# Proof: 0.999... (with 9s repeating forever) is strictly less than 1.

- **Generated**: 2026-03-28
- **Verdict**: DISPROVED
- **Audit trail**: [proof_audit.md](proof_audit.md) | [proof.py](proof.py)

## Key Findings

- 0.999... (with 9s repeating forever) equals **exactly 1**, not less than 1.
- Four independent mathematical methods (algebraic, geometric series, fraction identity, numerical convergence) all confirm the value is 1.0.
- The strict inequality 0.999... < 1 is **false** in every standard and extended number system.
- No peer-reviewed mathematical literature disputes this result.

## Claim Interpretation

**Natural language claim**: "0.999... (with 9s repeating forever) is strictly less than 1."

**Formal interpretation**: The repeating decimal 0.999... (defined as the limit of the sequence 0.9, 0.99, 0.999, ...) is compared to 1 using strict inequality (<). The claim asserts this value is strictly less than 1.

**Operator choice**: The "<" operator is taken directly from the claim ("strictly less than"). We work in the standard real number system. The notation "0.999... with 9s repeating forever" is interpreted as the infinite series sum_{k=1}^{infinity} 9 * 10^{-k}, which is the standard mathematical meaning of a repeating decimal.

## Evidence Summary

| ID | Fact | Verified |
|----|------|----------|
| A1 | Algebraic proof: if x = 0.999... then x = 1 | Computed: 1.0 (equals 1 exactly) |
| A2 | Geometric series proof: sum of 9/10^k for k=1..inf equals 1 | Computed: 1.0 (equals 1 exactly) |
| A3 | Fraction proof: 1/3 = 0.333..., so 3 * (1/3) = 0.999... = 1 | Computed: 1.0 (equals 1 exactly) |
| A4 | Numerical convergence: partial sums approach 1 with zero gap | Computed: True (limit converges to 1) |

## Proof Logic

The proof disproves the claim by demonstrating that 0.999... = 1, using four mathematically independent approaches:

**Method 1 -- Algebraic (A1)**: Let x = 0.999... Then 10x = 9.999... Subtracting: 10x - x = 9, so 9x = 9, thus x = 1. This is verified computationally using Python's exact `Fraction(9, 9)` which equals 1 exactly.

**Method 2 -- Geometric Series (A2)**: The repeating decimal 0.999... = 9/10 + 9/100 + 9/1000 + ... is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since |r| < 1, the sum converges to a/(1-r) = (9/10)/(9/10) = 1. Verified using exact `Fraction` arithmetic.

**Method 3 -- Fraction Identity (A3)**: Since 1/3 = 0.333... (repeating), multiplying both sides by 3 gives 3/3 = 0.999... But 3/3 = 1, therefore 0.999... = 1. Verified via `Fraction(1, 3) * 3 = 1`.

**Method 4 -- Numerical Convergence (A4)**: The partial sums S_n = 1 - 10^{-n} converge to 1 as n approaches infinity. Using high-precision decimal arithmetic (60 digits), the gap between S_n and 1 shrinks to zero. By the Archimedean property of the reals, a number whose distance from 1 is less than every positive real number is equal to 1.

All four methods independently confirm that 0.999... = 1.0 (A1, A2, A3, A4). Since 1 is not strictly less than 1, the claim is disproved.

## Counter-Evidence Search

- **Alternative number systems**: Investigated hyperreals, surreals, and p-adic numbers. In all of these, the notation "0.999... with 9s repeating forever" (one 9 for every natural number) still equals 1. Only a non-standard truncation with a hypernatural number of digits could differ, but that's not what the claim describes.
- **Algebraic proof validity**: Confirmed that multiplying infinite repeating decimals by 10 is rigorous when grounded in epsilon-delta limits.
- **Geometric series convergence condition**: Verified |r| = 1/10 < 1, so the formula applies.
- **Literature search**: No peer-reviewed mathematics paper disputes 0.999... = 1. Pedagogical research documents why students resist the equality, but the mathematical community is unanimous.

## Conclusion

**DISPROVED.** The claim that 0.999... (with 9s repeating forever) is strictly less than 1 is false. Four independent mathematical methods -- algebraic manipulation, geometric series summation, fraction identity, and numerical convergence analysis -- all confirm that 0.999... = 1 exactly. The strict inequality fails because 1 is not less than 1. This is a well-established theorem in real analysis with no credible counter-arguments in the mathematical literature.

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Generated by [proof-engine](https://github.com/yaniv-golan/proof-engine) v0.10.0 on 2026-03-28.
