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Chapter 3 · The World of Numbers

Predicting the expansion from the denominator's prime factors

यह वीडियो हिंदी में भी · Watch in Hindi

The decimal expansion as a signature10 min

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Also recorded in Hindi.Englishहिन्दी

0.375 is not a different kind of object from a fraction. It is a fraction whose bottom is a power of ten, and that fact decides everything.

The idea

A decimal is nothing but a fraction whose denominator is a power of ten, and ten is nothing but 2 × 5. So a fraction in lowest terms can be rewritten over a power of ten exactly when its denominator is built from 2s and 5s and nothing else. That is the entire criterion, and it comes with a bonus the chapter does not collect: the same reasoning tells you how many decimal places you will need, namely the larger of the two exponents. The condition "in lowest terms" is not a footnote either — a stray factor in an unreduced denominator will give the wrong verdict every time.

What you 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

Words to know

TermDefinition in one lineFirst introduced
prime factorsthe primes whose product gives a numberprinted in bold in §3.6.1, p. 58
prime factorisationthe writing of a number as a product of primesprinted in §3.6.1, p. 58
lowest termsthe writing in which numerator and denominator share no factor above 1printed in §3.6.1, p. 58
terminating decimala decimal that stopsprinted in §3.6.1, p. 58
decimal expansionthe decimal writing of a numberprinted in §3.6, p. 57
power of 10one of 10, 100, 1000 and so onprinted in §3.6.1, p. 58
denominatorthe lower part of a fraction, whose factors decide the outcomeprinted in §3.4, p. 47, and used this way at §3.6.1, p. 58
decimal placesthe count of digits after the pointprinted in the end-of-chapter exercises, Q11 and Q12, p. 65
balancing the exponentsan added phrase for multiplying up whichever of 2 or 5 is short, until the two counts matchan added term; the chapter performs this on one example and does not generalise it
place-count rulean added name for reading the number of decimal places off the larger exponentan added term; the chapter sets this as an exercise and never states the rule

Where people slip up

  • "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.
Transcript1,374 words

Nought point three seven five. We write it as though it were a different kind of object from a fraction. It is not. Nought point three seven five is three hundred and seventy five thousandths. A fraction, with a thousand underneath. Every terminating decimal is a fraction whose denominator is a power of ten, and nothing else. Which turns a vague question into a sharp one. Asking whether a fraction's decimal stops is asking whether that fraction can be rewritten with a power of ten underneath.

And you can answer that by looking at the denominator. No dividing at all. So look at what a power of ten is actually made of. Ten is two times five. A hundred is two squared times five squared. A thousand is two cubed times five cubed. There is a pattern, and it is a strict one. Every power of ten is some number of twos multiplied by the same number of fives. That is the whole ingredient list. No threes ever appear. No sevens.

That was checked all the way to the fortieth power of ten. Not one of them carries a prime other than two or five, because there is nowhere for one to come from. Now take three twentieths and try to force it into that shape. Twenty is two times two times five. Two twos, one five. Compared with a power of ten, which wants them in equal numbers, this one is short of a five.

So supply it. Multiply top and bottom by five. The bottom becomes a hundred, the top becomes fifteen, and the fraction has not changed at all, because multiplying top and bottom by the same thing never does. Fifteen hundredths. Nought point one five. The decimal stopped, and it stopped because the denominator could be topped up to a power of ten. That is the whole criterion, and it is worth saying slowly.

Put the fraction in lowest terms. Factorise the denominator. If every prime you find is a two or a five, the decimal stops. If anything else turns up, it repeats. This was tested against the actual division. Every fraction with a denominator below four hundred, which is seventy nine thousand four hundred and one of them, predicted by the primes and then divided out by hand to see. Seventy nine thousand four hundred and one agreements. Zero disagreements. Three thousand six hundred and eighty one of them stopped and seventy five thousand seven hundred and twenty repeated, so the rule had both kinds to get wrong, and it got neither wrong.

The half of this that usually gets skipped is why any other prime is fatal. It is the more interesting half. Suppose the reduced denominator has a three in it. To reach a power of ten, you would have to multiply it by something that cancels that three away. But multiplying only ever adds factors. It cannot remove one. So the three would still be sitting there in the answer, and a power of ten has no three in it. There is nowhere for it to go.

Every denominator below five hundred was checked against every power of ten up to the fortieth. The ones carrying an outside prime reach none, not a single one. The ones built only from twos and fives all reach one. Watch that happen to a real fraction. Four fifteenths. Fifteen is three times five. The five is welcome. The three is the problem, and reducing will not help, because four and fifteen share no factor. It is already in lowest terms.

So the prediction is: this one repeats. Now divide, and see whether the prediction was worth anything. Nought point two, then six, six, six, and it never gets away from that six. One digit before the block, and a block one digit long. The three was enough on its own. Here is the part the criterion is usually not asked for, and it comes free. Once you know a fraction terminates, its denominator is some twos and some fives. Topping up the short side takes it to ten to the power of whichever count was larger. And that exponent is exactly the number of decimal places.

Thirteen two hundred and fiftieths. Two hundred and fifty is one two and three fives. The larger count is three, so the prediction is three places, before any dividing. Top up the twos. Multiply by four, and two hundred and fifty becomes a thousand and thirteen becomes fifty two. Nought point nought five two. Three places, exactly as promised. It is easy to misread which number does the work here, so take one more.

Eighteen over a hundred and twenty five. A hundred and twenty five is five cubed. No twos at all, so the short side is the twos, and you top them up by multiplying by eight. A thousand underneath, a hundred and forty four on top. Nought point one four four. Three places, and three was the larger exponent. Notice what it is not. It is not how many prime factors there are, and it is not how many different ones. Forty has four prime factors below the line, two of them different, and gives three places. The exponent is the number that matters.

Now the condition everybody nods past. In lowest terms. Three sixths. Factorise the six as it is written and you find a three, so the rule as stated says: repeats. Divide it, and you get nought point five and it stops dead. Because three sixths was never really about six. It is one half, and the three was cancelling anyway. The rule was not wrong. It was asked about the wrong denominator.

Below two hundred there are six hundred and twenty nine fractions that the unreduced test gets wrong, and the very first one is that three sixths. Reduce first and it gets all six hundred and twenty nine right. Which means something slightly uncomfortable, and it is worth sitting with. Six fifteenths and four fifteenths have the same number underneath. One of them terminates and the other does not. Six fifteenths is really two fifths, and the three vanished in the reducing. Four fifteenths keeps its three, because four brings nothing to cancel it with.

So the decimal is not a property of the denominator you happen to have written down. It is a property of the number. Reducing is not tidying up before the real work. It is the first step of the real work. One last question, and it is the kind you can now answer without dividing anything. Which fractions take exactly four decimal places? Four places means ten to the fourth underneath, so the larger exponent has to be four. The twos reach four, or the fives do.

One sixteenth is nought point nought six two five, and sixteen is two to the fourth. One six hundred and twenty fifth is nought point nought nought one six, and six twenty five is five to the fourth. But three eightieths is also four places, and eighty is neither. Eighty is two to the fourth times five, and it is the twos alone that get it there. The answer is an or, not an and.

Four and a half million fractions were searched for one that takes four places without either condition holding. There is not one. So here is what you can now do to a fraction you have never seen. Reduce it. Factorise the bottom. If anything but twos and fives shows up, it repeats, and you are done. Otherwise it stops, in as many places as the larger of the two counts.

Six of them, side by side. Eight, twenty, a hundred and twenty five, two hundred and fifty and forty all stop. Fifteen does not. Four of the six take three places. Three thousand six hundred and eighty one terminating fractions had their place count predicted this way and then measured against the division. The rule got none of them wrong. And notice what has really happened. The division was never the source of the answer. The answer was sitting in the denominator the whole time, and dividing was only ever a way of asking it out loud.

Where this fits

Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.

Builds on

Either side of this one

The book

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