Is 0.999... equal to 1 — prove it

Question

Is $0.\overline{9}$ (0.999…, with the 9s repeating infinitely) equal to 1? Prove your answer using at least two different methods.

Solution — Step by Step

Most students initially say “0.999… is almost 1, but not quite.” This intuition is wrong. $0.\overline{9}$ is exactly equal to 1 — not approximately, not “infinitely close to” — precisely equal. The three proofs below all reach the same conclusion, and none of them has any logical flaw. The discomfort comes from thinking of infinite decimals as “going on forever and never finishing” — but mathematically, $0.\overline{9}$ represents a completed infinite sum with a definite value.

Let $x = 0.\overline{9} = 0.9999...$

Multiply both sides by 10:

10x = 9.9999...

Subtract the first equation from the second:

10x - x = 9.9999... - 0.9999...

9x = 9

x = 1

Therefore $0.\overline{9} = 1$ . $\square$

This works because when we subtract, all the 9s after the decimal cancel perfectly — the result is exactly 9 with no remainder.

We know that $\frac{1}{3} = 0.\overline{3}$ (this is accepted as a standard decimal conversion).

Multiply both sides by 3:

3 \times \frac{1}{3} = 3 \times 0.\overline{3}

1 = 0.\overline{9}

Therefore $0.\overline{9} = 1$ . $\square$

This is perhaps the simplest proof — it follows directly from a fact most students already accept.

$0.\overline{9}$ is the infinite series:

0.\overline{9} = 0.9 + 0.09 + 0.009 + \cdots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots

This is a geometric series with first term $a = \frac{9}{10}$ and common ratio $r = \frac{1}{10}$ .

Since $|r| = \frac{1}{10} < 1$ , the series converges. Its sum is:

S = \frac{a}{1 - r} = \frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1

Therefore $0.\overline{9} = 1$ . $\square$

If $0.\overline{9} \neq 1$ , there must be a real number between them (by the density property of real numbers — between any two distinct reals, there is another real).

Can you name any number $x$ such that $0.\overline{9} < x < 1$ ?

Try any decimal: $0.9999...5$ doesn’t make sense (you can’t put something after an infinite string of 9s). There is no such number. Therefore, they cannot be different. $0.\overline{9} = 1$ . $\square$

Why This Works

This result reveals something deep about the real number system: every real number has a unique decimal representation, except that numbers whose decimal terminates can also be written with an infinite string of 9s. So 1 = 1.0000… = 0.9999…, just as 0.5 = 0.4999… The “two representations” look different but are the same number.

The key concept is that repeating decimals represent limits of infinite series, not unfinished approximations. $0.\overline{9}$ is not “approaching” 1 — it IS the limit of the sequence 0.9, 0.99, 0.999, … which is exactly 1.

Alternative Method — Using Limits Explicitly

0.\overline{9} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{9}{10^k} = \lim_{n \to \infty} \left(1 - \frac{1}{10^n}\right) = 1 - \lim_{n \to \infty} \frac{1}{10^n} = 1 - 0 = 1

Since $\lim_{n \to \infty} \frac{1}{10^n} = 0$ (powers of 10 grow without bound), the limit equals exactly 1.

Common Mistake

The most common objection is: “But there’s always a difference of $0.000...01$ between $0.\overline{9}$ and 1.” This is a misconception. The string $0.000...01$ is only meaningful with a finite number of zeros — you can always put the 1 at the end. With infinitely many zeros, there is no “end” to put the 1 at. The value $0.000...01$ with infinitely many zeros is exactly 0, not a small positive number. This is precisely what makes $1 - 0.\overline{9} = 0$ .

In board exams, the algebraic proof (Proof 1) is the expected method — set $x = 0.\overline{9}$ , multiply by 10, subtract. This is the standard NCERT approach for converting repeating decimals to fractions. Geometric series proof is for Class 11 students.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Limits Explicitly

Common Mistake

Try These Next