quick floating point check please!

riddler · 22 Nov 2006 at 01:57

i. (0.1*100)+(0.1011*100) = 100+1011 = 1111
ii. (0.10111*101)-(0.10011*101) = 100111 – 10011 = 100
iii. (0.11*100)*(0.1001*100) = 1100 * 1001 = 1101100
iv. (0.11*101)(0.1*11) = 11000 100

are all of these correct? thanks in advance

riddlermarc · 22 Nov 2006 at 10:40

/wonders why no-one has answered this...

I'm not sure what the issue is here, surely a calculator will give you the answer if you're not happy doing it in your head? Take each bracket separately, do them first..

i. (0.1*100)+(0.1011*100) = 100+1011 = 1111
= 10 + 10.11 = 20.11
ii. (0.10111*101)-(0.10011*101) = 100111 – 10011 = 100
= 10.21211 - 10.11111 = 0.101
iii. (0.11*100)*(0.1001*100) = 1100 * 1001 = 1101100
= 11 * 10.01 = 110.11
iv. (0.11*101)(0.1*11) = 11000 100
= (11.11)(1.1)

Should this not be in GD? :confused:

Sic · 22 Nov 2006 at 10:43

anything that resembles something that you might need some sort of remote brain for gets moved here.

riddlermarc · 22 Nov 2006 at 10:56

Sic said:
anything that resembles something that you might need some sort of remote brain for gets moved here.

lol, the HG&P Geek Squad® strikes again

xyphic · 22 Nov 2006 at 12:39

I'm guessing (due to the lack of anything that isn't a 0 or 1) these are binary expansions, so:

i. (0.1*100)+(0.1011*100) === (1/2 * 4) + ((1/2 + 1/8 + 1/16) * 4) = 2 + 2.75 = 4.75 === 100.11

Doing the binary arithmetic:

0.1 * 100 = 10
0.1011 * 100 = 10.11

10 + 10.11 = 100.11

Given that that doesn't match your result I'd guess one of us is doing something wrong! Could be my assumption of the binary though...

Eriedor · 22 Nov 2006 at 12:50

Um i checked yer first equation and it seems wrong. Not sure if im getting the wrong end of the stick as its been awhile since i did all this but i read it as:

Code:

 (0.1*100) + (0.1011*100)
=(0.5 * 4) + (0.6875 * 4)
=     2    +    2.75
=        4.75
=       100.11

/edit - beaten to it, didn't refresh before i posted ;p