A little help with some binary please...

Leeum

Leeum

Leeum

Associate
Joined
19 Sep 2005
Posts
1,242
Hi,

Doing some binary multipication and I'm a little stuck.

My working is as follows:

binarymult.jpg


The problem comes when adding it all up. I understand that for adding four 1's, it equals 0 carry 2. When I come to add up the next set of digits, I have three 1's and 2 already being carried. What happens then? :confused:

Any help appreciated.
 
You need to understand exactly what binary means, then it will be much clearer to you. Try turning it into a real number and doing the addition that way, then put that back into binary so you can see what the result should look like.
 
Hxc said:
Binary multiplication? Is that actually possible with a simple method?
Click to expand...

Why not? It's just the same as decimal multiplication, but you only use 1s and 0s!

Back to the OP, you mention that 4 1s added up is 0 with 2 carried.
IMO you should think all in terms of binary, i.e. 1 + 1 + 1 + 1 = 100, so it's 0 and you're carrying 10.
So, in the next column it's 0 and then 1 is the one after this.
 
64 + 32 + 00 + 8 + 4 + 0 + 1
x 32 + 0 + 8 + 0 + 2 + 1

I think...haven't done binary mathematics in a couple of years...used to do full silly maths with binary, hex, oct etc.

Edit: I'm quite likely wrong. :p :D

InvG
 
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I'd have thought (can't remember much of my BSc CompSci) but I'd have though that carrying 2 would equate to carrying 10. You'd effectively carry 0 to the first left, then 1 to the next left

001
001
001
001

You add the ones which gives 4, which equals 100 - this would carry like this:

10 (the carry)
001
001
001
001
--0

The end result would give you an answer of 100 - which is 4 :)

(Think Haircut above put it more gracefully)
 
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Just bothered to work it out manually (because Start->Programs->Accessories is too much like hard work).
1001001001111
Leeum said:
The problem comes when adding it all up. I understand that for adding four 1's, it equals 0 carry 2. When I come to add up the next set of digits, I have three 1's and 2 already being carried. What happens then? :confused:
Click to expand...
Well think of it like having 5 1s.

Start with the first 2
1 + 1 = 0 carry 1
Then the next 2,
0 + 1 + 1 = 0 carry 2
Then the final 1,
0 + 1 = 1 carry 2

So if you carry 2 x 1, then add 3 x 1 the 2 x1 end up rolling over and you end up with 1 in that position.
 
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Think the answer is:

100100100111

hope this comes out ok, but the carry's look like:

Code: 
---1-1-------
-11101011----
_____________
-----1101101
----1101101-
---0000000--
--11-11-1---
-0000000----
11-11-1-----
_____________
1001001001111

Edit to follow... :)
 
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You certainly were. Surely my gratitude is prize enough? :D

Haircut/ Poweredge... any chance I could add you to MSN? Just started my CS degree and maths isn't my strong point... especially when it's taught by a ukrainian lecturer.
 
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