Number systems: Questions and Answers

70 number systems questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Imagine you are living thousands of years ago in a small agricultural settlement along the banks of the Saraswati river. You have a herd of cattle. Every morning, they go out into the dense forests to graze, and every evening, they return. How do you ensure that a calf has not wandered off?Answer: Use one-to-one correspondence: put one pebble in a pot for every animal that leaves in the morning and take one pebble out for every animal that returns in the evening. If the pot is empty, the herd is complete; if pebbles are left, that many animals are missing.
• 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?Answer: 90 copper ingots (12 bags = 6 groups of 2 bags, and 6 × 15 = 90).
• 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.Answer: They are all prime numbers (in fact, all the primes between 10 and 20). The next three primes are 23, 29 and 31.
• 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.Answer: No. For example, 3 − 5 = −2 and 4 − 4 = 0, and neither −2 nor 0 is a natural number.
• 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?Answer: 4 fingers × 3 joints = 12 on one hand. Counting naturally goes up to 12 before you start again, so it suits a base-12 (duodecimal) system: one full hand = one dozen.
• Brahmagupta did not stop at zero. He realised that if subtraction of a number from itself can result in zero (5 − 5 = 0), then what would happen if we subtracted a larger number from a smaller one (3 − 5 = □)?Answer: 3 − 5 = −2. Subtracting a larger number from a smaller one goes below zero and gives a negative number.
• Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt.
  (Hint: If someone takes away (−) four of your debts that are each worth ₹3 (that is, −3), you are effectively ₹12 richer! Therefore, (−3) × (−4) = +12.)Answer: A negative number is a debt, and multiplying by a negative means removing debts. Removing 4 debts of ₹3 each makes you ₹12 better off, so (−3) × (−4) = +12: removing debt is a gain.
• 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?Answer: 4 − 15 = −11 °C (11 degrees below zero).
• 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.Answer: (−850) + 1200 + (−450) = −100. He is left with a debt of ₹100.
• Calculate the following using Brahmagupta’s laws:Answer: (i) −60 (ii) 56 (iii) 14 (iv) −5
• 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).Answer: Suppose your worth is ₹10 because you hold ₹15 but owe ₹5: 15 + (−5) = 10. If the ₹5 debt is cancelled (subtract −5), you hold ₹15 and owe nothing, so your worth is ₹15: 10 − (−5) = 10 + 5 = 15. Taking away a debt is the same as being given money.
• As society grew more complex, measuring became just as important as counting. If a farmer divides a field of wheat among his three children, how much does each get? If a recipe calls for half a cup of ghee, how do we represent that mathematically?Answer: Each child gets 13 of the field (if it is shared equally), and half a cup is 12 cup. Both are fractions: numbers that stand for parts of a whole.
• Can you explain why we need q ≠ 0 in the definition of a rational number?Answer: Because pq means p ÷ q, and division by 0 has no meaning: no number times 0 gives a non-zero p, and for p = 0 every number would work. So p0 is not a number at all.
• 1. 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.Answer: 1. Rewrite each fraction as an equivalent fraction whose denominator is a common multiple of both denominators (best: the LCM). E.g. 14 + 16 = 312 + 212 = 512. 2. With p = 12, q = 23, r = −14: both p(q + r) and pq + pr equal 524, so the law holds.
• Prove that the following rational numbers are equal:Answer: Use the rule ab = cd if ad = bc. (i) 2 × 6 = 12 = 3 × 4 (ii) 5 × 8 = 40 = 4 × 10 (iii) (−3) × 10 = −30 = 5 × (−6) (iv) 9 × 1 = 9 = 3 × 3. Each pair is equal.
• Find the sum:Answer: (i) 710 (ii) 2924 (iii) −514
• Find the difference:Answer: (i) 712 (ii) 58 (iii) −19
• Find the product:Answer: (i) 15 (ii) 3588 (iii) −1049
• Find the quotient:Answer: (i) 209 (ii) 5655 (iii) −85
• Show that: (12 + 34) × 83 = 12 × 83 + 34 × 83.Answer: LHS = 54 × 83 = 103 and RHS = 43 + 2 = 103, so LHS = RHS (the distributive law).
• Simplify the following using the distributive property: 79(67 − 34).Answer: 79 × 67 − 79 × 34 = 23 − 712 = 112.
• Find the rational number x such that: 56(x + 35) = 56x + 12.Answer: Opening the bracket gives 56x + 12 = 56x + 12, which is true for every value of x. So there is no single answer: every rational number x satisfies the equation (for example x = 0).
• Try and represent 85 and −74 on a number line.Answer: 85 = 135: divide the gap from 1 to 2 into 5 equal parts and mark the 3rd point after 1. −74 = −134: divide the gap from −1 to −2 into 4 equal parts and mark the 3rd point to the left of −1 (one part to the right of −2).
• The absolute value of a rational number x, written as |x|, represents its distance from 0 on the number line. |53| = 53, |−53| is also equal to 53 and |0| = 0.Answer: |53| = 53, |−53| = 53 and |0| = 0, because 53 and −53 are both 53 units from 0, and 0 is at 0 itself.
• Try to explain why the average of two rational numbers a and b, which equals (a + b)2, is always a rational number between a and b.Answer: It is rational because sums and quotients (by non-zero numbers) of rational numbers are rational. It lies between them because, if a < b, then a + b2 − a = b − a2 > 0 and b − a + b2 = b − a2 > 0: the average is exactly halfway from a to b.
• This means that there are infinitely many rational numbers between any two points. It feels as though the rational numbers must completely fill the number line, leaving no gaps whatsoever. But do they?Answer: No. Some points on the number line are not rational. The diagonal of a square of side 1 has length √2, and √2 can be marked on the line, but √2 cannot be written as pq. Such points are irrational numbers.
• Represent the rational numbers 23, −54 and 112 on a single number line.Answer: 23: 2 of 3 equal parts from 0 to 1. 112 = 32: the midpoint of 1 and 2. −54 = −114: 1 of 4 equal parts to the left of −1.
• Find three distinct rational numbers that lie strictly between −12 and 14.Answer: Write −12 = −48 and 14 = 28. Three numbers between them: −38, 0, 18 (others such as −14 or −18 are also correct).
• Simplify the expression: (−14) + (512).Answer: −312 + 512 = 212 = 16
• A tailor has 1534 metres of fine silk. If making one kurta requires 214 metres of silk, exactly how many kurtas can he make?Answer: 1534 ÷ 214 = 634 × 49 = 7 kurtas, with no silk left over.
• Find three rational numbers between 3.1415 and 3.1416.Answer: 3.14151, 3.14153 and 3.14155 (each is a terminating decimal, so rational; e.g. 3.14151 = 314151100000).
• Can you think of other way(s) to find a rational number between any two rational numbers?Answer: Yes. Besides the average a + b2: (1) write both with a common denominator and scale it up (×10, say) until whole numerators fit between; (2) use decimals and add a digit; (3) step part of the way: a + b − an for any whole number n ≥ 2; (4) for positive fractions ab < cd, the ‘mediant’ a + cb + d lies between them.
• Can √2 be written as a rational number pq?Answer: No. If √2 = pq in lowest terms, then p² = 2q², which forces p to be even and then q to be even too. That contradicts ‘lowest terms’, so √2 is irrational.
• Try to prove the irrationality of √3 using the approach of proof by contradiction. Will the same approach work for √5, √7, or √10?Answer: If √3 = pq in lowest terms, then p² = 3q², so 3 divides p, then 3 divides q: a contradiction. Yes, the same approach works for √5, √7 (use 5, 7 in place of 3) and √10 (use the factor 2 of 10), because none of 5, 7, 10 is a perfect square.
• We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?Answer: Use right triangles. On the number line take OA = 1, draw AB = 1 perpendicular at A; then OB = √(1² + 1²) = √2. An arc with centre O and radius OB cuts the number line at P, so OP = √2.
• Try 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.Answer: √3: at the point √2 on the number line draw a perpendicular of length 1; the hypotenuse from O is √(2 + 1) = √3. √5: take legs 2 and 1, hypotenuse √(4 + 1) = √5. In general, if √(n − 1) is already marked, a perpendicular of 1 at that point gives √(n − 1 + 1) = √n; repeating step by step gives every √n.
• We know what it means to add 2, 100, or even a lakh terms. What does it mean to add an infinite number of terms?Answer: It means finding the value that the running totals get closer and closer to as we add more and more terms, starting from the first. For Mādhava’s series 4 × (1 − 13 + 15 − 17 + …) the running totals 4, 2.667, 3.467, 2.895, 3.340, … close in on π = 3.14159….
• It terminates: The division eventually leaves a remainder of 0. The decimal stops. 38 = 0.375. (Can you tell for which rational numbers the decimal will be terminating?)Answer: 3 ÷ 8: remainders 6, 4, 0, so 38 = 0.375 and the division stops. The decimal terminates exactly when the denominator (in lowest terms) has no prime factors other than 2 and 5; here 8 = 2³.
• It repeats: The division never reaches a remainder of 0, but the sequence of digits begins to loop infinitely. 511 = 0.454545 … = 0.45.Answer: 5 ÷ 11 gives remainders 6, 5, 6, 5, … and digits 4, 5, 4, 5, …, so 511 = 0.454545… = 0.45, a non-terminating repeating decimal with repeating block 45.
• Try to find the decimal expansions of 103 and 1112. What do you observe about the repetition of the digits after the decimal point?Answer: 103 = 3.333… = 3.3 and 1112 = 0.91666… = 0.916. In 103 the digit 3 repeats right from the first decimal place; in 1112 the digits 9 and 1 do not repeat and only then does 6 repeat for ever. Neither decimal ends.
• Why do some rational numbers have repeating decimal representations? Imagine calculating 17 using long division. You are dividing by 7. What are the possible remainders at each step? They can only be 1, 2, 3, 4, 5, or 6 (why not 0?).Answer: Remainders on dividing by 7 are less than 7. They cannot be 0, because then 7 would divide 10 × 10 × … × 10, whose only prime factors are 2 and 5. So only 1–6 are possible, a remainder must repeat within 6 steps, and the digits loop: 17 = 0.142857.
• The decimal expansion of pq will be terminating precisely when the prime factors of q are only 2, only 5 or both 2 and 5. Can you explain why?Answer: If q = 2^m × 5ⁿ, multiplying top and bottom by enough 2s or 5s makes the denominator a power of 10, so the decimal terminates. Conversely, a terminating decimal is N10^k, and in lowest terms its denominator divides 10^k = 2^k5^k, so it has no prime factors other than 2 and 5.
• Convert 0.35 into the form pq.Answer: 0.35 = 35100 = 720
• Convert 0.6 into the form pq.Answer: Let x = 0.6. Then 10x = 6.6, so 9x = 6 and x = 69 = 23.
• Convert 0.45 into the form pq.Answer: Let x = 0.45. Then 100x = 45.45, so 99x = 45 and x = 4599 = 511.
• Convert 0.16 into the form pq.Answer: Let x = 0.16. 10x = 1.6 and 100x = 16.6. Subtracting, 90x = 15, so x = 1590 = 16.
• Convert 2.357 into the form pq.Answer: Let x = 2.357. 100x = 235.7, 1000x = 2357.7. Subtracting, 900x = 2122, so x = 2122900 = 1061450.
• Convert 2.4537 into the form pq.Answer: Let x = 2.4537. 100x = 245.37, 10000x = 24537.37. Subtracting, 9900x = 24292, so x = 242929900 = 60732475.
• Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 720, 415 and 13250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.Answer: 720 (20 = 2² × 5) terminates: 0.35. 415 (15 = 3 × 5) repeats: 0.26. 13250 (250 = 2 × 5³) terminates: 0.052.
• Perform the long division for 113. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 213? Now compute 313, 413, etc. What do you notice?Answer: 113 = 0.076923 (block 076923). 213 = 0.153846 is not a rotation of 076923, so 076923 is not fully cyclic. 313 = 0.230769 and 413 = 0.307692 are rotations of 076923. The twelve fractions k13 split into two families: rotations of 076923 (k = 1, 3, 4, 9, 10, 12) and rotations of 153846 (k = 2, 5, 6, 7, 8, 11).
• Classify the following numbers as rational or irrational:
  Find the explicit fractions in case they are rational.Answer: (i) rational, 91 (ii) irrational (iii) rational, 13 (iv) rational, 411533333 (v) irrational (vi) rational, 58900464030599686975325000000000000000000
• 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.Answer: x = 0.9, 10x = 9.9. Subtracting, 9x = 9, so x = 1. The two decimals 0.999… and 1 name the same number.
• We have seen that the repeating block of 17 is a cyclic number. Try to find more numbers (n) whose reciprocals (1n) produce decimals with repeating blocks that are cyclic.Answer: n = 17, 19, 23, 29 (also 47, 59, 61, 97, …). For example 117 = 0.0588235294117647, and 217 = 0.1176470588235294 is the same block rotated. These are the primes n for which the repeating block of 1n has the full length n − 1.
• Consider 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 number line.Answer: No real number squares to −1: a positive times a positive is positive, a negative times a negative is positive, and 0 × 0 = 0. So √−1 is not on the real number line. Mathematicians created a new number i with i × i = −1, the start of the imaginary numbers.
• 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:Answer: (i) 350 = 0.06 (terminating) (ii) 29 = 0.222… = 0.2 (non-terminating repeating)
• Prove that √5 is an irrational number.Answer: Suppose √5 = pq in lowest terms. Then p² = 5q², so 5 divides p; writing p = 5k gives q² = 5k², so 5 divides q. Then 5 is a common factor of p and q: a contradiction. Hence √5 is irrational.
• Convert the following decimal numbers in the form of pq.Answer: (i) 635 (ii) 3250 (iii) 1511495 (iv) 1223990 (v) 2399 (vi) 3718 (vii) 1913900 (viii) 2813900 (ix) 216239999
• Locate the following rational numbers on the number line.Answer: Use successive magnification. (i) 0.532: zoom into 0.5–0.6, then 0.53–0.54 (in thousandths) and mark the 2nd mark. (ii) 1.15 = 1.1555… = 5245: zoom into 1.1–1.2, then 1.15–1.16 (in thousandths); it lies just past 1.155, 59 of the way from 1.15 to 1.16.
• Find 6 rational numbers between 3 and 4.Answer: 3 = 217 and 4 = 287, so 227, 237, 247, 257, 267, 277 (or 3.1, 3.2, 3.3, 3.4, 3.5, 3.6).
• Find 5 rational numbers between 25 and 35.Answer: Multiply top and bottom by 6: 1230 and 1830. Five numbers between: 1330, 715, 12, 815, 1730.
• Find 5 rational numbers between 16 and 25.Answer: LCM of 6 and 5 is 30: 16 = 530, 25 = 1230. Five numbers between: 15, 730, 415, 310, 13.
• If x3 + x5 = 1615, find the rational number x.Answer: 5x + 3x15 = 1615, so 8x = 16 and x = 2.
• Let a and b be two non-zero rational numbers such that a + 1b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.Answer: a = −1b, so ab = −1b × b = −1. Hence ab is negative.
• 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 p10⁴, 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.Answer: Multiplying by 10⁴ moves all four decimal digits in front of the point, giving an integer p whose last digit is the non-zero 4th decimal digit, so 10 ∤ p. Yes: since 10 ∤ p, p cannot be divisible by both 2 and 5, so at least one of 2⁴, 5⁴ cannot cancel and stays in the lowest-form denominator. (It need not be divisible by both, e.g. 0.0005 = 12000 = 12⁴ × 5³.)
• Without performing division, determine whether the decimal expansion of 18125 is terminating or non-terminating. If it terminates, state the number of decimal places.Answer: 18125 is in lowest terms and 125 = 5³, so it terminates: 18 × 2³5³ × 2³ = 1441000 = 0.144, with 3 decimal places.
• A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.Answer: 3 decimal places. p2³ × 5 = 25p10³, and because p is odd and not a multiple of 5 (lowest form), 25p ends in 5, so the third decimal digit is non-zero.
• Let a = 712 and b = 56. Express both a and b in the form k₁m and k₂m 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.Answer: With m = 36: a = 2136, b = 3036 (k₂ − k₁ = 9 > 6). Five numbers: 2236, 2336, 2436, 2536, 2636. Between k₁ and k₂ there are k₂ − k₁ − 1 whole numbers, so getting n of them needs k₂ − k₁ − 1 ≥ n; the stated condition k₂ − k₁ > n + 1 guarantees this (strictly, k₂ − k₁ ≥ n + 1 is already enough).
• 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.Answer: x² + y² + z² = (x + y + z)² − 2(xy + yz + zx) = 0 − 0 = 0. A sum of squares of rational numbers is 0 only if each square is 0, so x = y = z = 0.
• Show that the rational number (a + b)2 lies between the rational numbers a and b.Answer: Take a < b. Adding a to both sides gives 2a < a + b; adding b gives a + b < 2b. So 2a < a + b < 2b, and halving: a < a + b2 < b.
• Find the lengths of the hypotenuses of all the right triangles in the figure which is referred to as the square root spiral.Answer: The hypotenuses are √2, √3, √4 = 2, √5, √6, √7, √8, √9 = 3, √10, √11 (ten triangles).

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