How to perform logical inclusive OR operation on hexadecimal numbers? In logic there is so called OR operation that is quite clear to me as long as it is in the binary system. For example, if I want to OR such binary values as "101" (which corresponds to decimal "5") and "110" (which corresponds to decimal "6"), you would do it this way:
101
+
110
= 
111
The logic here is quite clear: if there is at least one "1", then the result must also be "1".
However, I have no idea how this operation can be performed on hexadecimal numbers.
For example, if I needed to OR such hexadecimal values as "1A" (which corresponds to decimal "26") and "1F" (which corresponds to decimal "31"), how would i do that than?
1A
+
1F
=
??
 A: To expand on copper.hat’s answer just a little, observe that every hexadecimal digit corresponds to a string of four bits:
$$\begin{array}{c|c}
\text{Hex}&\text{Bin}\\ \hline
0&0000\\
1&0001\\
2&0010\\
3&0011\\
4&0100\\
5&0101\\
6&0110\\
7&0111\\
8&1000\\
9&1001\\
A&1010\\
B&1011\\
C&1100\\
D&1101\\
E&1110\\
F&1111
\end{array}$$
Thus, any hexadecimal number converts very easily to binary: just convert the individual digits. Hex $B94A$, for instance converts to $1011\;1001\;0100\; 1010$. (That conversion is practically hard-wired, since I earned my spending money in college writing IBM 360 assembler language programs!)
A: Treat the hex number as binary. $\text{1A}_{16} = 11010_2$, $\text{1F}_{16} = 11111_2$, so the 'or' is clearly $\text{1F}_{16}$.
A: The other answers here are based on converting from hexadecimal to binary, then computing a bitwise OR. I would like to propose a different method without converting to a different base.
Where my idea comes from: (not-tested)
A standard diode in electronics only allows electricity to go one way. A simple OR gate in electronics  is two diodes pointing into each other. If you were to put a high voltage through one side (F) the output would be that same voltage (F). Even if you added a lower voltage (A) to the other input, the output would remain the same as the output is already at a higher voltage.
So, our current OR gates really end up being "Which Ever Is Greater" gates. That would mean your example of 1A+1F = 1F. Also, 5B+A0=AB. This is very different than a binary bitwise OR, but that is because it is a "true" hexadecimal bitwise OR.
