A decimal expansion is a series: 0.d₁d₂d₃… is shorthand for d₁/10 + d₂/100 + d₃/1000 + … . When a block of digits repeats, that series is geometric with |r| < 1, so it has a sum and the sum is a fraction. The same argument settles 0.999… = 1, which turns out to be a fact about notation rather than a paradox.
Core ideas
Nothing new is proved here. The infinite geometric sum is applied to the notation everyone has been using for numbers since primary school.
0.333… is not a number waiting to be finished; it is the series 3/10 + 3/100 + 3/1000 + … , and its value is the number the partial sums 0.3, 0.33, 0.333, … approach. Finite decimals are the same statement with all but a few digits equal to 0.
If the block is p digits long, the ratio is 1/10ᵖ, and 1/10ᵖ is less than 1 for every p ≥ 1. So the convergence test never fails for a repeating decimal: the sum exists, whatever the digits are, and it is a ratio of whole numbers.
The two strings of symbols denote the same real number, in the way that 1/2 and 2/4 do. There is no last 9, so there is no gap of 0.000…1 — that string does not name a real number, because a positive number smaller than every 1/10ⁿ does not exist.
The material
Place value written as a series, the block conversion in both directions, and the one number with two decimal names.
For digits d₁, d₂, d₃, … each in 0 to 9, the notation 0.d₁d₂d₃… denotes the sum of the series d₁/10 + d₂/100 + d₃/1000 + … , that is, the number its partial sums approach. The expansion terminates if every digit from some point on is 0, and repeats if from some point on a fixed block of digits recurs for ever.
This is what place value has always meant. 0.25 is 2/10 + 5/100 = 1/4, a sum of two terms, and 0.333… is the same kind of sum with no last term. The only difference is that the second one needs the partial sums to settle before it denotes anything, and the infinite geometric series is exactly the tool that says when they do. Notation for the repeating block varies: a bar over the block, a dot over each end of it, or an ellipsis after two or three copies. All three are in use and the choice between them is arbitrary. This course writes the block out two or three times and then … , so 0.454545… means the 45 goes on for ever.
Take 0.454545… . The block 45 occupies the first two decimal places, then the next two, then the next two, so the series is 45/100 + 45/10000 + 45/1000000 + … . Each term is the one before divided by 100.
Suppose the expansion repeats a block B of p digits, starting immediately after the decimal point. Then the number is B/(10ᵖ - 1): the series has a₁ = B/10ᵖ and r = 1/10ᵖ, and |r| < 1 holds for every p ≥ 1.
If a few fixed digits come first, the repeating tail is divided by a further power of ten and the fixed part is added back, which changes the arithmetic but not the conclusion.
With a₁ = B/10ᵖ and r = 1/10ᵖ, the sum formula gives S∞ = (B/10ᵖ)/(1 - 1/10ᵖ). Multiply numerator and denominator by 10ᵖ: the numerator becomes B and the denominator becomes 10ᵖ - 1, which is the string of p nines.
Both B and 10ᵖ - 1 are whole numbers and 10ᵖ - 1 is not zero, so the sum is a ratio of whole numbers. Adding the fixed head, itself a terminating decimal and so a fraction, keeps it one.
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The school method — call the number x, multiply by a power of ten, subtract — is this proof compressed. Multiplying 0.454545… by 100 shifts the whole tail one block to the left, so 100x and x have identical tails and everything cancels except the block that falls out in front. It is the Sₙ - rSₙ manoeuvre with the shift run the other way: there the second line was r times the first, here it is 1/r times.
By the formula: a₁ = 9/10, r = 1/10, so S∞ = (9/10)/(9/10) = 1. By the block rule: B = 9 and p = 1, so the value is 9/9 = 1. By thirds: 0.333… = 1/3 is uncontroversial, and multiplying both sides by 3 gives 0.999… = 1.
The objection people raise is that the partial sums 0.9, 0.99, 0.999, … are all less than 1, so the total must be too. Every one of those partial sums is indeed less than 1. The sum of the series is not one of them; it is the number they approach. The gap 1 - 0.99…9 with n nines is exactly 1/10ⁿ, and no positive number is smaller than every 1/10ⁿ, so a number differing from 1 by less than all of them differs from it by 0. A consequence is that decimal names are not quite unique: every terminating decimal has a second name ending in repeating nines, 0.25 = 0.24999… , since 24/100 + (9/1000)/(1 - 1/10) is 6/25 + 1/100 = 1/4. This is the only ambiguity there is — a number with no terminating expansion has exactly one decimal name.
Dividing a whole number by q by hand produces a remainder at every step, and each remainder is one of 0, 1, …, q - 1. Either 0 turns up, and the expansion terminates, or within q steps some earlier remainder recurs — and once a remainder recurs the digits after it repeat exactly as they did before, because the same remainder produces the same next digit. So every fraction terminates or repeats, with a block shorter than q; there is no third behaviour.
1/7 = 0.142857142857… , a block of six, which is the longest a denominator of 7 allows. 1/13 = 0.076923076923… , also six. 3/40 = 0.075 exactly, because 40 = 2³·5 and a denominator built only from 2s and 5s can be turned into a power of ten: 3/40 = 75/1000.
Putting the two directions together: a number is rational exactly when its decimal expansion terminates or repeats. Foundations of Algebra sorted the reals into rational and irrational and asserted this; the geometric series is what turns the assertion into an argument. It also gives a way to write down an irrational number on demand — 0.101001000100001… , with one more zero in each gap, never repeats and so cannot be a fraction.
Interactive
Summed exactly, printed as a fraction, and divided back out as a check
Method
The work is in the first step. Once the head and the tail are separated the rest is the sum formula.
In 0.2454545… the head is 0.2 and the tail is the block 45 starting at the third decimal place. Count the digits in each: one fixed digit and a block of two.
Its first term is the block over 10 to the power of the position it ends at, and its ratio is 1/10ᵖ where p is the length of the block. For 0.2454545… that is a₁ = 45/1000 and r = 1/100.
S∞ = a₁/(1 - r), then add the head as a fraction. Clear the compound fraction by multiplying numerator and denominator by 10ᵖ rather than reaching for decimals.
Long division of the fraction you produced must reproduce the digits you started from. It is the one check that tests the calculation end to end, and it catches a misplaced power of ten immediately.
One fixed digit and then a two-digit block. Both routes are worked in full because the second is the first one compressed, and it is worth seeing that they are the same calculation.
Both routes were carried out and both produced 27/110, which is one instance checked rather than a rule established; what makes the method correct in general is the cancellation in the proof above.
The denominator explains itself. 990 = 99·10: the 99 is the two nines belonging to a two-digit block, and the 10 is the single fixed digit delaying it. After cancelling, 110 = 2·5·11, and the 11 is the reason this decimal repeats at all — a denominator carrying any prime other than 2 and 5 cannot be written over a power of ten.
For a faded conversion, use 0.18333…. The supplied split is the fixed head 18/100 plus a tail whose first term is 3/1000 and whose ratio is 1/10. Sum the tail, add the head, reduce the fraction, and then reproduce the result by shifting and subtracting. Long-divide your fraction far enough to recover the fixed digits and repeat before opening the quiz.
You should be able to convert either way between a repeating decimal and a fraction, in both the plain and the delayed-block cases, and defend 0.999… = 1 without appealing to authority. Being able to say why 1/7 repeats and 3/40 does not is part of it.
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