In this note I am responding to some questions on the ``slash notation'' that were asked after class on 02/09/2002. Let's look at the address space defined by w.x.y.z/n . Popularly: This is the set of adresses, or class of addresses, that have the same first n bits as w.x.y.z . (The FIRST n bits: count from left). More accurately: This is the the class of addresses that if we bitwise AND them with the mask that consists of n ones followed by (32 - n) zeros gives w.x.y.z . From the second (correct!) definition we see that w.x.y.z/n makes sense only if the last (32 - n) bits in w.x.y.z are zero. The first n bits can be anything. (The first definition is ``sloppy'', and by definition a sloppy definition is not a definition!). Example: 128.235.0.0/18 . Since 128 = 1000 0000 235 = 1110 1011 (128 + 64 + 32 + 8 + 2 + 1) , 128.235.0.0/18 is the class of addresses of which the first 18 bits are 1000 0000 1110 1101 00 and the last 14 bits can be anything. For example, try 128.235.42.170 . Is this in 128.235.0.0/18 ? 128.235.42.170 = 1000 0000 1110 1011 0010 1010 1010 1010 mask = 1111 1111 1111 1111 1100 0000 0000 0000 (18 ones, 14 zeros) AND: 1000 0000 1110 1011 0000 0000 0000 0000 YES! 128.235.42.170 is in 128.235.0.0/18 . Next, try 128.235.170.170 128.235.170.170 = 1000 0000 1110 1011 1010 1010 1010 1010 MASK = 1111 1111 1111 1111 1100 0000 0000 0000 AND: 1000 0000 1110 1011 1000 0000 0000 0000 NO! 128.235.170.170 is NOT in 128.235.0.0/18 . (popularly: the 17-th bit is wrong, is one, must be zero) Another example: is 128.236.42.42 in 128.235.0.0/18 ? 128.236.42.42 = 1000 0000 1110 1100 0010 1010 0010 1010 MASK = 1111 1111 1111 1111 1100 0000 0000 0000 AND: 1000 0000 1110 1100 0000 0000 0000 0000 NO! (popularly: the 14-th bit is one, must be zero, and the 15-th and 16-th bit are zero, must be one). This could have been seen immediately! (Since 235 is not equal to 236, there must be a difference somewhere in the 9-th through 16-th bits). --- For mathematically inclined students: Consider the class w.x.y.z/n . suppose w.x.y.z has a (at least one) one among its last (32 - n) bits. Prove that the class of adresses w.x.y.z/n is empty. ---