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Challenge^2: Floating Point without Errors

11 2020-10-18 02:12

>>10 Btw there is "pathological example" for 0.999...!=1
1. assume 1=0.999... , transform right parts to series
2. 1=9/10+9/10^2+9/10^n+.... ,substracting equals results in 0
3. 1-9/10-9/10^2-9/10^n-...=0 ,transform into series of divisions
4.10/10 -9/10 -> 10/100 -> -9/100-> 1/10 -> 1/10/10/10/10/10...=0
5. 1/10/10/...=1/10^n=0 , now multiply both sides by 10^n
6 1=0*10^n=0 , i.e. 1=0

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