Geogebra afrunder tal med for mange decimaler.

Filip Binderup shared this problem 2 years ago

Not a Problem

Danish:

Hvis et tal har mere end 9 decimaler afrunder den til nærmeste 8. decimal. dette vil sige at 1-(10^(-9)) vil den afrunde til 1, dette vil sige at hvis jeg har brug for at runde 1-(10^(-12)) ned, ved hjælp af floor() kommandoen, vil jeg stadig få 1, selvom det rigtige svar er 0.

English:

If a number has more than 9 decimals, it will automatically be rounded to the nearest 8th decimal. This meas that 1-(10^(-9)) will be rounded to 1. I need to round 1-(10^(-12)) down using the floor() command, I would get 1 when the value I should get, and that I need, is actually 0.

The same problem

Comments (4)

Michael Borcherds ● 2 years ago

You mean with a==b? Yes, that's deliberate

Please post your .ggb file so we can help

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Filip Binderup ● 2 years ago

The problem is that floor(1-10^(-9))=1 when it should be 0

Here's the ggb file, but beware, it's in danish.

it's a unit converter, where you can turn a number into either binary, hexadecimal and more, if you try to convert a number with more than 9 digits, the code breaks and more often than not it'll give back an incorrect answer.

Files: here.ggb

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Michael Borcherds ● 2 years ago

Well, you can try a trick like floor(1-x*1000000)/1000000 but you will always run into limitations at some point with floating point numbers. You might have more luck using the CAS which keeps numbers exact

https://docs.oracle.com/cd/...

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Michael Borcherds ● 2 years ago

ps https://wiki.geogebra.org/e... https://wiki.geogebra.org/e...

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