# Math E-301: Homework 4 Notes Due 10/5/09

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2 (c) We know that 1 in Z has a multiplicative inverse, namely 1 itself. Do any other elements of Z[i] have multiplicative inverses? Find all such elements or explain why there are none. The element 1 and 1 are their own inverses, and i and i are inverses of eachother since i( i) = 1. That s it, in fact. Terminology note: The elements of a ring (like Z[i]) which have multiplicative inverses are called units. So in Z[i], the units are ±1, ±i. To show this is all the units, we have to get a bit more involved. It s not expected that you would have come up with this on your own, but it s a nice thing to think about now. Let s say we have an element a + bi which is a unit. Then there s a multiplicative inverse for a + bi, call it c + di. So (a + bi)(c + di) = 1, and by part (b), we can see this means ac bd = 1 and ad + bc = 0. This limits the possibilities for a and b, but it s not at all obvious from this that only ±1 (in other words, a = ±1, b = 0) and ±i (in other words, a = 0, b = ±1) can be units. One way to show this is to use our geometric perspective on Z[i]. Every element a + bi in Z[i] can be viewed as a point (a, b) on a 2D plane, and we can think of it s distance from the origin, a 2 + b 2. The norm of a + bi is the square of this distance: we write, N(a + bi) = a 2 + b 2. Let s say we have an element a + bi which is a unit. Then there s a multiplicative inverse for a + bi, call it c + di. So (a+bi)(c+di) = 1. If we take the norms of both sides, we can check that we get (a 2 +b 2 )(c 2 +d 2 ) = 1 (since the norm of 1 is 1). So if a + bi is a unit, then a 2 + b 2 must divide 1, since it s an integer. But, this means a 2 + b 2 = ±1. Now it s easy to see this can only happen in the cases mentioned above. Whew! 2. In class, we discussed the Civil War number puzzle: Start with any number N. Partition it into two numbers, A and B with A + B = N and compute A B. Next split A and B into two summands each (i.e. W, X with W + X = A, and Y, Z with Y + Z = B), and compute W X and Y Z. Keep going until each number has been split into 1 s. Now go back and sum all the products you computed at each step. No matter how you broke things down you should get the same sum. For example, for the number 8, we might write 8 = so the first product is 15. Then write 5 = 2+3 and 3 = 1+2 and we get the products 6 and 2. Next write 2 = 1+1, 3 = 1+2, 2 = 1+1 and get products 1, 3, 1. Now there is only one 2 left to break up: 2 = so we get one last product, 1. The sum of all these is 28. (a) Try this out for N = 4, 5, 6, 7. For each N, break up the number in several ways to confirm that you get the same final sum of products no matter how you break it down. (b) What do you notice about the sequence S N, where S N is the sum of all the products when you break down N? Make a conjecture and generate more data to test it. You should get S 4 = 6, S 5 = 10, S 6 = 15, S 7 = 21. This sequence 6, 10, 15, should remind you of triangular numbers. In particular, it appears that: Conjecture: S N = T N 1, where T N 1 is the (N 1)st triangular number. We can check this works for N = 2, 3, 9,... until we re feeling pretty convinced. Then it s time to look for an answer. (c) (Extra) Try to prove your conjecture using the following idea: Think of N cities, with every city connected by a road. Each time you partition a number, for example 8 = 5 + 3, think of this as separating the group of 3 cities from the group of 5 cities by destroying all roads between them. How many roads are destroyed?

4 (c) If you square an integer, what are the possibilities for its last digit? See previous part. 5. In class, we asked whether in a commutative ring (R, +, ) the cancellation property holds true for multiplication. In other words, if A, B, C are non-zero elements, does A B = A C imply that B = C? (a) Explore this question in the commutative rings Z m, for m = 5, 6, 7, 8, 9, 10, 11. Does it ever fail to hold true? It fails in Z 6, Z 8, Z 10. For example, 2 3 = 4 3 in Z 6, since both 6 and 12 are equivalent to 0 modulo 6, but 2 4 in Z 6. (b) In a commutative ring, we call a non-zero element A a zero divisor if there is another element B 0 such that A B = 0. For each Z m for m = 5, 6, 7, 8, 9, 10, 11, find all zero divisors, if there are any. What do you notice about the connection between zero divisors and the cancellation property? In Z 6, the zero divisors are 2,3. In Z 8, 2, 4, 6. In Z 10, 2 and 5. We can notice that the zero divisors of Z m are those numbers A which share a common factor with m. Why does this make sense? When a zero divisor like A is involved in a multiplication problem AB AC mod m, we can t necessarily cancel (as in the example above). (c) Give an argument using the properties of commutative rings to show: If a commutative ring R has no zero divisors, then the cancellation property holds true. We start by assuming that R has no zero divisors. We want to show if AB = AC in R where A, B, C are not zero, then B must equal C. Rewrite AB = AC as AB AC = 0, by adding AC, the additive inverse of AC, to both sides. By distributive property, A(B C) = 0. Since there are no zero divisors, either A or B C must be zero. (Otherwise we d have two non-zero elements multiplying together to be 0 in R.) But A is not zero by our assumption, so B C must be zero. Adding C to both sides, shows that B = C. Ta da! (or the more traditional, QED.) (d) Extra Is the other direction true? In other words, if the cancellation property holds in R, does this imply that R has no zero divisors? Yes it is! So together with what we know from (c), we see that having no zero divisors is the same as having the cancellation property for a ring. Try showing this by contraposition. In other words, assume that R is a ring with a zero divisor, call it A and show that the cancellation property must fail for some elements of R. Please feel free to talk to me if you d like another hint. 6. Let s start looking at higher powers in Z m. (a) Make a chart showing the first 10 powers of each element in Z 5. For example, for the element 2 in Z 5, the list would be 2 1 2, 2 2 4, (since 2 3 = 8 has remainder 3 when divided by 5) and so forth.

5 We did this in class, but here it is again: Z (b) Now write out a table of the first 12 powers of the numbers 2 and 3 in Z 7 and the first 12 powers of 2 and 3 in Z 10. What do you notice? From this data, make some observations about the patterns that you re seeing in these various rings/fields. Z Z Notice that 2 4 = 16 6 mod 10. So when we compute 2 5 we can do mod 10. After this point, we start to cycle back through the powers. We can notice that neither 2 nor 3 cycle through all the elements of Z 10, and there is a power of 3 that gets us 1, but no such luck for 2. But in Z 7 and Z 5, we always seem to get 1 as the result of some power. Hmm... why does this make sense? 7. We ended class by solving some linear equations Ax B in Z 7 and Z 8 and we saw situations where there was exactly one solution, more than one solution or no solutions. To explore the question How many solutions will Ax B have in Z m? solve the following linear equations in each of the various Z m listed. Then make some observations/conjectures about the number of solutions Ax B will have in Z m, depending on the values of A, B and m. Feel free to generate some more data of your own by solving other linear equations too. All these questions can be answered by looking at the multiplication table for Z m. For example, to solve 6x = 3 in Z 11, we look for 3 in the 6-row and then look to see the number x whose column it s in, in this case x = 6. Check? 6 6 = 36 3 mod 11. There are many observations to be made, many based on the greatest common divisors of A and m. In class, we mentioned that since Z p is a field, (still to be proved), there will always be exactly 1 solution to Ax B mod p, since A has a multiplicative inverse in Z p so we can divide both sides by A to solve for x. See HW 5 #9 for some observations- We ll be exploring these more later in the course. (a) 2x 4 in Z 5, Z 6, Z 7, Z 8 In Z 5, x = 2. In Z 6, x = 2, 5. In Z 7, x = 2. In Z 8, x = 2, 6. (b) 5x 1 in Z 6, Z 7, Z 10, Z 11 In Z 5, x = 5. In Z 7, x = 3. In Z 10, there are no solutions. In Z 11, x = 9. (c) 5x 2 in Z 6, Z 7, Z 10, Z 11 In Z 6, x = 4, in Z 7, x = 6, in Z 10, no solutions, in Z 11, x = 7. (d) 5x 5 in Z 6, Z 7, Z 10, Z 11 In Z 6, x = 1, in Z 7, x = 1, in Z 10, x = 1, 3, 5, 7, 9, in Z 11, x = 1.

6 (e) 6x 3 in Z 8, Z 9, Z 10, Z 11 In Z 8, no solutions, in Z 9, x = 2, 5, 8, in Z 10, no solutions, in Z 11, x = Through much of this problem set, you ve been working with the multiplication tables for Z 5 through Z 11. Please make at least two of your own observations about patterns in these multiplication tables. The multiplication tables for these seven different systems should give you plenty of data to work with as you look for patterns. Then try to come up with an explanation for why the patterns you observe exist. Depending on what pattern you notice, this may be difficult to do with just the tools we have so far, but do your best to come up with some type of plausible reason. There are plenty of patterns to note. For instance, in all the rings Z m when m is even, there is a middle row that alternates a0a0a0... where a equals m 2. There seems to be a frame around all the multiplication tables going clockwise from the upper left corner consisting of 1, 2, 3,...m1, which then descends back to 1, then ascends back up to m1, then back down to 1. Think about why this might be true. An explanation is here 1. Note too that the main diagonal is always a palindrome (for instance in Z 11, its ). Explanation here 2. (Extra questions to explore, research, ponder...) These are some of the other open questions we left class with... What is Z 1? How can you show associativity in Z[i] using associativity of Z?

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