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copied!<p>I'm not sure whether this will help clarify things, because I'm not sure what you consider to be the problem.</p> <p>Here is a test program that uses BigDecimal, as previously suggested, to display the values of the intermediate answers. At the final step, adding the third copy of 1.0/3 to the sum of two copies, the exact answer is half way between 1.0 and the next double lower than it. In that situation the round-to-even rounding rule picks 1.0.</p> <p>Given that, I think it <strong>should</strong> round to 1.0, contradicting the question title.</p> <p>Test program:</p> <pre><code>import java.math.BigDecimal; public class Test { public static void main(String[] args) { final double oneThirdD = 1.0/3; final BigDecimal oneThirdBD = new BigDecimal(oneThirdD); final double twoThirdsD = oneThirdD + oneThirdD; final BigDecimal twoThirdsBD = new BigDecimal(twoThirdsD); final BigDecimal exact = twoThirdsBD.add(oneThirdBD); final double nextLowerD = Math.nextAfter(1.0, 0); final BigDecimal nextLowerBD = new BigDecimal(nextLowerD); System.out.println("1.0/3: "+oneThirdBD); System.out.println("1.0/3+1.0/3: "+twoThirdsBD); System.out.println("Exact sum: "+exact); System.out.println("Rounding error rounding up to 1.0: "+BigDecimal.ONE.subtract(exact)); System.out.println("Largest double that is less than 1.0: "+nextLowerBD); System.out.println("Rounding error rounding down to next lower double: "+exact.subtract(nextLowerBD)); } } </code></pre> <p>Output:</p> <pre><code>1.0/3: 0.333333333333333314829616256247390992939472198486328125 1.0/3+1.0/3: 0.66666666666666662965923251249478198587894439697265625 Exact sum: 0.999999999999999944488848768742172978818416595458984375 Rounding error rounding up to 1.0: 5.5511151231257827021181583404541015625E-17 Largest double that is less than 1.0: 0.99999999999999988897769753748434595763683319091796875 Rounding error rounding down to next lower double: 5.5511151231257827021181583404541015625E-17 </code></pre>

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