Is 0.999 repeating exactly equal to 1 or just approximately equal explain the proof

The key insight is that $0.\overline{9}$ is not a "process" that approaches 1 — it IS 1, exactly. **The confusion comes from thinking of $0.\overline{9}$ as a number with "infinitely many 9s followed by a last digit." There is no last digit. Infinity is not a number.** Proof using the definition of real numbers: A real number is defined by its equivalence class of Cauchy sequences. The decimal $0.\overline{9}$ and $1.000...$ are different representations of the same equivalence class. **For the "isnt there a number between them?" argument:** If $a < b$, then $(a+b)/2$ is strictly between them. But theres no number strictly between $0.\overline{9}$ and $1$. Try to construct one: $0.\overline{9} + \epsilon$ for any $\epsilon > 0$ gives a number $0.\overline{9} + \epsilon > 1$. Why? Because $1 - 0.\overline{9} = 0$ exactly. If you accept that $1/3 = 0.\overline{3}$, then multiply both sides by 3: $1 = 0.\overline{9}$. Its that direct.