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The finite decimal 0.999 equals 999/1000, so it is smaller than one by 0.001. The recurring decimal 0.999…, when the notation means nines continue forever, equals one exactly in the ordinary real-number system. The dots change the mathematical object being described.
This is a useful reading distinction rather than a trick. Every finite string of nines stops at a decimal place and leaves a positive gap. The infinite recurring decimal represents the value determined by the whole expansion, not one very long finite string with a final nine.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–30
Sections 31–33
SECTION 1 OF 33
Quick read: finite prefixes and the recurring value have different roles
The numbers 0.9, 0.99 and 0.999 are each below one. Their respective gaps are 0.1, 0.01 and 0.001. Adding another finite nine reduces the gap by a factor of ten but does not make that finite decimal equal one.
The notation 0.999… means there is no last decimal place in the repeating pattern. Its value is the limit of those finite prefixes, which is one. A limit can equal a value that none of the finite terms equals.
The most useful first check is therefore what the notation promises. Does it show exactly three decimal places, explicitly repeat the digit nine forever, or merely show a shortened calculator display whose omitted digits are unspecified? Those are different statements and require different conclusions.
Fractions, decimals, exact values and approximation appear across secondary mathematics. This guide connects them through one carefully defined example. The formal limit discussion is an optional explanation of the notation, rather than a claim that every subject level requires the same advanced terminology.
For a broader route, read eduKate’s explanation of exact answers and decimal approximations. That guide distinguishes exact representations from rounded reports. Here the narrower question is how a finite decimal differs from a fully specified recurring expansion.
The SEAB 2027 G2 Mathematics syllabus provides the relevant wider number-and-algebra context. This article does not claim that the particular 0.999… proof is a separately prescribed examination topic.
If the result feels surprising, start with what the learner already knows about place value and fractions. The difficulty often concerns the meaning of an infinite representation, not an inability to subtract decimals. A calm sequence of labelled examples makes the distinction easier to inspect.
The written decimal 0.999 has exactly three digits after the decimal point. Its value is nine tenths plus nine hundredths plus nine thousandths, or 999/1000. That is an exact fraction.
Subtracting from one gives 1000/1000 − 999/1000 = 1/1000. Thus 0.999 is exactly 0.001 below one. It is not equal to one merely because it is close or might round to one at a lower precision.
The same reasoning applies to 0.9999, whose exact gap is 1/10000. Each finite terminating decimal can be expressed with a power of ten as its denominator, making the gap explicit.
This is an important foundation: the recurring-decimal result does not erase the distinctions between ordinary finite decimals. If the question writes 0.999 without a recurring indication, use its finite value unless another instruction explicitly changes the interpretation.
In 0.999…, the intended convention here is that every position after the decimal point contains a nine. There is no final nine followed by zeros. The ellipsis describes an infinite repeating pattern.
A bar over the recurring digit can express the same idea more explicitly. In plain text, a phrase such as zero point nine recurring avoids ambiguity. The notation must communicate the entire repeating pattern, not simply suggest that more unknown digits exist.
For example, 0.999123… with unspecified continuation is not automatically nine recurring. Its early digits already differ from the all-nines expansion, and the omitted continuation has not been fully defined by a repeating-nine convention.
Read what is actually specified. Dots after a number can indicate a known recurring expansion or an informal abbreviation. The mathematical conclusion depends on that meaning, so name the pattern before applying a familiar identity.
| Representation | Exact finite fraction where applicable | Gap below one |
|---|---|---|
| 0.9 | 9/10 | 1/10 |
| 0.99 | 99/100 | 1/100 |
| 0.999 | 999/1000 | 1/1000 |
| 0.9999 | 9999/10000 | 1/10000 |
| n finite nines | 1 − 1/10ⁿ | 1/10ⁿ |
| 0.999… with nine recurring forever | 1 | 0 |
The finite rows all have a last position and a positive gap. The final row refers to the value of the infinite expansion. It is not obtained by choosing an unusually large finite n and pretending its remaining gap is zero.
This table is useful because it preserves both true statements: every finite prefix is below one, and the recurring value equals one. The statements concern different objects, so they do not conflict.
Let sₙ be the decimal containing exactly n nines after the point. Then sₙ = 1 − 10⁻ⁿ. For n = 3, this gives 1 − 0.001 = 0.999. For n = 4, it gives 0.9999.
The formula follows from writing a number with n nines as the integer 10ⁿ − 1 divided by 10ⁿ. Expanding the fraction gives one minus one over 10ⁿ.
This expression is more useful than writing a very long string of digits. It records the exact value and exact gap for every finite length at once, including lengths too large to display conveniently.
The condition finite n remains important. You cannot substitute an ordinary number called infinity into the finite formula and perform routine arithmetic with it. Instead, examine what happens to the values and gaps as n grows without bound.
The gap 10⁻ⁿ becomes smaller than any chosen positive tolerance when n is sufficiently large. If the tolerance is one millionth, choose enough nines to make the gap less than that value. Smaller positive tolerances can be met by taking more positions.
That is the meaning of the gaps tending to zero. It is not a claim that one particular finite gap has already become zero. Each finite gap remains positive, while their limit is zero.
Therefore the finite values sₙ tend to one. By the usual definition of the infinite decimal as the limit of its finite place-value sums, 0.999… denotes one.
The definition supplies the exact value. We are not deciding that an inconveniently small gap may be ignored for practical purposes. Practical rounding would be an approximation; this recurring-decimal identity is exact within the stated real-number interpretation.
The sequence one half, three quarters, seven eighths and fifteen sixteenths approaches one, while every displayed term is smaller than one. Its limit is nevertheless one. This is the same type of distinction as the finite nines and their recurring value.
There is no requirement that a sequence reach its limit at a finite stage. Some sequences do, such as a constant sequence of ones. Others approach a limit without any finite term equalling it.
Saying the nines never reach one is therefore accurate if it refers to the finite-prefix sequence. It does not show that the infinite decimal’s value differs from one, because that value is defined by the limit rather than by a final finite term.
When discussing the result, keep the nouns clear: a term of the sequence, the sequence as a whole, and its limiting value. Confusion often arises when all three are referred to only as the decimal.
Let x denote the real value of the all-nines recurring decimal. It is at most one and at least every finite prefix sₙ. Suppose, for contradiction, that x is smaller than one by a positive amount d.
Choose enough nines that 10⁻ⁿ < d. Then sₙ = 1 − 10⁻ⁿ is greater than 1 − d, which equals x. But x must be at least that finite prefix. The two inequalities contradict one another.
Therefore no positive real gap can separate x from one. Since x is at most one, it must equal one. The argument uses a fixed hypothetical positive gap and a finite prefix that would exceed it.
This proof is helpful when someone imagines a tiny leftover amount. Any positive real amount they propose can be made larger than the remaining finite-prefix gap, so it cannot be the gap of the complete recurring value.
The fraction one third has recurring decimal 0.333…, with three repeating forever. Multiplying the fraction by three gives one exactly. Multiplying its recurring expansion by three gives 0.999… under the same infinite-decimal interpretation.
Thus 0.999… equals one. This is a familiar route for learners comfortable with recurring fractions, and it connects a surprising-looking decimal statement to an ordinary exact fraction identity.
Keep finite approximations separate. Three times 0.333 is 0.999, which is smaller than one. The finite decimal 0.333 is not exactly one third, so the finite calculation does not contradict the recurring calculation.
The equivalent-fractions guide develops the broader idea of different representations naming one value. Decimal notation can also have more than one representation for the same real number.
Write x = 0.999… with the repeating pattern specified. Then 10x = 9.999…. Subtracting x from 10x gives 9x = 9, because the recurring fractional parts agree exactly. Hence x = 1.
This argument works with the values represented by the infinite expansions. It is not simply erasing an unknown tail on a calculator screen. The identical recurring parts must be established before the subtraction is justified.
For a finite three-nine decimal, x = 0.999 and 10x = 9.99. Their fractional parts are not identical: one has three nines after the point and the other two. Subtraction gives 9x = 8.991, not nine.
The finite comparison shows exactly where an invalid imitation of the proof would fail. The recurring notation carries an infinite pattern that survives the decimal shift; a fixed finite string carries a different ending after that shift.
For sₙ = 1 − 10⁻ⁿ, multiplying by nine gives 9sₙ = 9 − 9 × 10⁻ⁿ. The positive missing amount remains visible for every finite n.
At n = 3, nine times 0.999 is 8.991. At n = 4, nine times 0.9999 is 8.9991. Both are below nine, with the exact difference determined by the number of finite places.
Their limits equal nine, matching the recurring proof. The finite and infinite calculations agree when each is interpreted according to its own defined quantity.
This is a useful general lesson about algebraic appearances. Two lines that look almost identical can encode different boundary or ending conditions. Before copying a manipulation, check which property makes that manipulation valid.
The exact decimal 0.5 also has the representation 0.4999…, with nines continuing forever after the four. The recurring tail 0.0999… equals 0.1, so adding it to 0.4 gives 0.5.
Likewise, 2.34999… equals 2.35, and 0.24999… equals 0.25. These are exact alternate expansions, not merely statements that the numbers round to the same display.
The terminating version is usually simpler to write, so it is commonly preferred. That preference does not make the recurring version a different number. A real value can have both representations at these decimal boundaries.
Do not generalise that any recurring decimal equals a nearby terminating one. The all-nines tail is the specific pattern responsible here. A decimal such as 0.444… equals four ninths, which is a different rational number.
The finite decimal 0.999 rounds to 1.00 to two decimal places under ordinary rounding. That reporting step does not make 0.999 exactly equal to one. The original value and its rounded report remain different.
The recurring decimal 0.999… equals one before any rounding instruction is applied. This is an exact identity, so it has a different reason from the finite number’s rounded report.
Many values can round to 1.00 at two decimal places. A shared rounded display therefore does not establish exact equality of the underlying values. The instruction specifies a reporting precision, not a reversible way to recover the original number.
This links to the broader exact-answer and approximation guide. Keep exact identities, finite approximations and rounded presentation as separate claims that can each be checked.
Question: compare A = 0.999, B = 0.9999 and C = 0.999… with nine recurring forever. Find each gap from one and decide which values equal one exactly.
A is 999/1000, so its gap is 1/1000. B is 9999/10000, so its gap is 1/10000. Both are strictly smaller than one, and B is closer to one than A is.
C is the limit of the all-nines finite prefixes. Their gaps 10⁻ⁿ tend to zero, so C equals one and has gap zero. The final recurring value does not contain a last positive finite gap.
If all three are reported to two decimal places, they produce 1.00. That shared report does not erase A’s and B’s exact differences. State both the exact comparison and the requested rounded presentation when a question asks for them.
Treat a decimal without a recurring indication as finite. Whenever dots are used below, the accompanying words specify the repeated pattern. Work with exact fractions where possible before giving rounded reports.
Question A. Write 0.999 as a fraction and find its gap from one. Question B. Repeat for 0.99999. Question C. Express a finite decimal with n nines as one minus a power of ten.
Question D. State the value of 0.999… when nine repeats forever. Question E. Compare three times 0.333 with three times 0.333… where three repeats forever. Question F. Evaluate 0.4999… with an infinite all-nines tail.
Question G. Evaluate 2.34999… with nines recurring after the four. Question H. Round finite 0.999 to two decimal places and state whether that proves exact equality with one. Question I. Explain why a calculator display of 0.999999 does not by itself mean nine recurring forever.
Question J. If x is the recurring all-nines value, explain why 10x − x = 9. Question K. Calculate the same subtraction for finite x = 0.999. Question L. Explain why every finite prefix being below one is compatible with the recurring value equalling one.
Answer A. 0.999 = 999/1000 and its gap is 1/1000. Answer B. 0.99999 = 99999/100000 with gap 1/100000. Answer C. The finite value is 1 − 10⁻ⁿ for a positive integer n.
Answer D. The recurring value equals one exactly. Answer E. Three times finite 0.333 equals 0.999, while three times the recurring expansion of one third equals one. The two starting values differ.
Answer F. 0.4999… equals 0.5. Answer G. 2.34999… equals 2.35. The complete recurring-nine tails supply the exact decimal-boundary values.
Answer H. The rounded report is 1.00. It does not prove exact equality because finite 0.999 remains 0.001 below one. Answer I. A screen shows a finite representation or report; its omitted information is not established as an infinite sequence of nines merely by appearance.
Answer J. Multiplying the defined recurring expansion by ten gives 9.999…, whose fractional recurring part equals that of x. Their difference is nine. Answer K. For finite x, 9.99 − 0.999 = 8.991.
Answer L. The finite prefixes are terms of a sequence, while the recurring decimal denotes its limit. A sequence can approach a value without any finite term equalling it. The exact formula 1 − 10⁻ⁿ makes both statements visible.
An objection sometimes imagines a last nine followed by a tiny gap in the next position. But the specified recurring expansion has no last nine. For every finite decimal position, another position follows and also contains nine.
The proposed next position after all infinitely many places is not an ordinary finite decimal place. Writing a one there does not define a positive real gap in standard decimal notation.
This is why a picture of a very long finite string can mislead. However many digits are drawn, the picture ends. The mathematical notation describes a repeating rule that continues through every finite position.
Use the exact finite-gap formula to keep the distinction clear. Every finite stopping point gives a positive 10⁻ⁿ gap. The complete recurring value is defined through the limit, whose gap is zero.
Suppose someone proposes that the gap is one trillionth. A finite prefix with more than twelve nines has a smaller gap than that. The proposed amount therefore cannot remain as the gap of the complete recurring value.
If they propose a smaller positive number, choose a correspondingly longer finite prefix. The argument does not rely on a particular tiny threshold. It works for any positive real gap that is suggested.
The conclusion concerns ordinary real numbers. Within that system, there is no smallest positive real number waiting after the repeating-nine expansion. Between zero and any positive real d lies another positive real, such as d/2.
That observation supports the earlier contradiction proof. Exact equality is not achieved by declaring a sufficiently small positive difference irrelevant; the proof shows that no positive real difference can satisfy the defining conditions.
SECTION 20 OF 33
19. Increasing prefixes and a fixed value are different descriptions
The sequence 0.9, 0.99, 0.999 and so on has changing terms. The recurring decimal 0.999… names one fixed real value. It is not a number physically moving along the number line as someone writes digits.
The writing process can illustrate the sequence, but the notation describes its mathematical limit. Distinguishing a process used to approximate a value from the value being named removes much of the apparent paradox.
The same distinction appears with π. A person can write increasingly accurate finite decimal approximations, while π itself remains one fixed number. A finite approximation process does not imply the intended real value is undefined until a writer finishes.
The particular value of the all-nines recurring expansion happens to be one. The general lesson is about representations and limits, not a claim that every endless decimal approaches an adjacent integer.
The recurring-nine value can be written as the infinite sum 9/10 + 9/100 + 9/1000 + …. Its finite partial sums are exactly the prefixes already studied. The infinite sum means their limiting value.
The geometric-series relationship gives a first term nine tenths and a common ratio one tenth. Its sum is (9/10)/(1 − 1/10) = 1. This provides another compact exact calculation for readers who have met geometric series.
The result relies on the common ratio having magnitude less than one, which makes the partial sums converge. It is not a licence to assign arbitrary finite values to every infinite sum or subtract unrelated divergent expansions.
This is optional connected reading. The finite-gap argument already explains the equality without requiring a geometric-series formula. Choose the representation that makes the reasoning clearest at the learner’s current stage.
For distinct finite decimals, aligning place values and comparing the first differing digit works. For 0.999 and 1.000, the integer part already shows the first is smaller. Their exact difference confirms it.
At the boundary between a terminating expansion and an equivalent all-nines expansion, there are two representations of one value. A superficial claim that the first visible digit differs therefore the values must differ overlooks that representational exception.
Using the usual terminating or trailing-zero representation consistently avoids that ambiguity in ordinary comparison work. The existence of an alternative all-nines expansion does not make place-value comparison unreliable; it requires understanding which representations are being used.
For teaching, demonstrate the exact identity first and then choose a standard representation for calculations. The goal is clarity about the value, not repeated switching of notation merely to make familiar comparisons appear confusing.
A calculator can show only finitely many visible digits. A displayed 0.999999 may be the exact finite input, a rounded or truncated report, or the result of finite-precision arithmetic, depending on the calculation and device.
The display alone does not establish that the intended value has nines in every later decimal position. To decide exact equality, inspect the original expression or a valid symbolic argument.
If the input was exactly 999999/1000000, the mathematical value is below one by one millionth. If the input was an exact expression known to simplify to one, a near-one display may reflect the numerical procedure rather than a different exact result.
Keep those possibilities separate instead of treating every row of nines as either automatically one or automatically evidence of a mathematical error. The source expression and representation rules determine what the digits mean.
SECTION 24 OF 33
23. Exact identity and numerical tolerance serve different purposes
In practical numerical work, a task may accept values within a stated tolerance. Then two distinct numbers can be close enough for that purpose. A tolerance comparison does not assert that their exact mathematical values are equal.
Finite 0.999 may be sufficiently close to one for a particular approximation task, but its exact difference remains 0.001. The recurring value 0.999… equals one with no positive real difference at all.
State which claim is being made. Approximately equal at a chosen precision, acceptable within a tolerance and exactly equal are different relations with different evidence.
This connects the example to school answer presentation. A correct rounded report follows the requested accuracy. An exact identity follows a mathematical derivation. Neither should be used as an unexplained substitute for the other.
Suppose a student says every finite string is below one, so the recurring value must also be below one. The finite observation is correct. The unsupported step transfers a property of every term to the limit without examining whether that property is preserved.
If another student says 0.999 equals one because it rounds to one, the error is different. The student has transferred a rounded-report agreement into an exact equality of the original values.
If a student treats an unspecified ellipsis as proof that nines recur forever, the issue is notation interpretation. Ask what pattern the dots establish before applying any arithmetic identity.
These distinctions help target practice. The learner may need a finite-gap calculation, a limit explanation, or an exact-versus-rounded comparison. Repeating the algebraic proof alone may not repair the particular meaning change that caused the objection.
Begin with finite decimals of increasing length and calculate their exact gaps as fractions. This establishes the place-value pattern without using infinity. Ask the learner to state why each finite gap is positive.
Then define the recurring notation explicitly and introduce the limiting-value meaning. Use the formula 1 − 10⁻ⁿ to connect the finite sequence with its limit. Ask whether a limit must be one of its finite terms.
Next compare finite 0.333 multiplied by three with recurring one third multiplied by three. Finally compare exact equality with a shared rounded display, so the two common sources of confusion are addressed separately.
For independent explanation, ask for one proof and one distinction: why the recurring value equals one, and why finite 0.999 still differs from one. Both belong in a complete understanding of the example.
First identify whether the representation is finite, explicitly recurring or merely abbreviated. For a finite decimal, write its exact fraction or place-value sum. For a recurring expansion, identify the repeating block and its intended infinite continuation.
Distinguish a finite prefix from the value of the whole expansion. Use an exact fraction argument, a partial-sum formula or an appropriate limit argument to establish the recurring value.
If the task requests rounding, perform that reporting step separately and mark approximation where appropriate. Do not turn a common rounded display into an exact identity without a reason.
The mathematics learning hub connects place value, fractions, exactness, approximation and proof. The useful habit is to preserve the meaning of the notation through every calculation, so a small mark such as an ellipsis does not silently change the question.
SECTION 28 OF 33
27. A finite prefix leaves a bounded range of possible continuations
If a nonnegative decimal begins 0.999, its first three digits contribute exactly 999/1000. Any further nonnegative decimal tail contributes at most 1/1000, because the largest possible tail has nines in every later position.
Thus knowing only the first three digits does not determine a unique value. A continuation of zeros gives finite 0.999. A continuation of all nines gives one. Other specified tails give values between those endpoints.
The particular recurring pattern matters. The statement begins with three nines supplies less information than nine repeats forever. Treating those descriptions as equivalent silently fills unknown digits with an assumed pattern.
This is useful when reading a shortened calculation or a textbook’s dots. Ask whether the continuation is defined by a repeating rule or merely omitted from the display. An exact identity needs the former kind of complete mathematical specification.
The argument also supplies an error bound for truncation: after n decimal positions, the omitted nonnegative fractional tail is at most 10⁻ⁿ. The all-nines example reaches that boundary and explains why finite-prefix approximations can be close while still distinct.
In binary place value, the first fractional position is one half, the next one quarter and the next one eighth. The binary expansion 0.111… with ones recurring forever therefore means one half plus one quarter plus one eighth and so on.
Its finite n-place prefix equals 1 − 2⁻ⁿ. The gaps tend to zero, so the recurring binary value equals one. A finite prefix still lies below one, just as finite strings of decimal nines do.
This optional extension shows that the phenomenon comes from positional representation, not a peculiar trick involving the decimal digit nine. In base ten, the largest digit is nine; in base two, it is one.
Keep the base label explicit when using the example. Binary 0.111 is seven eighths, whereas decimal 0.111 is one hundred and eleven thousandths. The same printed digits can represent different values when their place-value bases differ.
The extension is useful for readers connecting mathematics with computing, but it is not required for the decimal proof. The familiar finite-gap formula already contains the central reasoning in the base used throughout the rest of this guide.
SECTION 30 OF 33
29. Percentage notation preserves the finite-versus-recurring distinction
The finite percentage 99.9% equals 0.999 and is smaller than one, or one hundred per cent, by 0.1 percentage points. Its finite decimal representation has the same exact gap one thousandth on the zero-to-one scale.
If 99.999…% is explicitly defined with nines recurring forever after the decimal point, its value is exactly one hundred per cent. Dividing that percentage by one hundred gives the recurring decimal 0.999…, which equals one.
Do not use this identity to change an ordinary recorded percentage such as 99.9% into 100%. A finite recorded percentage and an explicitly infinite recurring notation are different information, just as in the original decimal comparison.
If a report rounds 99.9% to a whole percentage point, it may display 100%. That is a reporting decision, not proof that the unrounded finite value was exactly complete. Keep the underlying value and its chosen precision distinct.
This transfer makes the lesson practical. A small notation difference can change whether a statement claims a finite shortfall, an exact boundary value or a rounded display. The arithmetic should preserve whichever claim was actually supplied.
Let y = 0.272727…, with the block twenty-seven repeating forever. Multiplying by one hundred gives 100y = 27.272727…. The matching recurring fractional parts subtract to give 99y = 27.
Therefore y = 27/99 = 3/11. This familiar conversion uses the same pattern-shift reasoning as the all-nines proof, with a two-digit block instead of one digit.
For finite y = 0.2727, the same subtraction does not remove an identical infinite tail. The shifted number has a different finite ending, and the difference is not exactly twenty-seven. The recurring condition is doing real mathematical work.
This example can make the nine-recurring result feel less isolated. Recurring decimals are exact rational representations governed by a consistent algebraic method. The all-nines value happens to simplify to an integer rather than a noninteger fraction.
Check the fraction independently: three divided by eleven generates the repeated block twenty-seven under ordinary long division. A correct conversion preserves the full repeating pattern, while a finite prefix remains an approximation to that recurring rational number.
Compare finite 0.99999 with finite 0.999. Their gaps below one are 0.00001 and 0.001. Subtracting the values gives 0.00099, which is the difference between those two positive gaps in reverse order.
This calculation confirms that adding two more finite nines increases the value while leaving a smaller positive shortfall. No finite digit count has been treated as an infinite expansion, so the exact inequalities remain intact.
If a question asks how many nines are needed to make the gap less than 0.0001, use 10⁻ⁿ < 10⁻⁴. A finite positive integer n must exceed four, so five nines suffice. Four nines give a gap equal to the threshold rather than less than it.
The strict inequality matters. A question asking a gap at most 0.0001 would allow four nines. This is another example of reading the condition before calculating, now within the same finite-prefix model.
The exercise connects place value, powers of ten and boundary language without requiring a formal calculus course. It also keeps approximation targets distinct from exact recurring equality: meeting a positive tolerance is a finite task, while the all-nines identity names the limiting value.
Before answering, write the exact notation being discussed. Finite 0.999, recurring 0.999… and a display rounded to 1.00 can appear in one conversation, but moving between them requires an explicit reason. A proof about the recurring expansion should not be used to erase the finite number’s positive gap.
Likewise, a calculation showing that a finite prefix is smaller than one does not decide the value of the entire expansion. Keep the represented object fixed while evaluating a claim, and identify the definition or reporting step whenever it changes.
These answers use the ordinary real-number interpretation and an explicitly recurring-nine pattern. A finite display or an unspecified continuation needs its own reading before the same conclusion can be applied.
Is 0.999 exactly one?
No. With exactly three decimal places, it is 999/1000 and is smaller than one by 1/1000. It can round to one at a stated precision without being exactly equal to one.
Is 0.999… exactly one when nine repeats forever?
Yes. The finite prefixes equal 1 − 10⁻ⁿ, and their limit is one. The infinite recurring decimal denotes that limiting value. There is no positive real gap left in the exact identity.
Where is the last nine?
There is no last nine in the specified recurring expansion. Every finite position is followed by another nine. A very long finite written string has an ending, but the recurring notation describes a different mathematical representation.
What should I remember first?
Read the continuation. A finite decimal, an infinite recurring expansion and a rounded calculator report are different kinds of representation. Establish which one is meant, then use the corresponding exact calculation or approximation rule.
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