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Gecko:Image Snapping and Rendering: Difference between revisions
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Gecko:Image Snapping and Rendering (view source)
Revision as of 00:07, 14 November 2008
, 14 November 2008→Requirement 8
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Since F1' < F2' and they're both integers, we must have F2' - 0.5 > F1 (otherwise F1 would have rounded up to F2'). Therefore F2' - C > F1. Therefore S.(F2' - C - D1) > S.(F1 - D1). Therefore S.(F2' - C - D1) > I1. Now S.(F2' - C - D1) = S.(F2' - P' + D1 + P/S - D1) = S.F2' - S.P' + P = S.F2' + T. So S.F2' + T > I1. | Since F1' < F2' and they're both integers, we must have F2' - 0.5 > F1 (otherwise F1 would have rounded up to F2'). Therefore F2' - C > F1. Therefore S.(F2' - C - D1) > S.(F1 - D1). Therefore S.(F2' - C - D1) > I1. Now S.(F2' - C - D1) = S.(F2' - P' + D1 + P/S - D1) = S.F2' - S.P' + P = S.F2' + T. So S.F2' + T > I1. | ||
Similarly we must have F1' + 0.5 <= F2 (or F2 would have rounded down to F1'). Therefore F1' - C < F2. Therefore S.(F1' - C - D1) < S.(F2 - D1). Therefore S.(F1' - C - D1) < I2. Now S.(F1' - C - D1) = S.(F1' - P' + D1 + P/S - D1) = S.F1' + T. So S.F1' + T < I2. | |||
Hence the ranges have non-empty intersection. | |||