Adding two binary numbers: a carry of $1$ and two bits that are both $1$.

Question

I understand the trivial cases of binary addition when we have something like, for example:

$101$

$+$

$110$

(stacked intentionally)

The computation proceeds as follows, from right to left (LSB to MSB):

$1 + 0 = 1 \text{carry} 0$

$0 + 1 + \text{carry0} = 1 \space\text{carry0}$

$1 + 1 + \text{carry0} = 0 \space\text{carry1} $

But what I don't understand is what we carry when we have something like this:

$011$

$+$

$011$

After evaluating the first set of rightmost bits, we get:

$1 + 1 = 0 \space\text{carry1}$

And then we have:

$1 + 1 + \text{carry1} = ???$

Here, $1 + 1 + 1 = 3$, which is $110$ in binary. Does that mean we would carry a bit to the next left column AND the second column over to the left from the current one?

Answer 1

You're wrong about 3 being written 110 in binary, it is written 11. In fact:

$3= 2 + 1 =1*2^1 + 1*2^0 = (11)_2$

So, to come back to your addition, 1 + 1 + carry1 = 1carry1

Answer 2

In your notation we have $1 + 1 + \text{carry}1 = 1\text{carry}1$, and you can just continue as normal.