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binary division discrepancy

Hello, I am trying to understand the following modulo 2 long division.

I understand that 1101 goes into 1111, so you write a 1 on the top.

The next stage after 'XOR'ing is to bring down the 1.

Now, 1101 does not go into 101, so you write a 0 at the top. So far so so good.

Now you bring down another 1 and 1101 still does not go into 1011, so how come they wrote a 1 in the third position at the top ? Shouldn't that be a zero ?

If someone could please explain where I am going wrong, I would most grateful.

Re: binary division discrepancy

It is wrong. In the second part of the division they take 1101 away from 1011 incorrectly.

Re: binary division discrepancy

sorry, I have explained myself more clearly in the edited first post.

Still looking for an answer if anyone can help.

Re: binary division discrepancy

Ok I thought you just wanted confirmation that it was wrong.

Quote:

so how come they wrote a 1 in the third position at the top ? Shouldn't that be a zero ?

Yes they made a big mistake there.

It should look like this

http://puu.sh/2kPgM

Re: binary division discrepancy

$\displaystyle \text{Hello, fran1942!}$

$\displaystyle \text{The problem is: }\:2026 \div 13 \:=\:155, r11$

Code:

1 0 0 1 1 0 1 1

-----------------------

1 1 0 1 | 1 1 1 1 1 1 0 1 0 1 0

1 1 0 1

-------

1 0 1 1 0

1 1 0 1

---------

1 0 0 1 1

1 1 0 1

--------

1 1 0 0 1

1 1 0 1

---------

1 1 0 0 0

1 1 0 1

---------

1 0 1 1

I suspect a typo in the problem.

$\displaystyle \text{If the dividend were }11,\!111,\!101,\!{\color{red}10}\;\!0_2$

. . $\displaystyle \text{the problem would be: }\:2028 \div 13 \:=\:156$