<![CDATA[Forced rounding create precision troubles]]>
https://help.geogebra.org/topic/forced-rounding-create-precision-troubles
Tue, 27 Mar 2018 04:48:59 +0000Mon, 26 Mar 2018 20:41:56 +0000Zend_Feed<![CDATA[calculate with 32/64 bit in decimal is alway an approximation. Usual is with rounding and in the most applications this is more exact (see attachment). If you want cut instead of round for your division you can use floor(n / 9 (10¹⁵)) / 10¹⁵ .]]>
https://help.geogebra.org/topic/forced-rounding-create-precision-troubles#comment-236642
Tue, 27 Mar 2018 00:22:59 +0000<![CDATA[Good morning, Dear @rami. I've understood what's the explanation: it's difference of numerical representation with finite precision and infinite precision. Example: - with FINITE precision we'll always have: 1/9 + 8/9 = 0.999... until maximum precision resolution and when we do rounding, we compensate that finite precision resolution. - with INFINITE precision we'll always have 1/9 + 8/9 = 1 ]]>
https://help.geogebra.org/topic/forced-rounding-create-precision-troubles#comment-236644
Tue, 27 Mar 2018 05:46:24 +0000