Question 1

A 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?

Accepted Answer

Unitary method — scale a fixed rate. The exchange rate is fixed: 15 ingots for every 2 bags. So the answer only depends on how many lots of 2 bags he brings. 12 bags ÷ 2 bags per lot = 6 lots Each lot is worth 15 ingots, so 6 × 15 = 90 ingots. Check by a second route (rate per single bag). One bag fetches 15/2 = 7.5 ingots, so 12 bags fetch 12 × 15/2 = 180/2 = 90. The two routes agree. Notice that the rate per bag is not a whole number, yet the total is — because 12 is a multiple of 2, the halves pair up. He will leave with 90 copper ingots.

Question 2

Look 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.

Accepted Answer

Test each number for factors. Check whether each number has any divisor other than 1 and itself. It is enough to test the prime divisors up to the square root of the number: a factor bigger than the square root must be paired with one smaller, and any composite factor is itself built from smaller primes. All four numbers here are below 25 = 5^2, so only 2 and 3 need to be tried. (The digit-sum test: a number is a multiple of 3 exactly when its digits add to a multiple of 3.) So every one of them is a prime number. In fact these four are all the primes between 10 and 20 — that is what makes this column of the Ishango bone so striking. (They also form two "twin" pairs, 11, 13 and 17, 19, each pair two apart.) Continuing the pattern. Take the numbers after 19 in turn: - 20, 22 — even, so divisible by 2. - 21 = 3 × 7. - 23 — not divisible by 2 or 3, and 5^2 = 25 23, so it is prime. - 24, 26, 28 — even. 25 = 5^2. 27 = 3^3. - 29 — not divisible by 2, 3, 5, and 6^2 = 36 29, so it is prime. - 30 — even. - 31 — not divisible by 2, 3, 5, and 6^2 = 36 31, so it is prime. All four are prime numbers. The next three primes are 23, 29 and 31.

Question 3

We 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.

Accepted Answer

Closure is broken by a single counterexample. "Closed under an operation" means: take any two members of the set, apply the operation, and the result must land back inside the same set. Because the claim is about every pair, one failing pair is enough to disprove it. The natural numbers are 1, 2, 3, 4, … Example 1. 3 and 8 are both natural numbers, but 3 - 8 = -5, and -5 is not a natural number — you cannot reach it by counting. Example 2. 7 and 7 are both natural numbers, but 7 - 7 = 0, and 0 is not a natural number either (counting starts at 1 ). Why it fails in general. For natural numbers a and b, the difference a - b is again natural only when a b. Since subtraction is allowed to take the numbers in either order, the set cannot be closed. This is exactly the gap that pushed mathematicians to invent 0 and the negative numbers; enlarging the set to the integers …, -2, -1, 0, 1, 2, … makes subtraction always possible, because every difference then lands back inside the set. (Your own counterexamples may differ from the ones above — any pair with a ≤ b works equally well. The conclusion, though, is the same for everyone.) No — the natural numbers are not closed under subtraction.

Question 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?

Accepted Answer

Count the joints, not the fingers. The thumb is doing the pointing, so it is the counter, not something counted. That leaves four fingers, and each of those has 3 joints (the creases where the finger bends): 4 × 3 = 12. You can count up to 12 on one hand. How this relates to base-12 (duodecimal) counting. A base is just the size of the "bundle" you use before you start again. Counting on the finger joints of one hand naturally produces bundles of 12, not 10 — which is why so many old trading systems counted in dozens: 12 items to a dozen, 12 months in a year, 12 inches in a foot, 12 hours on a clock face. Twelve also happens to be a very convenient base for a merchant, because it divides evenly in many ways: 12 = 2 × 6 = 3 × 4, so a dozen can be split into halves, thirds, quarters and sixths without breaking anything. Ten only splits into halves and fifths. And how base 60 appears. Use the other hand to keep track of how many complete dozens you have counted. Five fingers on that hand means five dozens: 5 × 12 = 60. That is one reason the Babylonian base-60 system took hold, and its traces are still with us: 60 seconds in a minute, 60 minutes in an hour, and 360 = 6 × 60 degrees in a full turn. One hand gives 12 counts — the origin of the dozen and of base-12; with the second hand tallying dozens you reach 60, the origin of base-60.

Question 5

The 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?

Accepted Answer

A drop is the addition of a negative number. "Drops by 15^° C" means the change is -15, so the midnight temperature is 4 + (-15) = 4 - 15. Working it out on the number line. Start at 4 and move 15 steps to the left. The first 4 steps bring you down to 0. That uses up 4 of the 15 steps, leaving 15 - 4 = 11 steps still to take, and those 11 steps go below zero: 4 - 15 = -11. Check. Reverse the change: if it is -11^° C at midnight and the day was 15^° warmer, then -11 + 15 = 4^° C. Correct. The midnight temperature is -11^° C (11 degrees below zero).

Question 6

A 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.

Accepted Answer

Debt is negative, fortune is positive. This is exactly Brahmagupta's language: a debt ( rina ) is written with a minus sign, a fortune ( dhana ) with a plus sign. Each event becomes one signed number: His final standing is the sum of all three: (-850) + 1200 + (-450) Step 1. (-850) + 1200 = 350. (The fortune is larger than the debt, so what is left is a fortune of 1200 - 850 = 350.) Step 2. 350 + (-450) = -100. (Now the debt is larger, so what is left is a debt of 450 - 350 = 100.) (-850) + 1200 + (-450) = -100 Check by grouping differently. Add the two debts first: (-850) + (-450) = -1300; then -1300 + 1200 = -100. Same answer, as it must be, since addition of integers is associative. His final standing is -100; that is, he ends up ₹100 in debt.

Question 7

Calculate the following using Brahmagupta's laws: (i) (-12) × 5 (ii) (-8) ×(-7) (iii) 0 - (-14) (iv) (-20) ÷ 4

Accepted Answer

Brahmagupta's sign laws. In the Brahmasphutasiddhanta (628 CE) Brahmagupta stated the rules we still use, in terms of fortunes (positive) and debts (negative): - fortune × fortune = fortune, and debt × debt = fortune; - fortune × debt = debt; - a debt subtracted from zero is a fortune; - the same sign rules hold for division. (i) (-12) × 5 — a debt times a fortune is a debt. Multiply the sizes, 12 × 5 = 60, then attach the minus sign: (-12) × 5 = -60 (Sense check: taking on a debt of 12 five times over is a debt of 60.) (ii) (-8) × (-7) — a debt times a debt is a fortune. Multiply the sizes, 8 × 7 = 56: (-8) × (-7) = +56 (iii) 0 - (-14) — a debt subtracted from zero becomes a fortune. Removing a debt of 14 leaves you 14 better off: 0 - (-14) = +14 (iv) (-20) ÷ 4 — a debt shared into 4 equal parts gives debts. Divide the sizes, 20 ÷ 4 = 5, and keep the minus sign: (-20) ÷ 4 = -5 (Check by multiplying back: (-5) × 4 = -20.) (i) -60 (ii) 56 (iii) 14 (iv) -5.

Question 8

Explain, 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 ).

Accepted Answer

Subtracting means removing — and removing a debt makes you richer. A ledger example. Suppose your accounts read like this: So your net worth today is ₹10. Now your friend says, "Forget the loan." That entry of -5 is removed from your books — in symbols, it is subtracted from your net worth: 10 - (-5). Nothing was handed to you, but the burden dragging your total down is gone. Your accounts now hold only the cash: new net worth = 15. Comparing the two lines, 10 - (-5) = 15 = 10 + 5. Cancelling a debt of 5 has exactly the same effect on you as being given a gift of 5. That is the whole meaning of the rule. A second way to see it — continue the pattern. Watch what happens to 10 - n as n walks down through zero: 10 - 3 = 7, 10 - 2 = 8, 10 - 1 = 9, 10 - 0 = 10, 10 - (-1) = 11, 10 - (-2) = 12, …, 10 - (-5) = 15. Each time the number being subtracted goes down by 1, the answer goes up by 1. For the pattern to stay unbroken as we cross zero, 10 - (-5) has to be 15. A third way — subtraction as "what must be added". By definition, 10 - (-5) is the number that must be added to -5 to get back to 10: (-5) + = 10 ⇒ = 15, since -5 + 15 = 10. (Your own real-world example may be different — a cancelled fine, a refunded penalty or a reversed bank charge all illustrate the same rule, and any of them is an equally good answer.) All three arguments agree: 10 - (-5) = 10 + 5 = 15. Subtracting a negative number is the same as adding the corresponding positive number, because removing a debt is a gain.

Question 9

Prove 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

Accepted Answer

Cross-multiplication test. Two rational numbers a/b and c/d (with b, d ≠ 0 ) are equal exactly when a × d = b × c. Why this works: multiplying both fractions by bd turns a/b into ad and c/d into bc, and multiplying both sides of an equality by the same non-zero number cannot change whether it is true. Equivalently, multiplying the numerator and denominator of a fraction by the same non-zero number does not change its value. (i) 2/3 and 4/6. 2 × 6 = 12, 3 × 4 = 12. Equal. Directly: 4/6 = (2 × 2)/(2 × 3) = 2/3. (ii) 5/4 and 10/8. 5 × 8 = 40, 4 × 10 = 40. Equal. Directly: 10/8 = (2 × 5)/(2 × 4) = 5/4. (iii) -3/5 and -6/10. (-3) × 10 = -30, 5 × (-6) = -30. Equal. Directly: -6/10 = -(2 × 3)/(2 × 5) = -3/5. (iv) 9/3 and 3. Write the whole number as a fraction, 3 = 3/1. Then 9 × 1 = 9, 3 × 3 = 9. Equal. Directly: 9/3 = (3 × 3)/(3 × 1) = 3/1 = 3. In all four cases the cross products match, so each pair names the same rational number.

Question 10

Find the sum: (i) 2/5+3/10 (ii) 7/12+5/8 (iii) -4/7+3/14

Accepted Answer

Common denominator, then add the numerators. Fractions can only be added once the pieces are the same size, so first rewrite each pair with the LCM of the denominators. (i) 2/5 + 3/10. LCM of 5 and 10 is 10. Multiply the first fraction above and below by 2: 2/5 = (2 × 2)/(5 × 2) = 4/10. 4/10 + 3/10 = (4 + 3)/10 = 7/10. 7 and 10 share no factor, so this is in lowest terms. (ii) 7/12 + 5/8. 12 = 2^2 × 3 and 8 = 2^3, so the LCM is 2^3 × 3 = 24. 7/12 = (7 × 2)/(12 × 2) = 14/24, 5/8 = (5 × 3)/(8 × 3) = 15/24. 14/24 + 15/24 = 29/24. 29 is prime, so no cancelling is possible. (As a mixed number, 15/24 — sensible, since 7/12 and 5/8 are each a little over a half.) (iii) -4/7 + 3/14. LCM of 7 and 14 is 14. -4/7 = -(4 × 2)/(7 × 2) = -8/14. -8/14 + 3/14 = (-8 + 3)/14 = -5/14 = -5/14. The answer is negative because the negative part is larger in size, which matches the estimate -0.571 + 0.214 ≈ -0.357. (i) 7/10 (ii) 29/24 (iii) -5/14.

Question 11

Find the difference: (i) 5/6-1/4 (ii) 11/8-3/4 (iii) -7/9-(-2/3)

Accepted Answer

Common denominator, then subtract the numerators. (i) 5/6 - 1/4. 6 = 2 × 3 and 4 = 2^2, so the LCM is 2^2 × 3 = 12. 5/6 = (5 × 2)/(6 × 2) = 10/12, 1/4 = (1 × 3)/(4 × 3) = 3/12. 10/12 - 3/12 = 7/12. Check by adding back: 7/12 + 3/12 = 10/12 = 5/6. Correct. (ii) 11/8 - 3/4. LCM of 8 and 4 is 8. 3/4 = (3 × 2)/(4 × 2) = 6/8. 11/8 - 6/8 = 5/8. (iii) -7/9 - (-2/3). Subtracting a negative is adding the positive, so -7/9 - (-2/3) = -7/9 + 2/3. LCM of 9 and 3 is 9, and 2/3 = 6/9: -7/9 + 6/9 = (-7 + 6)/9 = -1/9. Sense check: -7/9 ≈ -0.778 and -2/3 ≈ -0.667; the first is the smaller (further left), so the difference should be a small negative number, and -1/9 ≈ -0.111 fits. (i) 7/12 (ii) 5/8 (iii) -1/9.

Question 12

Find the product: (i) 2/3 × 3/10 (ii) 7/11 × 5/8 (iii) -4/7 × 5/14

Accepted Answer

Multiply numerators together and denominators together, cancelling common factors first. For rational numbers, a/b × c/d = (a × c)/(b × d). No common denominator is needed — that is only for addition. (i) 2/3 × 3/10. Cancel the 3 top and bottom, and cancel a factor 2 between 2 and 10: 2/3 × 3/10 = (2 × 3)/(3 × 10) = 6/30 = 1/5. (ii) 7/11 × 5/8. (7 × 5)/(11 × 8) = 35/88. 35 = 5 × 7 and 88 = 2^3 × 11 share no factor, so this is already in lowest terms. (iii) -4/7 × 5/14. A negative times a positive is negative. Multiply the sizes: (4 × 5)/(7 × 14) = 20/98, and 20 and 98 are both even, so divide top and bottom by 2: 20/98 = 10/49. Restoring the sign: -4/7 × 5/14 = -10/49. (i) 1/5 (ii) 35/88 (iii) -10/49.

Question 13

Find the quotient: (i) 2/3 ÷ 3/10 (ii) 7/11 ÷ 5/8 (iii) -4/7 ÷ 5/14

Accepted Answer

Dividing by a fraction means multiplying by its reciprocal. a/b ÷ c/d = a/b × d/c (c ≠ 0). Why: c/d × d/c = 1, so multiplying by d/c undoes multiplication by c/d — exactly what dividing is supposed to do. (i) 2/3 ÷ 3/10. 2/3 × 10/3 = 20/9. Check: 20/9 × 3/10 = 60/90 = 2/3. Correct. (ii) 7/11 ÷ 5/8. 7/11 × 8/5 = 56/55. 56 = 2^3 × 7 and 55 = 5 × 11 share no factor. (The answer is just over 1, which is right: 7/11 ≈ 0.636 is slightly bigger than 5/8 = 0.625.) (iii) -4/7 ÷ 5/14. -4/7 × 14/5 = -(4 × 14)/(7 × 5) = -56/35. Both 56 and 35 are divisible by 7: -56/35 = -8/5. Check: -8/5 × 5/14 = -40/70 = -4/7. Correct. (i) 20/9 (ii) 56/55 (iii) -8/5.

Question 14

Show that: (1/2+3/4) × 8/3=1/2 × 8/3+3/4 × 8/3.

Accepted Answer

Distributive property — evaluate each side separately and compare. Left-hand side. Do the bracket first. LCM of 2 and 4 is 4: 1/2 + 3/4 = 2/4 + 3/4 = 5/4. Then 5/4 × 8/3 = (5 × 8)/(4 × 3) = 40/12 = 10/3. Right-hand side. Work out the two products, then add. 1/2 × 8/3 = 8/6 = 4/3, 3/4 × 8/3 = 24/12 = 2. 4/3 + 2 = 4/3 + 6/3 = 10/3. Compare. LHS = 10/3 = RHS. What this shows. Multiplication distributes over addition for rational numbers, just as it does for whole numbers: a × (b + c) = a × b + a × c. Here a = 8/3, b = 1/2, c = 3/4, and the two sides came out equal — you may either add first and multiply once, or multiply twice and add. (In decimals both sides are 3. 3, another quick confirmation.) Hence (1/2+3/4) × 8/3=1/2 × 8/3+3/4 × 8/3 = 10/3, as required.

Question 15

Simplify the following using the distributive property: 7/9(6/7-3/4)

Accepted Answer

Distributive property: spread the 7/9 over both terms. 7/9(6/7-3/4) = 7/9 × 6/7 - 7/9 × 3/4 First product. The 7 s cancel: 7/9 × 6/7 = 6/9 = 2/3. Second product. 7/9 × 3/4 = 21/36 = 7/12 (dividing top and bottom by 3 ). Subtract. LCM of 3 and 12 is 12, and 2/3 = 8/12: 8/12 - 7/12 = 1/12. Check the other way round (bracket first). 6/7 - 3/4 = 24/28 - 21/28 = 3/28, 7/9 × 3/28 = 21/252 = 1 12. Both routes give the same value, which is the distributive property in action. 1/12.

Question 16

Find the rational number x such that: 5/6(x+3/5)=5/6 x+1/2.

Accepted Answer

Expand the bracket first, then compare the two sides. Apply the distributive property to the left-hand side: 5/6(x+3/5) = 5/6x + 5/6 × 3/5. Work out that constant term — the 5 s cancel: 5/6 × 3/5 = 15/30 = 1/2. So the left-hand side is 5/6x + 1/2, which is exactly the right-hand side, whatever x is. Subtracting 5/6x + 1/2 from both sides of the given equation leaves 0 = 0, a statement that is true for every value of x and contains no information about x. Verification with a few values. Every value works. So there is no single value to "find". The equation is simply the distributive property written out for 5/6, x and 3/5, and it holds only because the constant works out exactly right: 5/6 × 3/5 = 1/2. (Had the right-hand side read, say, 5/6x + 1/3, the equation would have reduced to 1/2 = 1/3 and had no solution at all.) The equation is an identity, not a condition on x: every rational number x satisfies it.

Question 17

Represent the rational numbers 2/3,-5/4 and 1 1/2 on a single number line.

Accepted Answer

Put all three numbers over one common denominator, then use that as the size of the small divisions. The three numbers use different denominators, so first make them comparable. Note 11/2 = 3/2. The LCM of 3, 4 and 2 is 12, so express each as twelfths: 2/3 = (2 × 4)/(3 × 4) = 8/12, -5/4 = -(5 × 3)/(4 × 3) = -15/12, 11/2 = 3/2 = 3 × 6 2 × 6 = 18 12. How to draw it. 1. Draw a straight line, mark a point 0 on it, and step off equal unit lengths to mark -2, -1, 0, 1, 2. (You need to reach -2 on the left because -5/4 lies beyond -1, and 2 on the right because 3/2 lies beyond 1.) 2. Divide every unit interval into 12 equal parts. Each small step is now worth 1/12, and all three numbers land exactly on one of these marks — this is the whole point of choosing the common denominator 12. 3. Count the small steps from 0, to the right for a positive number and to the left for a negative one: A quicker way to place each one, if you draw them separately. - 2/3: cut the stretch from 0 to 1 into 3 equal parts and take the 2 nd mark. - -5/4 = -11/4: go to -1, then cut the stretch from -1 to -2 into 4 equal parts and take 1 step further left. - 11/2: the midpoint of 1 and 2. Order on the line (left to right). -5/4 < 0 < 2/3 < 11/2, since as twelfths these are -15, 0, 8, 18. In decimals: -1.25, 0. 6, 1.5 — which confirms the order and the positions above. So on a single line marked in twelfths, -5/4 sits 15 small steps to the left of 0, 2/3 sits 8 small steps to the right, and 11/2 sits 18 small steps to the right (exactly midway between 1 and 2 ).

Question 18

Find three distinct rational numbers that lie strictly between -1/2 and 1/4.

Accepted Answer

Rewrite both numbers with the same denominator, then the numbers in between become easy to read off. The LCM of 2 and 4 is 4, but a denominator of 4 leaves only -1/4 and 0 strictly in between — enough here, but it is safer to make room. Multiply up to eighths instead: -1/2 = -4/8, 1/4 = 2/8. Now the question reads: which eighths lie strictly between -4/8 and 2/8? The numerators strictly between -4 and 2 are -3, -2, -1, 0, 1, giving -3/8, -2/8 = -1/4, -1/8, 0, 1/8. Choose any three of these, for instance -3/8, -1/8, 1/8. Check (convert to decimals and confirm the order). -1/2 = -0.5 < -3/8 = -0.375 < -1/8 = -0.125 < 1/8 = 0.125 < 1/4 = 0.25. All three lie strictly inside, and all three are distinct. Other answers are equally valid. There are in fact infinitely many rational numbers between any two distinct rational numbers: given any two of them you can always take their average (the midpoint), which lies strictly between, and repeat forever. For example the midpoint of -1/2 and 1/4 is 1/2(-1/2 + 1/4) = 1/2 × (-1/4) = -1/8, and repeating the trick generates as many as you like. So -1/4, 0, 1/5 would be just as correct an answer as the one above. Three such numbers: -3/8, -1/8, 1/8 (any three rationals strictly between -0.5 and 0.25 are acceptable).

Question 19

Simplify the expression: (-1/4)+(5/12).

Accepted Answer

Making the denominators the same. Two fractions can only be added when their pieces are the same size, so the first job is to rewrite both over a common denominator. The LCM of 4 and 12 is 12. -1/4 = -(1× 3)/(4× 3) = -3/12 (Multiplying top and bottom by the same number does not change a fraction's value — it only cuts each piece into 3 smaller ones.) Now the two fractions can be added by adding numerators: -3/12+5/12 = (-3+5)/12 = 2/12 Reduce to lowest terms by dividing numerator and denominator by their HCF, which is 2: 2/12 = (2÷ 2)/(12÷ 2) = 1/6 Check with decimals. - 14=-0.25 and 5/12=0.41 6; their sum is 0.1 6 = 16. Answer: 1/6.

Question 20

A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of silk, exactly how many kurtas can he make?

Accepted Answer

Division of mixed fractions. "How many lengths of 2 14 m fit inside 15 34 m?" is a division question: total length ÷ length per kurta. First turn both mixed fractions into improper fractions: 153/4=(15× 4+3)/4=63/4 m, 21/4=(2× 4+1)/4=9/4 m Divide by multiplying by the reciprocal: 63/4÷9/4=63/4×4/9=63/9=7 Why the answer is exact. Both lengths have denominator 4, i.e. both are a whole number of quarter-metres: the tailor has 63 quarter-metres and each kurta uses 9 of them. Since 9× 7=63 exactly, there is no silk left over — so "exactly how many" has a clean answer with no remainder to worry about. Check. 7× 2 14 = 7× 94=63/4=15 34 m. Answer: 7 kurtas, using up all the silk exactly (nothing left over).

Question 21

Find three rational numbers between 3.1415 and 3.1416.

Accepted Answer

Insert one more decimal place. The two numbers agree up to the fourth decimal place and differ only there. Write both with one extra decimal place, which does not change their values: 3.1415 = 3.14150, 3.1416 = 3.14160 Now the fifth decimal digit is free to be anything from 1 to 9, and every such choice gives a number squeezed between the two. Pick three of them: 3.14151, 3.14152, 3.14153 Why these are rational. Each one terminates, so each is a whole number over a power of ten: 3.14151=314151/100000, 3.14152=314152/100000, 3.14153=314153/100000 Check the order. 3.14150 < 3.14151 < 3.14152 < 3.14153 < 3.14160 The choice is not unique — 3.14155, 3.141592, and infinitely many others also work. (Note that 3.141592 is rational even though it looks like π: it stops, and π does not.) Answer: for example 3.14151, 3.14152, 3.14153, i.e. 314151/100000, 314152/100000, 314153/100000. Other answers are equally correct.

Question 22

Can you think of other way(s) to find a rational number between any two rational numbers?

Accepted Answer

Yes — several. This question is open: the methods below are all correct, and other correct methods exist. Take two rational numbers a<b. Way 1 — the average. The midpoint (a+b)/2 lies exactly halfway between them, so a<(a+b)/2<b. It is rational because adding two rationals and dividing by 2 can never leave the rationals. Way 2 — weighted points. Instead of halving, cut the gap into three (or any number of) equal parts: (2a+b)/3, (a+2b)/3 More generally (ma+nb)/(m+n) is rational and lies strictly between a and b for any positive integers m,n. This one method produces as many in-between numbers as you like in a single step. Way 3 — scale up the common denominator. Write a=p/q and b=r/q over the same denominator. If p and r are consecutive integers there is no room yet, so multiply top and bottom of both by 10: a=10p/10q, b=10r/10q Now nine integers sit strictly between 10p and 10r, giving nine fractions between a and b. Way 4 — decimal insertion. Write both as decimals and change a digit far enough to the right that the number stays inside the gap — the method used in Question 5. Way 5 — the mediant. If a=p/q and b=r/s with q,s 0, then the "wrong-looking" fraction (p+r)/(q+s) always lies strictly between them. For example between 14 and 5/12 the mediant is (1+5)/(4+12)=6/16= 38, and indeed 0.25<0.375<0.41 6. (Warning: this is a special trick for the mediant only — you may never add fractions this way.) One consequence. Whichever method you choose, it can be applied again to the new, smaller gap, and again, without ever running out. So between any two distinct rational numbers there are infinitely many rational numbers. Answer: yes — the average (a+b)/2, weighted means (ma+nb)/(m+n), scaling to a larger common denominator, inserting decimal digits, or the mediant (p+r)/(q+s). All are valid, so this answer is not unique.

Question 23

Without 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 and expressing these rational numbers as decimals.

Accepted Answer

Look at the prime factors of the denominator. Reduce the fraction to lowest terms first, then factorise the denominator. A decimal is just a fraction whose denominator is a power of ten, and 10=2× 5. So a fraction in lowest terms can be rewritten with a denominator 10^n — and therefore terminates — exactly when its denominator is built only from the primes 2 and 5. If any other prime ( 3,7,11,13,… ) survives in the denominator, no amount of multiplying can turn it into a power of ten, the long division never gives remainder 0, and the decimal repeats forever. Predictions. (All three are already in lowest terms: gcd (7,20)=1, gcd (4,15)=1, and 13 is prime and does not divide 250.) Now the long divisions. 7/20: 20 into 70 goes 3 times ( 60 ), remainder 10; 20 into 100 goes 5 times ( 100 ), remainder 0 — stop. 7/20=0.35 Shortcut check: 7/20=(7× 5)/(20× 5)=35/100=0.35. 4/15: 15 into 40 goes 2 ( 30 ), remainder 10; 15 into 100 goes 6 ( 90 ), remainder 10. The remainder 10 has come back, so the digit 6 must repeat for ever. 4/15=0.2666…=0.2 6 13/250: 250 into 130 goes 0, remainder 130; 250 into 1300 goes 5 ( 1250 ), remainder 50; 250 into 500 goes 2, remainder 0 — stop. 13/250=0.052 Shortcut check: 13/250=(13× 4)/(250× 4)=52/1000=0.052. Every prediction matched the division. Answer: 7/20=0.35 (terminating), 4/15=0.2 6 (non-terminating repeating), 13/250=0.052 (terminating).

Question 24

Perform 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?

Accepted Answer

Long division, watching the remainders. The remainders are what tell you when the block closes. Dividing 1 by 13: The remainder is 1 again — the number we started with — so from here every step repeats exactly. 1/13=0. 076923 repeating block 076923, length 6 Why the length is 6. Dividing by 13 can only ever leave remainders 1,2,…,12, and never 0 (no power of 10 is a multiple of 13 ). So the block can be at most 12 digits long. Here the remainders that actually appeared were 1,10,9,12,3,4 — six of the twelve — and then the cycle closed. Now 2/13: 20÷13→ 1 r 7; 70÷13→ 5 r 5; 50÷13→ 3 r 11; 110÷13→ 8 r 6; 60÷13→ 4 r 8; 80÷13→ 6 r 2 — back to the start. 2/13=0. 153846 Is 153846 a rotation of 076923? The rotations of 076923 are 076923, 769230, 692307, 923076, 230769, 307692 and 153846 is not among them. So 13 does not behave quite like 7. All twelve: What you notice. The behaviour is cyclic, but it splits into two families rather than one: - rotations of 076923: k=1, 3, 4, 9, 10, 12 - rotations of 153846: k=2, 5, 6, 7, 8, 11 No block from one family is a rotation of a block in the other, and the two families together use all twelve values of k. Compared with 17. There 1/7=0. 142857 has block length 6=7-1, so the single cycle uses up every possible remainder, and all six of 1/7,…,6/7 are rotations of one block. For 13 the block is only 6 long while there are 12 possible remainders, so the remainders have to split into two separate cycles of six. Answer: 1/13=0. 076923, repeating block 076923 of length 6; 2/13=0. 153846. The fractions k/13 are cyclic, but in two families of six rotations each — 1,3,4,9,10,12 rotate 076923 and 2,5,6,7,8,11 rotate 153846 — unlike 1/7, where all six form a single cycle.

Question 25

Classify the following numbers as rational or irrational: (i) √ 81 (ii) √ 12 (iii) 0.33333... (iv) 0.123451234512345... (v) 1.01001000100001... (Notice the pattern: Is it repeating a single block?) (vi) 23.560185612239874790120 Find the explicit fractions in case they are rational.

Accepted Answer

Test: a number is rational exactly when its decimal either stops or eventually repeats a fixed block. Anything non-terminating and non-repeating is irrational. (i) √ 81. Since 9× 9=81, √ 81 =9. An integer is a rational number. Rational, √ 81 =9=9/1. (ii) √ 12. First, √ 12 =√ 4× 3 =2√3, and 9<12<16 gives 3<√ 12 <4, so it is not a whole number. Suppose it were rational, say √ 12 = pq in lowest terms. Then p^2=12q^2, so 3 p^2, and since 3 is prime, 3 p. Write p=3m: then 9m^2=12q^2, i.e. 3m^2=4q^2. So 3 4q^2, and as 3 4 we get 3 q^2, hence 3 q. Now 3 divides both p and q, contradicting "lowest terms". Irrational ( ≈ 3.4641016… ). No fraction exists. (iii) 0.33333… — the block 3 repeats. Let x=0. 3. Then 10x=3. 3, and subtracting, 9x=3, so x= 39. Rational, =1/3. (iv) 0.123451234512345… — the block 12345 repeats, length 5. Let x=0. 12345. Then 100000x=12345. 12345, so 99999x=12345 and x=12345/99999=(12345÷ 3)/(99999÷ 3)=4115/33333 Rational, =4115/33333. (Check: 4115÷ 33333=0.123451234512345… ) (v) 1.01001000100001… — look at the runs of zeros: 1 zero, then 2, then 3, then 4, and so on. Is a single block repeating? No. If some block of length L repeated, the pattern would be periodic from some point on; but the runs of zeros keep growing and eventually exceed L, so no such block can exist. The decimal is non-terminating and non-repeating. Irrational. No fraction exists. (vi) 23.560185612239874790120. As printed this decimal stops after 21 places (there is no " … "), and every terminating decimal is rational: put the digits over the matching power of ten. 23.560185612239874790120=23560185612239874790120/(10^21) Both parts are divisible by 40 (the numerator ends in 120 ), giving =589004640305996869753/25000000000000000000 which is in lowest terms because the numerator is odd and does not end in 0 or 5. Rational. (One caution: this is only true because the decimal is written as terminating. If the digits were meant to run on for ever with no repeating block, the number would be irrational.) Answer: (i) rational, 9=9/1; (ii) irrational; (iii) rational, 13; (iv) rational, 4115/33333; (v) irrational; (vi) rational (it terminates), 589004640305996869753/25000000000000000000.

Question 26

The number 0. 9 (which means 0.99999 … ) is a rational number. Using algebra (let x=0. 9, multiply by 10, and subtract), explain why 0. 9 is exactly equal to 1.

Accepted Answer

Algebra on a repeating decimal. The trick is to multiply by a power of ten that makes the endless tail line up with itself, so that subtraction wipes it out. Let x=0. 9 =0.9999… The repeating block is a single digit, so shift by one place — that is, multiply by 10: 10x=9.9999… Look carefully at what each side is. After the decimal point, 10x has.9999… and x has.9999… — the same endless tail, because pushing an endless string of 9s one place along still leaves an endless string of 9s. So when we subtract, the tails cancel completely: 9x=9 ⇒ x=1 A second way, to be sure. You already know 13=0.333…. Multiply both sides by 3: 3× 13=1 and 3× 0.333…=0.999… Two names for the result of the same multiplication, so 0.999…=1. A third way — how big is the gap? If 0.999… were less than 1, the difference 1-0.999… would be a positive number. But it is smaller than 1-0.9=0.1, smaller than 1-0.99=0.01, smaller than 0.001, and so on — smaller than every positive number of the form 10^-n. The only number that small is 0. So the gap is 0 and the two are the same number. What this does and does not say. It does not say 0.999… "creeps up to" 1 or "is almost" 1. 0.999… is a name for one fixed number, and that number is 1 — exactly as 24 and 12 are two names for one number. Answer: 0. 9 =1 exactly.

Question 27

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. Non-uniqueness of decimal representations. Just as 1=10/10=100/100, rational numbers can have two decimal forms. Any terminating decimal has an alternative with repeating 9 s: 1.000 …=0.999 …, 2.47000 … =2.46999 …. Is it not surprising that 0.999 …=1? Many would have guessed that it is slightly less than 1.

Accepted Answer

What makes 1/7 special. 17=0. 142857 has a block of length 6, and 6=7-1. That is the whole story: dividing by 7 can leave remainders 1,2,3,4,5,6, and this single cycle uses every one of them before returning to the start. So 2/7,3/7,…,6/7 simply enter the same cycle at a different point, and their blocks are rotations of 142857: 17=0. 142857, 27=0. 285714, 37=0. 428571, 47=0. 571428, 57=0. 714285, 67=0. 857142 So the search is: find n whose block length for 1/n is exactly n-1. ( n must be prime, and cannot be 2 or 5, since those give terminating decimals.) How to test one by hand. Divide 1 by n and count the steps until the remainder 1 reappears. If that count equals n-1, the number is cyclic. Results. Below 100 the complete list is n=7, 17, 19, 23, 29, 47, 59, 61, 97 Spot-check that 17 really is cyclic. 2/17=0. 1176470588235294 — start reading the block of 1 17, namely 0588235294117647, from the " 11 " near its end and wrap around: 1176470588235294. Same digits, rotated. Contrast with 13 (Question 2): its block is 6 long but there are 12 remainders, so the remainders split into two cycles and the twelve fractions k/13 fall into two families of rotations, not one. --- The note at the end: non-uniqueness of decimals. Yes — every terminating decimal has a second name ending in repeating 9s, and the algebra of Question 4 proves each case. For 2.47: let x=2.46 9; 100x=246. 9 =246+0. 9 =246+1=247, so x=2.47 Is it surprising? It is, the first time. The reason most people guess 0.999…<1 is that they picture the decimal being built digit by digit, and every finite stage — 0.9, 0.99, 0.999 — genuinely is less than 1. But 0.999… is not any one of those stages; it is the single number the whole endless list closes in on, and that number is 1. Note also that only terminating decimals get two names — 13=0.333… has just one. Answer: the required n are the primes whose 1/n has a full-length block n-1: below 100 these are 7, 17, 19, 23, 29, 47, 59, 61, 97. And yes, 0.999…=1; every terminating decimal has an alternative repeating-9 form.

Question 28

Convert 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

Accepted Answer

Long division, watching the remainder. The division stops when a remainder of 0 appears, and repeats when an earlier remainder comes back. (i) 3/50 Before dividing, look at the denominator: 50=2× 5^2 is built only from 2 s and 5 s, so we should expect a terminating decimal. 50 into 30 goes 0 times, remainder 30 ⇒ first decimal digit 0. 50 into 300 goes 6 times ( 6× 50=300 ), remainder 0 ⇒ next digit 6, and the division stops. 3/50=0.06 Check the other way: 3/50=(3× 2)/(50× 2)=6/100=0.06. Terminating. (ii) 2/9 Here 9=3^2 contains the prime 3, so expect a repeating decimal. 9 into 20 goes 2 times ( 18 ), remainder 2 ⇒ digit 2. But the remainder 2 is exactly the number we began with, so the very same step happens again, and again, for ever. 2/9=0.2222…=0. 2 Check: 0. 2 means x with 10x-x=2, so 9x=2 and x= 29. Non-terminating and repeating. Answer: (i) 3/50=0.06, a terminating decimal. (ii) 2/9=0. 2, a non-terminating repeating decimal.

Question 29

Prove that √ 5 is an irrational number.

Accepted Answer

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 p divides a^2, then p divides a. (Reason: the prime factorisation of a^2 is just the factorisation of a with every exponent doubled, so no new prime can appear in a^2 that was not already in a.) We use it with p=5. The proof. Suppose, for the sake of argument, that √5 is rational. Then it can be written as a fraction √5=p/q where p and q are integers, q≠ 0, and the fraction is in lowest terms — that is, p and q have no common factor other than 1. (Assuming lowest terms costs nothing, since every fraction can be reduced to lowest terms.) Square both sides: 5=(p^2)/(q^2) ⇒ p^2=5q^2 1 From (1), p^2 is 5 times an integer, so 5 divides p^2. By the fact above, 5 divides p. So we may write p=5m for some integer m Substitute this into (1): (5m)^2=5q^2 ⇒ 25m^2=5q^2 ⇒ q^2=5m^2 2 From (2), 5 divides q^2, and by the same fact, 5 divides q. The contradiction. We have now shown that 5 divides p and 5 divides q. So p and q share the common factor 5 — which flatly contradicts our statement that pq was in lowest terms. Every step after the assumption was valid, so the fault must lie in the assumption itself. Therefore √5 cannot be written as pq. Answer: √5 is irrational. (Its decimal begins 2.2360679… 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.)

Question 30

Convert the following decimal numbers in the form of p/q. (i) 12.6 (ii) 0.0120 (iii) 3.0 52 (iv) 1.2 35 (v) 0. 23 (vi) 2.0 5 (vii) 2.12 5 (viii) 3.12 5 (ix) 2. 1625

Accepted Answer

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

Question 31

Locate the following rational numbers on the number line. (i) 0.532 (ii) 1.1 5

Accepted Answer

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) 0.532 Step 1. Draw the segment from 0 to 1 and divide it into 10 equal parts: 0.1,0.2,…,0.9. The first decimal digit is 5, so the point lies in the strip between 0.5 and 0.6. Step 2. Magnify that strip — redraw 0.5 to 0.6 as a full-length segment — and divide it into 10 equal parts marked 0.50, 0.51, …, 0.60. The second digit is 3, so the point lies between 0.53 and 0.54. Step 3. Magnify 0.53 to 0.54 and divide into 10 parts marked 0.530, 0.531, …, 0.540. The third digit is 2, so the point is the 2nd mark, i.e. exactly 0.532. Because the decimal terminates, the process finishes: three magnifications land on the point precisely. (ii) 1.1 5 =1.15555… First pin down the exact value, so the position is certain rather than guessed. Let x=1.15555… 10x=11.5555…, 100x=115.5555… 90x=115-11=104 ⇒ x=104/90=52/45 Check: 52÷ 45=1.15555… So the point is exactly 52/45. Step 1. On the segment from 1 to 2 marked in tenths, the first decimal digit 1 puts the point between 1.1 and 1.2. Step 2. Magnify 1.1 to 1.2 into ten parts. The next digit 5 puts it between 1.15 and 1.16. Step 3. Magnify 1.15 to 1.16 into ten parts. The next digit 5 puts it between 1.155 and 1.156. Step 4. Magnify again: between 1.1555 and 1.1556 — and so on for ever, since the 5 s never stop. Why this still locates a single point. Each new strip is 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 52/45. In a practical drawing you mark it just past the middle of the 1.15 -to- 1.16 strip (it sits 59 of the way along it). Answer: (i) 0.532 is reached exactly after three magnifications, as the 2nd of ten marks inside the strip 0.53 to 0.54. (ii) 1.1 5 =52/45, located as the single point common to the endlessly shrinking strips 1.1 – 1.2, 1.15 – 1.16, 1.155 – 1.156, …

Question 32

Find 6 rational numbers between 3 and 4.

Accepted Answer

Give both numbers a common denominator big enough to leave room. Between the integers 3 and 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 6 numbers, so choose a denominator of 7 (one more than 6 ) — that leaves exactly 6 gaps: 3=(3× 7)/7=21/7, 4=(4× 7)/7=28/7 The fractions with denominator 7 lying strictly between 21/7 and 28/7 are the ones whose numerators are the integers strictly between 21 and 28: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 That is exactly 6 numbers, and each is rational (an integer over an integer). Check the two ends. 22/7≈ 3.143 3 and 27/7≈ 3.857<4, and since the numerators increase the whole list is in order. Other answers work just as well — for instance 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 3 and 4, so the answer is not unique. Answer: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7.

Question 33

Find 5 rational numbers between 2/5 and 3/5.

Accepted Answer

Scale up the common denominator. The two fractions already share the denominator 5, but their numerators 2 and 3 are consecutive integers, so there is no fraction with denominator 5 between them. Multiplying top and bottom of both by the same number does not change their values but does create room. We need 5 numbers in between, so multiply by 6 (leaving 6 gaps, which is more than enough): 2/5=(2× 6)/(5× 6)=12/30, 3/5=(3× 6)/(5× 6)=18/30 Now every numerator strictly between 12 and 18 gives a fraction strictly between the two: 13/30, 14/30=7/15, 15/30=1/2, 16/30=8/15, 17/30 Check in decimals. 0.4 < 0.4 3 < 0.4 6 < 0.5 < 0.5 3 < 0.5 6 < 0.6 (the outer two being 25=0.4 and 35=0.6 ). Other answers are equally correct — multiplying by 10 instead would give 21/50,22/50,…, and there are infinitely many possibilities. Answer: 13/30, 7/15, 1/2, 8/15, 17/30.

Question 34

Find 5 rational numbers between 1/6 and 2/5.

Accepted Answer

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 6 and 5 is 30: 1/6=(1× 5)/(6× 5)=5/30, 2/5=(2× 6)/(5× 6)=12/30 The integers strictly between 5 and 12 are 6,7,8,9,10,11 — six of them, and we need five. Taking the first five and reducing where possible: 6/30=1/5, 7/30, 8/30=4/15, 9/30=3/10, 10/30= 1 3 Check in decimals. 16≈ 0.167 < 0.2 < 0.2 3 < 0.2 6 < 0.3 < 0. 3 < 0.4= 25 The answer is not unique — 11/30 would also qualify, and multiplying by a larger factor (say using denominator 60 or 300 ) produces many more. Answer: 1/5, 7/30, 4/15, 3/10, 1/3.

Question 35

If x/3+x/5=16/15, find the rational number x.

Accepted Answer

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 3 and 5, which is 15. x/3=(x× 5)/(3× 5)=5x/15, x/5=(x× 3)/(5× 3)=3x/15 Add them: 5x/15+3x/15=(5x+3x)/15=8x/15 So the equation becomes 8x/15=16/15 Both sides now have the same denominator, 15. Two fractions with equal denominators are equal only when their numerators are equal, so 8x=16 ⇒ x=16/8=2 (The same thing, done in one line: multiply both sides of the original equation by 15 to clear all denominators, giving 5x+3x=16.) Check by substituting x=2 back into the original equation. 2/3+2/5=10/15+6/15=16/15 Answer: x=2.

Question 36

Let a and b be two non-zero rational numbers such that a+1/b=0. Without assigning any numerical values, determine whether a b is positive or negative. Justify your answer.

Accepted Answer

Rearrange until 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 a and b are non-zero rationals with a+1/b=0 ( b≠0 is exactly what makes 1b meaningful.) Move 1b to the other side: a=-1/b Now multiply both sides by b — allowed, since b≠ 0: ab=-1/b× b=-1 So ab is not merely negative; it is exactly -1, for every pair (a,b) satisfying the condition. And since -1<0, ab is negative. A second way to see the sign, without the algebra. The equation a=- 1b says that a is the opposite of 1b. 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 b and 1b agree in sign). Hence a has the opposite sign to 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): b=4⇒ a=- 14⇒ ab=-1; b=- 23⇒ a= 32⇒ ab= 32×(- 23)=-1. Answer: ab is negative — in fact ab=-1 always.

Question 37

A 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/(10^4), 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 2^4 or 5^4? Give reasons.

Accepted Answer

Multiplying by a power of 10 clears a terminating decimal. Let the rational number be 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 x looks like x = ± N.d 1d 2d 3d 4, where N is the whole-number part and d 1,d 2,d 3,d 4 are digits with d 4 ≠ 0. Multiplying by 10 moves the decimal point one place to the right, so multiplying by 10^4 moves it four places to the right — exactly enough to swallow all four decimal digits: 10^4 x = ± N d 1 d 2 d 3 d 4 ( now read as a whole number ). Call this integer p. Dividing both sides by 10^4, x = p/(10^4). Why p is not divisible by 10. The units digit of p is d 4, and d 4 ≠ 0 because the 4th decimal place holds the last non-zero digit. An integer is a multiple of 10 exactly when its units digit is 0. Since the units digit of p is not 0, 10 p. For example 0.0625 = 625/10^4, 12.3456 = 123456/10^4, 0.0001 = 1/10^4 — in each case the numerator ends in a non-zero digit. Second part: yes, it is necessary. Write 10^4 = 2^4 × 5^4, so x = p/(2^4 × 5^4). To reduce this to lowest form we cancel g = gcd (p, 10^4); the lowest-form denominator is (10^4)/g. Since 10^4 has only the primes 2 and 5, g = 2^a × 5^b with 0 ≤ a ≤ 4 and 0 ≤ b ≤ 4. Now use 10 p. If p had both a factor 2 and a factor 5, it would have the factor 2× 5 = 10 — which it does not. So at least one of " 2 p ", " 5 p " fails, giving two cases. Case 1: p is odd. Then no 2 can be cancelled, so a = 0 and g = 5^b. The lowest-form denominator is (2^4 × 5^4)/(5^b) = 2^4 × 5^ 4-b, which is divisible by 2^4 = 16. Case 2: p is not a multiple of 5. Then b = 0 and g = 2^a. The lowest-form denominator is (2^4 × 5^4)/(2^a) = 2^ 4-a × 5^4, which is divisible by 5^4 = 625. Every p with 10 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 p does not carry both factors. Some checks: 0.0625 = 625/10^4 = 1/16, 16 = 2^4 0.0016 = 16/10^4 = 1/625, 625 = 5^4 0.1234 = 1234/10^4 = 617/5000, 5000 = 2^3 × 5^4 0.0001 = 1/10^4, 10^4 = 2^4 × 5^4 (both) Answer. Such a number is x = p/(10^4) with p an integer whose units digit is the non-zero 4th decimal digit, so 10 p. And yes — in lowest form its denominator must be divisible by 2^4 or by 5^4 (by both when p is coprime to 10), because p cannot supply factors of 2 and of 5 at the same time.

Question 38

Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.

Accepted Answer

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

Question 39

A rational number in its lowest form has denominator 2^3 × 5. How many decimal places will its decimal expansion have? Explain your answer.

Accepted Answer

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

Question 40

Let a=7/12 and b=5/6. Express both a and b in the form (k 1)/m and (k 2)/m where k 1, k 2 and m are integers and k 2-k 1 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 k 2-k 1 n+1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.

Accepted Answer

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

Question 41

Three rational numbers x, y, z satisfy x+y+z=0 and x y+y z+z x=0. Show that all the rational numbers x, y, z must be simultaneously zero.

Accepted Answer

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 (x+y+z)(x+y+z) out: you get the three squares once each, and each of the products xy, yz, zx twice.) Now substitute what we are given. Since x+y+z = 0, the left side is 0^2 = 0; and since xy+yz+zx = 0, the last bracket is 0. So 0 = x^2 + y^2 + z^2 + 2 × 0, which gives x^2 + y^2 + z^2 = 0. Finish with the sign of a square. For any rational number t, t^2 ≥ 0 — a square is never negative. So x^2, y^2, z^2 are three numbers, none of them negative, whose total is 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, y^2 = 0, z^2 = 0, and therefore x = 0, y = 0, z = 0. Check. x=y=z=0 does satisfy both conditions: 0+0+0=0 and 0+0+0=0. It is worth seeing that the second condition is doing real work. The triple x = 1, y = -1, z = 0 satisfies x+y+z = 0, but xy+yz+zx = (1)(-1) + (-1)(0) + (0)(1) = -1 ≠ 0, so it is ruled out — exactly as the proof says it must be. Answer. x^2+y^2+z^2 = 0 follows from the two conditions, and a sum of squares vanishes only when every term does, so x = y = z = 0 is the only possibility.

Question 42

Show that the rational number ((a+b))/2 lies between the rational numbers a and b.

Accepted Answer

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 a ≠ b. There are two cases, and they are mirror images; suppose a < b (if instead b < a, just swap the names a and b — the expression (a+b)/2 is unchanged, since a+b = b+a ). It is bigger than a. (a+b)/2 - a = (a+b)/2 - 2a/2 = (a+b-2a)/2 = b-a/2. Because a < b we have b - a 0, and dividing a positive number by 2 keeps it positive. So (a+b)/2 - a 0 ⇒ (a+b)/2 a. It is smaller than b. b - (a+b)/2 = 2b/2 - (a+b)/2 = 2b-a-b/2 = b-a/2 0, by the same reasoning. So (a+b)/2 < b. Putting the two together, a < (a+b)/2 < b. It really is a rational number. If a and b are rational then a+b is rational (the sum of two rationals is rational), and dividing a rational number by the non-zero integer 2 again gives a rational number. So (a+b)/2 is a rational number lying strictly between a and b. Notice also that both differences came out equal to b-a/2: the number (a+b)/2 is the exact midpoint of a and b, the same distance from each. Example. With a = 7/12 and b = 5/6: (a+b)/2 = (7/12+10/12)/2 = (17/12)/2 = 17/24 = 0.708 3, and 0.583 3 < 0.708 3 < 0.833 3. (If a = b, then (a+b)/2 = 2a/2 = a = b: it equals them both, and nothing can lie strictly between two equal numbers. This is why we assumed a ≠ b.) Answer. For a ≠ b, (a+b)/2 - a = b - (a+b)/2 = b-a/2, so (a+b)/2 is a rational number lying exactly halfway between a and b.

Question 43

Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

Accepted Answer

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 O and mark P 1 with 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 1 unit perpendicular to it, and join its far end back to O. Each step makes a new right triangle whose two legs are "the previous hypotenuse" and "1 unit". Apply Pythagoras' theorem to each triangle in turn. First triangle OP 1P 2, right-angled at P 1, legs OP 1 = 1 and P 1P 2 = 1: OP 2^ 2 = 1^2 + 1^2 = 2 ⇒ OP 2 = √ 2. Second triangle OP 2P 3, right-angled at P 2, legs OP 2 = √ 2 and P 2P 3 = 1: OP 3^ 2 = (√ 2 )^2 + 1^2 = 2 + 1 = 3 ⇒ OP 3 = √ 3. Third triangle: OP 4^ 2 = (√ 3 )^2 + 1^2 = 3 + 1 = 4 ⇒ OP 4 = √ 4 = 2. Fourth triangle: OP 5^ 2 = 4 + 1 = 5 ⇒ OP 5 = √ 5, and so on. The pattern. Squaring removes the root, so the new leg of length 1 always contributes exactly 1^2 = 1: OP k+2^ 2 = OP k+1^ 2 + 1. Starting from OP 2^ 2 = 2, the squares run 2, 3, 4, 5, 6, … — the whole numbers in order. Hence the hypotenuse of the k -th right triangle, counting outwards from the first, is OP k+1 = √ k+1. So, reading the triangles outwards along the spiral, their hypotenuses are √ 2, √ 3, √ 4 =2, √ 5, √ 6, √ 7, √ 8, √ 9 =3, √ 10, … Fig. 3.14 shows only finitely many of them: if the spiral drawn there has N triangles, their hypotenuses are √ 2, √ 3, √ 4, …, √ N+1, taken in that order from the list above. Why the spiral is useful. Only the perfect-square entries give whole numbers — √ 4 =2, √ 9 =3, √ 16 =4, and so on. All the others, √ 2, √ 3, √ 5, √ 6, √ 7, √ 8, √ 10, …, 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 √ 2, √ 3, √ 4 =2, √ 5, √ 6, √ 7, √ 8, √ 9 =3, …; in general the k -th right triangle of the spiral has hypotenuse √ k+1.

Number	Divisible by \(\displaystyle 2\)?	Divisible by \(\displaystyle 3\)?	Verdict
\(\displaystyle 11\)	no (odd)	no (\(\displaystyle 1+1 = 2\))	prime
\(\displaystyle 13\)	no (odd)	no (\(\displaystyle 1+3 = 4\))	prime
\(\displaystyle 17\)	no (odd)	no (\(\displaystyle 1+7 = 8\))	prime
\(\displaystyle 19\)	no (odd)	no (\(\displaystyle 1+9 = 10\))	prime

Event	In Brahmagupta's words	As an integer (rupees)
Loan taken	a debt of \(\displaystyle 850\)	\(\displaystyle -850\)
Profit made	a fortune of \(\displaystyle 1200\)	\(\displaystyle +1200\)
Loss incurred	a debt of \(\displaystyle 450\)	\(\displaystyle -450\)

Item	Value in rupees
Cash in hand	\(\displaystyle +15\)
Debt owed to a friend	\(\displaystyle -5\)
Net worth	\(\displaystyle +15 + (-5) = 10\)

\(\displaystyle x\)	LHS \(\displaystyle =\frac{5}{6}\left(x+\frac{3}{5}\right)\)	RHS \(\displaystyle =\frac{5}{6}x+\frac{1}{2}\)
\(\displaystyle 0\)	\(\displaystyle \frac{5}{6}\times\frac{3}{5}=\frac{1}{2}\)	\(\displaystyle 0+\frac{1}{2}=\frac{1}{2}\)
\(\displaystyle 2\)	\(\displaystyle \frac{5}{6}\times\frac{13}{5}=\frac{13}{6}\)	\(\displaystyle \frac{10}{6}+\frac{3}{6}=\frac{13}{6}\)
\(\displaystyle -\frac{3}{5}\)	\(\displaystyle \frac{5}{6}\times 0 = 0\)	\(\displaystyle -\frac{1}{2}+\frac{1}{2}=0\)

Number	As twelfths	Direction from \(\displaystyle 0\)	Where it lands
\(\displaystyle \frac{2}{3}\)	\(\displaystyle \frac{8}{12}\)	\(\displaystyle 8\) steps right	between \(\displaystyle 0\) and \(\displaystyle 1\), two-thirds of the way
\(\displaystyle -\frac{5}{4}\)	\(\displaystyle -\frac{15}{12}\)	\(\displaystyle 15\) steps left	past \(\displaystyle -1\), a quarter of the way on to \(\displaystyle -2\)
\(\displaystyle 1\frac{1}{2}\)	\(\displaystyle \frac{18}{12}\)	\(\displaystyle 18\) steps right	exactly halfway between \(\displaystyle 1\) and \(\displaystyle 2\)

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

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

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

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

	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)	\(\displaystyle 2.\overline{1625}\)	\(\displaystyle \dfrac{21623}{9999}\)

Exercise Set 3.1 1–4 (part 1 of 7)

Exercise 1

Exercise 2

Exercise 3

Exercise 4