Predicting the expansion from the denominator's prime factors

What to assume they know

• Why long division must either stop or loop — that every fraction's decimal stops or repeats
• What "rational" means, and why the denominator cannot be zero — lowest terms, and reducing by a common factor
• Prime factorisation of a whole number
• Powers of ten, and that multiplying a fraction's top and bottom by the same number leaves it unchanged
• Reading a decimal as a fraction over 10, 100 or 1000

What they should be able to do

• Factorise a denominator into primes and read off whether only 2s and 5s appear
• Decide, without dividing, whether a given fraction terminates
• Convert a terminating fraction into a power-of-ten denominator, stating the multiplier used
• Explain why a denominator containing any prime other than 2 or 5 cannot be turned into a power of ten
• Explain why a denominator built only from 2s and 5s always can
• Predict the number of decimal places from the exponents of 2 and 5
• Explain why the fraction must be in lowest terms before the test is applied, and give a case where ignoring this misleads
• Check a prediction against an actual long division

Where it usually goes wrong

• "Big denominators repeat and small ones terminate." 13/250 terminates and 4/15 repeats. Size is irrelevant; only which primes appear matters.
• "You have to divide to find out." The entire point of the subsection is that you do not. Q1 makes the student predict first and then verify, in that order.
• "An odd denominator means a repeating decimal." 1/5 and 13/250 say otherwise. It is the presence of primes other than 2 and 5 that matters, not parity.
• "You can cancel the 3 out of the denominator to make it work." Not in lowest terms, because there is nothing in the numerator to cancel it against. This is the load-bearing reason and it is worth its own beat.
• "The test works on any writing of the fraction." It does not. 3/6 fails the test and terminates anyway. Reduce first, always.
• "The number of decimal places equals the number of prime factors in the denominator." It equals the larger of the two exponents. For 2³ × 5 that is 3, not 4, and Q12 tests exactly this.
• "3/80 ends at the fourth place, and 80 is not a power of 2 or of 5, so a fourth-place terminator need not have 2⁴ or 5⁴ below the line." This is the trap in Q10, and it is wrong twice over: the claim being tested asks only for divisibility, not for the denominator to be a bare power, and 80 = 2⁴ × 5 is divisible by 2⁴. Q10's answer is yes. Show why no counterexample can exist — place 4 forces the larger exponent to be 4 — rather than hunting for one.

Questions to check understanding

• Without dividing, classify a list of fractions as terminating or repeating, and justify each verdict by prime factorisation
• Convert a terminating fraction to a power-of-ten denominator and state the multiplier
• State how many decimal places a fraction with a given denominator will have, with reasons
• Explain why a denominator containing a 3 cannot yield a terminating decimal
• Explain why the fraction must be in lowest terms before the test is applied
• Given a decimal whose last non-zero digit is at a stated place, deduce what the denominator must and need not contain
• Predict and then verify by long division — the chapter's own two-part form

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated inputs. The chapter prints no answers.

• The chapter's worked case (§3.6.1, p. 58, under the printed subheading "Predicting the Type of Decimal Expansion"). Inputs as printed: the fraction 3/20; the prime factorisation of 20 given as 2² × 5; the chain in which top and bottom are multiplied by 5 so that the denominator becomes 100; and the result 0.15. The chapter says in as many words that the multiplier 5 was chosen to make the denominator 100. Verified: 3/20 = 15/100 = 0.15, two decimal places, and the exponents were 2 and 1 so the larger is 2 — the place count.
• The criterion (Think and Reflect, §3.6.1, p. 58). Printed: the expansion terminates precisely when the denominator's prime factors are only 2, only 5, or both 2 and 5, and the student is asked why. The chapter's answer, given in the sentence immediately after the box, is that the denominator can then be made a power of ten by multiplying top and bottom by a suitable number. That covers one direction; the other direction is the explanation's job.
• Why the other direction holds (not in the book). Suppose the fraction is in lowest terms and the denominator carries a prime other than 2 or 5. To make the denominator a power of ten you may only multiply, and 10ⁿ has no factor except 2s and 5s, so the offending prime would have to be cancelled instead — which lowest terms forbids, since the numerator has no factor in common with it. Verified as valid. The chapter does not run this half.
• The place-count rule (not in the book). If the reduced denominator is 2ᵃ × 5ᵇ, multiply top and bottom by 5^(a−b) when a exceeds b, or by 2^(b−a) when b exceeds a; the denominator becomes 10 raised to the larger of a and b, so that is the number of decimal places. Verified against the chapter's own case and its exercises: for 20 = 2² × 5¹ the larger exponent is 2 and 3/20 has two places; for 125 = 5³ it is 3; for 2³ × 5 it is 3.
• Exercise Set 3.5, Q1 (p. 61). Inputs: 7/20, 4/15 and 13/250, to be classified without dividing and then checked by dividing. Verified: 20 = 2² × 5, so 7/20 terminates, at 0.35, two places; 15 = 3 × 5, so 4/15 repeats, at 0.2666… with a one-digit block; 250 = 2 × 5³, so 13/250 terminates, at 0.052, three places. The three items are chosen to give one of each verdict plus a case where the larger exponent belongs to the 5.
• End-of-chapter Q11 (p. 65). Inputs: decide without dividing whether 18/125 terminates, and if it does, state how many places. Verified: 125 = 5³ and 18 is co-prime to 125, so it terminates in three places, at 0.144. This is precisely the place-count rule being examined.
• End-of-chapter Q12 (p. 65). Inputs: a rational in lowest terms whose denominator is 2³ × 5, asked how many decimal places its expansion has, with an explanation. Verified: the exponents are 3 and 1, so the larger is 3 and the answer is three places; the multiplier is 5² = 25, taking the denominator to 1000.
• End-of-chapter Q10 (p. 65). Inputs: a rational with a terminating expansion whose last non-zero digit falls in the fourth decimal place. Asked: show it can be written over 10⁴ with a numerator not divisible by 10; and decide whether the lowest-form denominator must be divisible by 2⁴ or by 5⁴, with reasons. Verified: it must — the answer is yes, and no counterexample exists. This follows from this topic's own place-count rule and should be derived from it. In lowest terms the denominator can only be 2^a × 5^b, and the last non-zero digit falls at place max(a, b). So a last digit at place 4 says max(a, b) = 4, which says a = 4 or b = 4 — that is, 2⁴ divides the denominator or 5⁴ does. Every example one might reach for obeys it rather than breaking it: 1/16 = 0.0625 has denominator 2⁴; 1/625 = 0.0016 has denominator 5⁴; and 3/80 = 0.0375 has 80 = 2⁴ × 5, which is divisible by 2⁴. Note the disjunction is inclusive and the divisibility need not be exact — 80 carries a spare 5 and still qualifies, which is the point most likely to be missed. This is the hardest item in the chapter and the place-count rule is what makes it tractable.
• The lowest-terms trap (the explanation's construction). Inputs: 3/6 and 6/15. Verified: 3/6 has a 3 in its denominator yet reduces to 1/2 and terminates at 0.5; 6/15 has a 3 in its denominator yet reduces to 2/5 and terminates at 0.4 — the 5 in 15 was always permitted, and it is the 3 that the reduction clears, which is the whole reason the test must be run on the lowest form. Meanwhile 4/15 does not reduce and repeats. Run the test on the unreduced forms and you get the wrong answer twice. The chapter states the lowest-terms condition and never demonstrates why it is needed.
• No figure accompanies this material. The subheading, the worked chain, the Think and Reflect box and the following sentence occupy p. 58 as text.

Figures to have open

• A factor-tree pair for 10, 100 and 1000 showing only 2s and 5s appearing. Standard schematic; this is the figure the whole criterion rests on.
• An exponent-balance graphic: two columns counting the 2s and the 5s, with the shorter column being topped up. This carries sections 6 and 8 and the chapter prints nothing like it.
• A prediction-then-verification pair of cards for section 10, matching the structure of Exercise Set 3.5, Q1.
• A small table of denominators against verdict and place count, filled for 8, 20, 15, 125, 250 and 2³ × 5. Built from the values above.
• No textbook figure exists for this material.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 3 on the world of numbers. Within §3.6.1 the chapter prints an unnumbered bold subheading, "Predicting the Type of Decimal Expansion", on p. 58, which can be named but not cited by number; the worked chain and the criterion sit beneath it.
• Think and Reflect, p. 58, stating the criterion and asking for the reason; the chapter's own one-sentence answer follows the box on the same page.
• Exercise Set 3.5, Q1, p. 61.
• End-of-chapter exercises Q10, Q11 and Q12, p. 65.
• Backward pointer: the bracketed question in Example 2 at §3.6.1, p. 57, is the question this subsection answers.