# Lesson 34 Solutions

1. Consider the set $\{0, 1, 2\}$ and define clock arithmetic on this set in the usual way.  Namely, 0 is the identity element so that anything plus 0 is the same original “anything” again, 1 plus 1 is 2 as usual, but then we “cycle back through” the end of the set.  For example, 2 plus 1 is 0.  The question is, then, what is 2 plus 2?  What about 2 plus 2 plus 2?  What is the inverse of each element in this group?

Solution: We note that this is addition modulo 3, so that any time we hit the number 3 we cycle back through to zero.  Thus, we can calculate 2 plus 2 by first adding 1 to 2, and then adding 1 again to 2.  But, when we add 1 to 2, we get to 3, which means we’re actually at 0, so that when we add 1 again we land at 1.  Thus, we get the strange (but awesome) result that $2+2=1 \ \ \ \ \ mod \ 3$

where the “mod 3” is there to remind us that this result holds in addition modulo 3.  Similarly, 2+2+2 can be computed by seeing that 2+2 is 1, as we found above, and then noting that 1 plus 2 is 3, which is actually 0.  Thus, we get the strange (but awesome) result that $2+2+2=0 \ \ \ \ \ mod \ 3$

Finally, let us find the inverse of each element of the group.  We know that the inverse of 0 is 0, since 0 plus 0 is 0.  Similarly, since 1 plus 2 is 0, we see that the inverse of 1 is 2, and the inverse of 2 is 1.  Thus, in some sense, when we’re working with addition modulo 3, we have that “1 is equal to negative 2” and that “2 is equal to negative 1.”  All of which are pretty cool results.

2.  Consider the set $\{0, 1, 2, 3, 4, 5, 6, 7\}.$  Using addition modulo 8 on this set (meaning we simply cycle back through the end of the set, as usual), determine what 5 plus 6 is.  Also determine 7 plus 7.  Finally, determine the inverse of 3.

Solution: We attack this problem the same way we did in the first exercise.  Namely, 5 plus 6 can be computed by adding 1 to 5 six times.  Let’s stay calm and do this one at a time.  5 plus 1 is 6, plus 1 is 7, plus 1 is 8 which is actually 0 (!!!) so that we have found 5 plus 3 equals 0.  We now need to still add three more to this, but 0 plus 3 is just 3.  Thus, we have that $5+6=3 \ \ \ \ \ mod \ 8$

Similarly, one finds that 7 plus 1 is 8 and is therefore actually 0, so that 7 plus 7 is equal to 6 plus 0 which is just 6.  Thus we have $7+7=6 \ \ \ \ \ mod \ 8$

Finally, let us compute the inverse of 3.  All we need to do is notice that 3 plus 5 equals 8 and therefore actually equals 0.  Thus, the inverse of 3 is 5, so that “5 equals negative 3” when we work modulo 8.

3.  Consider the set $\{0, 1, 2, 3, ..., N\}$ for some positive number $N.$  What is the inverse of the element $N-2$

Solution: We note that with the set given in this problem, we are working with addition modulo $N+1.$ (For example, in the main text of this lesson we had $N=4$ and that was addition modulo 5, in the first exercise above we had $N=2$ and that was addition modulo 3, and in the previous exercise we had $N=7$ which was addition modulo 8.)  Thus, once we get to the number $N+1$ we know we’re actually at 0.  Now, when we add 2 to the number $N-2$ we end up back at the number $N,$ so that if we add one more we end up at $N+1,$ which is actually 0.  Thus, the inverse of $N-2$ is 3, since $N-2$ plus $3$ is zero!  One is encouraged to check that this is all in line with the examples that we have discussed for particular values of $N$ in these exercises and in the main text of the lesson.

Back to Lesson 34

On to Lesson 35