4 ms·
Should probably read the articles first... I was really hoping for: a + b = c, where a = b = c I could only find: * a = b = c = 0 * a = b = c = k/inf (w
by bArray 7y ago
Should probably read the articles first... I was really hoping for:
a + b = c, where a = b = c
I could only find:
* a = b = c = 0
* a = b = c = k/inf (where k < inf and k > -inf)
Off-topic but are there other solutions? I checked in Wolfram Alpha and nothing...
- jmaa 7y agoNot sure where you get those weird infinity results from, but the "a = b = c = 0" solution is easily show: Note that "a + b = c" implies "b = c - a", and given our assumption that "a = b = c", we can rewrite as "a = a - a = 0". This obviously results in "a = b = c = 0". EDIT: Modifying the proof above, it becomes apparent that this is a property of groups (most number systems are groups): Again "a + b = c", iff "a + a = a" iff "a + a + (-a) = a + (-a)" which is "a = e". Note that addition over IEEE754 floats do not constitute a group, (in part) because "inf + inf == inf" evaluates to true.