- Thread starter flashylightsmeow
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Thanks. Don't think so. Certainly not an active one. I'll check it out.My Math Forum. Are you also a member there?

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)

I used the same method to convert 3.75, but when it comes to 0.703125 it's not as straightforward.

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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.

Sure

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.

Here we go

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Ok thanks. I've never seen that method before.

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).

eg

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.

EDIT: Ok had another thought:

You could eg write 0.703125 as 703125/1000000 which cancels down to 45/64 which in binary is 101101/1000000.

This is an easy division to do and get 0.101101 as before. (Easy division because denominator 64 is a power of 2). Nice method after all!

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.

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).

eg

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.

EDIT: Ok had another thought:

You could eg write 0.703125 as 703125/1000000 which cancels down to 45/64 which in binary is 101101/1000000.

This is an easy division to do and get 0.101101 as before. (Easy division because denominator 64 is a power of 2). Nice method after all!

Last edited:

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).

eg

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.

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Thanks! Can I trouble you with another question. Ex.16 below. I have answered it like this:I edited my previous post, but must have been after you read it. Have another look.

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My approach would be:

Suppose y>x and let y= x+k where k >0.

\(\displaystyle \frac{x+y}{2}=\frac{x+x+k}{2}=\frac{2x+k}{2} = x +\frac{k}{2} >x \) …. (if k>0, then k/2>0)

If y =x+k then x =y-k

\(\displaystyle \frac{x+y}{2} =\) ...you can fil this in …\(\displaystyle <y\).

Therefore …………...