I am confuse that how carry works while multiplying two binary numbers
1111 x 1111 = 011100001
but how to handle carries in calculation? Can anyone please explain?
I know how to handle partial sum method when there are three 1s to be added while working out final result of multiplication, but I don't know, how to add four 1s in partial sum?
$\endgroup$How carry works while adding four binary 1s in partial sum: (1)2 + (1)2 + (1)2 + (1)2
2 Answers
$\begingroup$Since $1_2+1_2+1_2+1_2=100_2$, you add a $1$ two columns to the left to the one you are working with.
$\endgroup$ 0 $\begingroup$Keep calm and carry on, so to speak.
When the carry into a given place value is at least $10$ (meaning two, of course), only the last bit stays put. The preceding bits in the carry must be carried further into the next place.
For this multiplication you carry $10$ from the fourth place into the fifth (counting units as the first place, twos as the second place, etc), having added $1$ to that place from a previous carry. But only the $0$ bit stays there; the $1$ bit in the $10$ goes over to the sixth place.
As you go along you will find more carries of $10$ or more, and ultimately you have to keep up the transfers all the way into the eighth place. This makes sense because, switching to base ten, $15×15=225$ versus $2^7-1=127$.
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