Thanks Debsta. Please see the page below.Can you please show how you approach Ex 13 for 3.75? The example above Ex 13 refers back to Ex 1 which you have not included.
"Regarding Ex. 13. I understand that the answers can be found by multiplying successively by 2 and taking the whole unit at each stage as 1 and 0 where there is none."
I don't understand what that is saying.
The way I would approach it is 3.75 = 2 + 1 + 0.5 + 0.25 = 1x2^1 +1x2^0 + 1x2^-1 +1x2^-2 = 11.11 (bicimal)
SureYou say: "Regarding Ex. 13. I understand that the answers can be found by multiplying successively by 2 and taking the whole unit at each stage as 1 and 0 where there is none."
Can you please show me how you do this for 3.75 (ie using the hint given) and then I can try the next one. I'm still unsure how this process works.
I don't see the preceding text as being of any use here. (see EDIT below)My question is, is there another way?
Given the preceding narrative on expressing decimals as quotients of two rationals I thought it may be possible to do so and convert both numerator and denominator to binary and go from there.
Interesting! Thanks very much Debsta. I wonder why both methods work!?Ok thanks. I've never seen that method before.
I don't see the preceding text as being of any use here.
The method I would use is repeated subtraction of powers of 2 (incl neg powers for after the "decimal" point ie 0.5, 0.25. 0.125, 0.0625, 0.03125, 0.015625 etc).
0.703125 - 0.50 = 0.203125 ….. .1 (put a one if you can subtract)
can't subtract 0.25 ……..………… .10 (put a zero if you can't)
0.203125 - 0.125 = 0.078125 …. .101
0.078125 - 0.0625 = 0.015625 …. .1011
can't subtract 0.03125 ……………….. .10110
0.015625 - 0.015625 =0 …………… .101101
which is basically what you would do with an integer (using positive powers of 2).
The "repeated multiplication" way seems more efficient.