# How to calculate this binary operation? [closed]

I found this binary operation, however It is the first time that I found this operator '.' :

That is the equation:

02 • 63 = 00000010 • 01100011 = 11000110

I don't understand how 0 • 0=1

I would be very grateful if you could help me.

## closed as off-topic by tylo, fgrieu, otus, e-sushiJul 21 '16 at 5:15

• This question does not appear to be about cryptography within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• This operator $\cdot$, like multiplication? Check the wiki for how binaray multiplication works. en.wikipedia.org/wiki/Binary_multiplier Though binary multiplication by 2 is simply a left shift. – Fleeep Jul 20 '16 at 14:12
• " 00000010 • 01100011 = 11000110 " stands for a variety of operators • . Examples include ordinary multiplication when values are in binary; ordinary multiplication modulo $2^8$; carry-less multiplication; and multiplication in $GF(2^8)$ for any polynomial. We have no way to guess which without context. Regarding " 0 • 0 = 1 ", that seems to be an error. – fgrieu Jul 20 '16 at 14:16
• I'm voting to close this question as off-topic because it has nothing to do with cryptography at all. It is just regular multiplication, where numbers are written in binary. – tylo Jul 20 '16 at 14:17
• @tylo, I found this operation in the description of an example of AES.iu.edu.jo/files/FacultyIT/Computer-Science/Courses/… page 7 – user6594048 Jul 20 '16 at 14:32
• I'd guess it's multiplication in the binary finite field GF(2^8). – CodesInChaos Jul 20 '16 at 14:48

This '.' is not 'AND' operation , its modular multiplication in the Galois Field $$GF(2^8)$$ and other operation $$\oplus$$ used in Mix Column Step of AES encrytion is modular addition in the Galois Field $$GF(2^8)$$. Here is a link to youtube channel by Christoff Paar where you can understand well, Watch lecture 7 and 8.