4} = \dfrac{63}{4}\times\dfrac{4}{9} = \dfrac{63}{9}=7\)
5Find three rational numbers between 3.1415 and 3.1416.
A number strictly between 3.1415 and 3.1416.
Any decimal numbers strictly between 3.1415 and 3.1416 will work — we can always find more by adding extra decimal digits.
Three such rational numbers: 3.14151, 3.14153 and 3.14158.
6*Can you think of other way(s) to find a rational number between any two rational numbers?
Besides taking the simple average \(\dfrac{a+b}{2}\), here are other approaches:
Common-denominator scaling: Convert both numbers to a common denominator with a much larger value (multiply numerator and denominator of both by the same large factor). This creates "room" between the numerators, and any integer numerator strictly between them (with the same large denominator) gives a valid rational number between the two.
Decimal insertion: Write both numbers to extra decimal places and insert any digit strictly between the corresponding digits (as done in Q5 above).
Weighted averages: Instead of the simple average, use \(\dfrac{ma+nb}{m+n}\) for any positive integers m, n — this also always lies between a and b (and gives a different value for each choice of m, n).
Think and Reflect
TRCan √2 be written as a rational number p/q?
No. As proved formally in the next subsection (3.5.1) using proof by contradiction, \(\sqrt{2}\) cannot be expressed in the form \(\dfrac{p}{q}\) for any integers p, q with \(q \neq 0\).
Assuming it could be written this way, and following the logical steps through, leads to a contradiction (both p and q would have to be even, violating the assumption that the fraction is in its lowest terms) — hence the assumption must be false.
TRTry to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?
Proof for √3:
Step 1: Assume \(\sqrt{3}\) is rational. Write \(\sqrt{3}=\dfrac{p}{q}\), where p, q are co-prime integers (no common factor other than 1) and \(q \neq 0\).
Step 2: Squaring both sides: \(3 = \dfrac{p^2}{q^2}\)
Step 3: Multiply both sides by \(q^2\): \(p^2 = 3q^2\)
Step 4: Since \(p^2\) is 3 times an integer, \(p^2\) is divisible by 3. Since 3 is prime, this means p itself must be divisible by 3. Let \(p=3k\) for some integer k.
Step 5: Substituting: \((3k)^2 = 3q^2 \Rightarrow 9k^2=3q^2 \Rightarrow q^2=3k^2\)
Step 6: This shows \(q^2\) is also divisible by 3, so q must be divisible by 3 as well.
Step 7 (Contradiction): Both p and q are divisible by 3, contradicting our assumption that they share no common factor other than 1.
Hence √3 is irrational.
Does the same approach work for √5, √7, √10?
Yes. The exact same proof structure works for \(\sqrt{5}\) and \(\sqrt{7}\), since 5 and 7 are also prime — the key step ("if a prime divides \(p^2\), it must divide p") applies directly to any prime number.
For \(\sqrt{10}\), 10 is not prime (\(10=2\times5\)), but the same idea still works: if \(p^2=10q^2\), then \(p^2\) is divisible by 2 (so p is even, \(p=2k\)); substituting gives \(4k^2=10q^2 \Rightarrow 2k^2=5q^2\), so \(q^2\) is even too, meaning q is even — again both p and q share the common factor 2, a contradiction. So \(\sqrt{10}\) is also irrational.
3.5.2 Construction of Length √n
Think and Reflect
TRWe have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?
We use the Baudhāyana–Pythagoras Theorem geometrically. If we construct a right triangle whose two legs both have rational lengths (for instance, both equal to 1 unit), the hypotenuse of that triangle will have length \(\sqrt{1^2+1^2}=\sqrt{2}\), which is irrational.
Since this hypotenuse is a real, physical line segment, we can pick it up with a compass (keeping one arm fixed at the origin) and swing an arc down onto the number line, marking exactly where a length of \(\sqrt{2}\) would fall.
TRTry to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
Constructing √3: Start from the √2 construction (O, A, B with OB = √2). At B, draw a new perpendicular to OB of length 1 unit, reaching a point C. By Pythagoras, \(OC^2 = OB^2+BC^2 = 2+1=3\), so \(OC=\sqrt{3}\). Swing this length onto the number line with a compass centred at O.
OA = AB = 1, so OB = √2; then BC = 1 perpendicular to OB gives OC = √3.
Constructing √5 (a quicker route): Draw OA = 2 units along the number line. At A, draw a perpendicular AB = 1 unit. Join OB. By Pythagoras, \(OB^2=2^2+1^2=4+1=5\), so \(OB=\sqrt{5}\). Transfer this length onto the number line with a compass centred at O.
General method — the Square Root Spiral: Repeat the process: at each stage, draw a new perpendicular segment of length 1 unit at the end of the previous hypotenuse, then join back to the origin O. If the previous hypotenuse had length \(\sqrt{k}\), the new hypotenuse has length \(\sqrt{k+1}\) (since \(\sqrt{k}^2+1^2=k+1\)). Starting from \(\sqrt{2}\) and repeating this \((n-2)\) more times produces a segment of length \(\sqrt{n}\) for any positive integer n — this is exactly the "square root spiral" shown later in Fig. 3.14.
3.6.1 Rational Decimals: Terminating and Repeating
Think and Reflect
TRTry to find the decimal expansions of 10/3 and 11/12. What do you observe about the repetition of the digits after the decimal point?
\(\dfrac{10}{3} = 3.3333\ldots = 3.\overline{3}\) — the digit 3 repeats immediately after the decimal point (pure repeating decimal).
\(\dfrac{11}{12}\): long division gives \(11 \div 12 = 0.91666\ldots = 0.91\overline{6}\) — here the digits "9" and "1" appear once (non-repeating), and only then does the "6" begin repeating.
Observation: \(\dfrac{10}{3}\) has denominator 3 (only the prime factor 3 — no 2s or 5s), so its decimal repeats immediately. \(\dfrac{11}{12}\) has denominator \(12=2^2\times3\) — it contains both a "terminating-friendly" factor (\(2^2\)) and a "repeating-causing" factor (3), so it first has some non-repeating digits (matching the power of 2) before settling into a repeating block (caused by the leftover factor of 3).
Predicting the Type of Decimal Expansion
Think and Reflect
TRThe decimal expansion of p/q will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?
A decimal terminates exactly when the fraction can be rewritten with a denominator that is a power of 10 (since \(\dfrac{k}{10^n}\) is simply the integer k with the decimal point moved n places).
Since \(10^n = 2^n \times 5^n\), a denominator can only be converted into a power of 10 by multiplying it by extra factors of 2 and/or 5 — this is only possible if the denominator's own prime factorisation already contains no primes other than 2 and 5.
If q contains any other prime factor (such as 3, 7, 11, ...), no amount of multiplying by 2s and 5s can ever cancel that prime out or turn the denominator into a pure power of 10. In long division, since the possible remainders are limited in number, a remainder must eventually repeat, and once it does, the digits loop forever — producing a repeating (non-terminating) decimal.
Exercise Set 3.5
1Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions.
\(\dfrac{7}{20}\): \(20 = 2^2\times5\), only prime factors 2 and 5 → terminating. Long division: \(7\div20 = 0.35\)
\(\dfrac{4}{15}\): \(15 = 3\times5\), has a factor of 3 → repeating. Long division: \(4\div15 = 0.2666\ldots = 0.2\overline{6}\)
\(\dfrac{13}{250}\): \(250 = 2\times5^3\), only prime factors 2 and 5 → terminating. Long division: \(13\div250 = 0.052\)
2Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?
Long division: \(\dfrac{1}{13} = 0.\overline{076923}\) — the repeating block is 076923 (6 digits).
\(\dfrac{2}{13} = 0.\overline{153846}\). Notice \(076923 \times 2 = 153846\) — the same 6 digits, cyclically shifted!
\(\dfrac{3}{13} = 0.\overline{230769}\), and \(076923 \times 3 = 230769\) — again the same digits, shifted.
\(\dfrac{4}{13} = 0.\overline{307692}\), and \(076923 \times 4 = 307692\) — the pattern continues.
3Classify the following numbers as rational or irrational: (i) √81 (ii) √12 (iii) 0.33333… (iv) 0.123451234512345… (v) 1.01001000100001… (vi) 23.560185612239874790120. Find the explicit fractions in case they are rational.
(i) \(\sqrt{81}=9\), an integer, which can be written as \(\dfrac{9}{1}\) — rational.
(ii) \(\sqrt{12}=2\sqrt{3}\); since 12 is not a perfect square, \(\sqrt{12}\) is irrational.
(iii) \(0.33333\ldots = 0.\overline{3}\) is a pure repeating decimal — rational. Let \(x=0.\overline{3}\): \(10x=3.\overline{3}\), so \(10x-x=3 \Rightarrow 9x=3 \Rightarrow x=\dfrac{3}{9}=\dfrac{1}{3}\).
(iv) \(0.\overline{12345}\) has a repeating block "12345" — rational. Let \(x=0.\overline{12345}\): \(10^5x - x = 12345 \Rightarrow 99999x=12345 \Rightarrow x=\dfrac{12345}{99999}=\dfrac{4115}{33333}\) (dividing both by 3).
(v) \(1.01001000100001\ldots\) — the number of zeros between successive 1s keeps increasing (1, 2, 3, 4, …), so there is no fixed repeating block — irrational.
(vi) \(23.560185612239874790120\) — this decimal terminates after a finite number of digits, so it is rational. Written as a fraction: \(\dfrac{23560185612239874790120}{10^{21}}\) (an integer over a power of 10, since it terminates).
4The number 0.9̄ (which means 0.99999…) is a rational number. Using algebra (let x = 0.9̄, multiply by 10, and subtract), explain why 0.9̄ is exactly equal to 1.
Let \(x = 0.\overline{9} = 0.99999\ldots\)
Multiply both sides by 10: \(10x = 9.99999\ldots = 9.\overline{9}\)
Subtract the first equation from the second: \(10x - x = 9.\overline{9} - 0.\overline{9} = 9\)
\(9x = 9 \Rightarrow x = 1\)
5*We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.
The cyclic property of \(\dfrac{1}{7}\) (block 142857) and \(\dfrac{1}{13}\) (block 076923, from Q2 above) occurs for special primes called full reptend primes — primes n for which the repeating block of \(\dfrac{1}{n}\) has the maximum possible length, \(n-1\) digits.
Other such primes include 17 (\(\dfrac{1}{17}=0.\overline{0588235294117647}\), a 16-digit cyclic block), 19 (an 18-digit cyclic block), and 23 (a 22-digit cyclic block).
Think and Reflect
TRConsider this puzzle: What is the square root of –1? We know that 1 × 1 = 1. We also know that (–1) × (–1) = 1. There is no Real Number that, when multiplied by itself, results in a negative number. Thus, √–1 cannot exist on the number line.
Squaring any real number always gives a non-negative result: a positive number times itself gives a positive result, and a negative number times itself also gives a positive result (as established in Section 3.3.1). Zero squared is zero.
Since no real number's square can ever be negative, there is no real number x satisfying \(x^2=-1\), so \(\sqrt{-1}\) does not exist anywhere on the real number line.
To work with such quantities, mathematicians stepped outside the real number line entirely and defined a new number, denoted \(i\), such that \(i^2=-1\). These are called Imaginary Numbers, and they underpin modern electrical engineering and quantum mechanics.
End-of-Chapter Exercises
1Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division: (i) 3/50 (ii) 2/9
(i) \(50=2\times5^2\), only prime factors 2, 5 → terminating. Long division: \(3\div50 = 0.06\)
(ii) \(9=3^2\), has factor 3 → repeating. Long division: \(2\div9 = 0.2222\ldots = 0.\overline{2}\)
2Prove that √5 is an irrational number.
Step 1: Assume \(\sqrt{5}\) is rational. Write \(\sqrt{5}=\dfrac{p}{q}\), where p, q are co-prime integers and \(q\neq0\).
Step 2: Squaring: \(5=\dfrac{p^2}{q^2}\)
Step 3: Multiplying by \(q^2\): \(p^2=5q^2\)
Step 4: Since \(p^2\) is a multiple of 5, and 5 is prime, p must be a multiple of 5. Let \(p=5k\).
Step 5: Substituting: \(25k^2=5q^2 \Rightarrow q^2=5k^2\)
Step 6: This shows \(q^2\) is a multiple of 5, so q is also a multiple of 5.
Step 7 (Contradiction): Both p and q share the common factor 5, contradicting the assumption that they are co-prime.
3Convert the following decimal numbers in the form of p/q: (i) 12.6 (ii) 0.0120 (iii) 3.052 (iv) 1.23̄5̄ (v) 0.2̄3̄ (vi) 2.05̄ (vii) 2.125̄ (viii) 3.125̄ (ix) 2.1̄6̄2̄5̄
(i) \(12.6\) terminates: \(12.6=\dfrac{126}{10}=\dfrac{63}{5}\)
(ii) \(0.0120\) terminates: \(0.0120=\dfrac{120}{10000}=\dfrac{3}{250}\)
(iii) \(3.052\) terminates: \(3.052=\dfrac{3052}{1000}=\dfrac{763}{250}\)
(iv) \(x=1.23\overline{5}\) (digits "23" non-repeating, "5" repeating). \(100x=123.\overline{5}\); \(1000x=1235.\overline{5}\). Subtracting: \(1000x-100x=1235.\overline{5}-123.\overline{5}=1112 \Rightarrow 900x=1112 \Rightarrow x=\dfrac{1112}{900}=\dfrac{278}{225}\)
(v) \(x=0.\overline{23}\) (pure repeating, 2 digits). \(100x=23.\overline{23}\). Subtracting: \(99x=23 \Rightarrow x=\dfrac{23}{99}\)
(vi) \(x=2.0\overline{5}\) ("0" non-repeating, "5" repeating). \(10x=20.\overline{5}\); \(100x=205.\overline{5}\). Subtracting: \(90x=185 \Rightarrow x=\dfrac{185}{90}=\dfrac{37}{18}\)
(vii) \(x=2.12\overline{5}\) ("12" non-repeating, "5" repeating). \(100x=212.\overline{5}\); \(1000x=2125.\overline{5}\). Subtracting: \(900x=1913 \Rightarrow x=\dfrac{1913}{900}\)
(viii) \(x=3.12\overline{5}\), same structure as (vii) with integer part 3. \(1000x-100x=3125.\overline{5}-312.\overline{5}=2813 \Rightarrow 900x=2813 \Rightarrow x=\dfrac{2813}{900}\)
(ix) \(x=2.\overline{1625}\) (pure repeating, 4-digit block, right after the decimal point). \(10000x=21625.\overline{1625}\). Subtracting: \(9999x=21623 \Rightarrow x=\dfrac{21623}{9999}\)
4Locate the following rational numbers on the number line: (i) 0.532 (ii) 1.1̄5̄
(i) \(0.532 = \dfrac{532}{1000}=\dfrac{133}{250}\). It lies between 0 and 1, just past the halfway mark. To locate it exactly, divide the unit segment from 0 to 1 into 1000 equal parts and mark the 532nd part from 0.
(ii) \(x=1.1\overline{5}=1.15555\ldots\). As a fraction: \(10x=11.\overline{5}\), \(100x=115.\overline{5}\), so \(90x=104 \Rightarrow x=\dfrac{104}{90}=\dfrac{52}{45}=1\dfrac{7}{45}\). It lies between 1 and 2. To locate it, divide the unit segment from 1 to 2 into 45 equal parts and mark 7 parts to the right of 1.
0.532 (=133/250) sits just past halfway between 0 and 1; 1.1̄5̄ (=52/45) sits just past 1, between 1 and 2.
5Find 6 rational numbers between 3 and 4.
Six rational numbers between 3 and 4.
Write 3 and 4 with denominator 7 (any denominator ≥ 7 works): \(3=\dfrac{21}{7}\), \(4=\dfrac{28}{7}\).
The 6 integers strictly between 21 and 28 give us 6 rational numbers between 3 and 4:
6Find 5 rational numbers between 2/5 and 3/5.
Five rational numbers between 2/5 and 3/5.
Write both with a larger common denominator, 50: \(\dfrac{2}{5}=\dfrac{20}{50}\), \(\dfrac{3}{5}=\dfrac{30}{50}\).
Choose 5 integers strictly between 20 and 30, e.g. 21, 22, 23, 24, 25:
7Find 5 rational numbers between 1/6 and 2/5.
Five rational numbers between 1/6 and 2/5.
LCM of 6 and 5 is 30: \(\dfrac{1}{6}=\dfrac{5}{30}\), \(\dfrac{2}{5}=\dfrac{12}{30}\).
Choose 5 integers strictly between 5 and 12, e.g. 6, 7, 8, 9, 10:
8If x/3 + x/5 = 16/15, find the rational number x.
LCM of 3 and 5 is 15: \(\dfrac{5x}{15}+\dfrac{3x}{15}=\dfrac{16}{15}\)
\(\dfrac{8x}{15}=\dfrac{16}{15}\)
\(8x=16 \Rightarrow x=2\)
9Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.
\(a+\dfrac{1}{b}=0 \Rightarrow a=-\dfrac{1}{b}\)
Multiplying both sides by b: \(ab = -\dfrac{1}{b}\times b = -1\)
10A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.
Since the decimal terminates exactly at the 4th place (with a non-zero digit there), the number equals some integer p divided by \(10^4=10000\), i.e. the number \(=\dfrac{p}{10^4}\).
If p were divisible by 10, its last digit would be 0, which would mean the decimal actually terminates earlier (at the 3rd place or sooner), contradicting the given condition. So p is not divisible by 10 — i.e. p is not simultaneously divisible by both 2 and 5.
Is the reduced denominator necessarily divisible by 2⁴ or 5⁴? Not always both, and not always the full 2⁴ or 5⁴ either — it depends on p:
Since \(10^4=2^4\times5^4\), when we simplify \(\dfrac{p}{10^4}\) to lowest terms, we cancel out any factors of 2 or 5 that p itself has. If p is odd and not divisible by 5, no cancellation happens, and the denominator stays exactly \(2^4\times5^4\). If p happens to be even (but not divisible by 5, since it can't be divisible by both), some factors of 2 cancel and the reduced denominator will have a smaller power of 2 (but keep the full \(5^4\)). Similarly, if p is divisible by 5 (but not 2), the reduced denominator keeps the full \(2^4\) but loses some 5s.
11Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.
\(125=5^3\), only prime factor 5 → the decimal terminates.
Multiply numerator and denominator by \(2^3=8\) to make the denominator a power of 10: \(\dfrac{18}{125}=\dfrac{18\times8}{125\times8}=\dfrac{144}{1000}=0.144\)
12A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
Denominator \(=2^3\times5^1\). To convert this into a power of 10 (\(=2^n\times5^n\)), we need to match the higher of the two exponents (3 and 1), which is 3 — so multiply by an extra \(5^2\) to get \(2^3\times5^3=10^3\).
13*Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ − k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ − k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
Currently \(a=\dfrac{7}{12}\) and \(b=\dfrac{5}{6}=\dfrac{10}{12}\), so with \(m=12\), \(k_1=7\), \(k_2=10\), giving \(k_2-k_1=3\), which is not \(>6\).
Scale both fractions up by multiplying numerator and denominator by 8, so \(m=12\times8=96\):
\(a=\dfrac{7}{12}=\dfrac{56}{96}\), \(b=\dfrac{5}{6}=\dfrac{80}{96}\). Now \(k_1=56\), \(k_2=80\), and \(k_2-k_1=24>6\).
Choosing 5 integers strictly between 56 and 80, e.g. 60, 64, 68, 72, 76:
Five rational numbers between a and b: \(\dfrac{60}{96}=\dfrac{5}{8}\), \(\dfrac{64}{96}=\dfrac{2}{3}\), \(\dfrac{68}{96}=\dfrac{17}{24}\), \(\dfrac{72}{96}=\dfrac{3}{4}\), \(\dfrac{76}{96}=\dfrac{19}{24}\)
Why is k₂ − k₁ > n + 1 necessary? The number of integers lying strictly between \(k_1\) and \(k_2\) is \(k_2-k_1-1\). To be able to pick n distinct integers in that gap, we need \(k_2-k_1-1 \geq n\), i.e. \(k_2-k_1 \geq n+1\). If the gap \(k_2-k_1\) is too small, there simply aren't enough whole numbers between \(k_1\) and \(k_2\) to choose n distinct numerators from — so the denominator m must first be scaled up enough to widen this gap before the method can work.
14*Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
Recall the algebraic identity: \((x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)\)
Substituting \(x+y+z=0\) and \(xy+yz+zx=0\):
\(0^2 = x^2+y^2+z^2+2(0)\)
\(x^2+y^2+z^2=0\)
Since the square of any rational number is always non-negative, and the sum of three non-negative numbers can only be zero if each one individually is zero:
\(x^2=0,\ y^2=0,\ z^2=0 \Rightarrow x=0,\ y=0,\ z=0\)
15*Show that the rational number (a + b)/2 lies between the rational numbers a and b.
a, the midpoint (a+b)/2, and b on a number line — the midpoint always sits between the two.
Without loss of generality, assume \(a<b\) (the argument is symmetric if \(a>b\); if \(a=b\), the average trivially equals both).
Adding a to both sides of \(a<b\): \(a+a<a+b \Rightarrow 2a<a+b \Rightarrow a<\dfrac{a+b}{2}\)
Adding b to both sides of \(a<b\): \(a+b<b+b \Rightarrow a+b<2b \Rightarrow \dfrac{a+b}{2}<b\)
Combining both results: \(a<\dfrac{a+b}{2}<b\)
16Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.
Fig. 3.14 — the square root spiral: each new right triangle has one leg of length 1, built onto the previous hypotenuse.
Each right triangle in the spiral has one leg of length 1 unit, and its other leg is the hypotenuse of the previous triangle. If the previous hypotenuse has length \(\sqrt{k}\), the new hypotenuse is \(\sqrt{(\sqrt{k})^2+1^2}=\sqrt{k+1}\).
Starting from the first triangle (legs 1 and 1, hypotenuse \(\sqrt{1^2+1^2}=\sqrt{2}\)), each subsequent triangle's hypotenuse increases the number under the root by exactly 1:
| Triangle | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th | 8th | 9th |
|---|---|---|---|---|---|---|---|---|---|
| Hypotenuse | √2 | √3 | √4 = 2 | √5 | √6 | √7 | √8 = 2√2 | √9 = 3 | √10 |
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