Re: math
- Posted by "Carl W." <euphoria at cyreksoft.yorks.com> Jun 14, 2002
- 559 views
Kat wrote: > Does anyone have a list of basic math questions and answers, using big > numbers, like: > > 12345678901234567890.123456 > +9991230000000000.56666666666688889999999 > = the answer the answer = 12355670131234567890.69012266666688889999999 > x - y = z 3.1415926535897932384626433832795 - -2.7182818284590452353602874713527 = 5.8598744820488384738229308546322 3.1415926535897932384626433832795 - 2.7182818284590452353602874713527 = 0.4233108251307480031023559119268 -3.1415926535897932384626433832795 - 2.7182818284590452353602874713527 = -5.8598744820488384738229308546322 -3.1415926535897932384626433832795 - -2.7182818284590452353602874713527 = -0.4233108251307480031023559119268 You can also use the double negatives as addition tests. > x * y = z 3.5121409 * 3.16362908763458525001406154038726382279 = 11.111111111111111111111111111111111111111111111 (The decimal part has 45 ones _but_no_more_) > x / y = z The division ones are likely to be the killers if your code has any problems in it: 355355355355 / 113113122717.85053892523746376986 = 3.1415926535897932384626433832794792408640017539... (this isn't the exact value of Pi in case you're wondering) This is a good one: 10 / 81 = 0.12345679012345679012345679012346.... This has powers of two every 4 digits: 10000 / 4999 = 2.000400080016003200640128025605121024204840968193... By adding zeroes to 10000... and adding nines to 4999... you can increase the number of digits each power of two fits into. If you can do factorial by multiplying repeatedly; 1*2*3*....*48*49*50 / 3.1415926535897932384626433832795 = 2617410836119049074025095693555308.086784253501579842420985109612694... To test your divisors, you can always remultiply by them afterwards and see how big the error is. Hope this aimless rambling helped in some way, :) Carl PS I used something called 'bc' to get some of the answers here. It's available on most Unixes like BSD and Linux.