MATH 3124: Problem Set 2Due on Tuesday, September 9, 2014

Brown 8:00am

Jonathan Ross

1

Jonathan Ross MATH 3124 (Brown 8:00am): Problem Set 2

Contents

Problem 1 3

Problem 2 3

(a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

(b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

(c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

(d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Problem 3 4

Problem 4 4

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Jonathan Ross MATH 3124 (Brown 8:00am): Problem Set 2

Problem 1

Find all x in the group Z12 of integers mod 12, under the operation of addition mod 12, that satisfy theequation (x x) (x x) = 0. Justify your answer.

In effect, the left side of the equation (xx)(xx) = 0 is equal to 4x (mod 12), because (xx)(xx)= 2x2x = 4x. Since x Z12 and the operation is defined under Z12, we must take 4x mod 12 to keepthe operation closed under Z12. The values of x that will satisfy the equation is where 4x is a multiple of

12, or in different notation, 4x mod 12 = 0. Therefore, only x = {0, 3, 6, 9} will satisfy the equation.

Problem 2

Let S = {(x, y) : x and y are real numbers with y 6= 0}. (Geometrically, this is the set of points in the planewith the x-axis - that is, the horizontal line y = 0 - removed. Define by (x, y) (z, w) = (z + xw, yw).

(a)

Is an operation on S? That is, is S closed under ? Justify your answer.

Yes, is an operation on S. With x, y, z, w R, we know that z+xw R. Additionally, because y, w 6= 0,yw R, and so yw 6= 0, the definition of will hold for all elements in S.

(b)

Find an identity element (e, f) for .

The identity element (e, f) for is (0, 1):

(x, y) (e, f) = (x, y) (0, 1)= (0 + 1x, 1y) = (x, y).

(c)

If (a, b) S, find an inverse (a, b) of (a, b).

An inverse (a, b) of (a, b) S will be defined such that (a, b) (a, b) = (0, 1) (the identity element). Theinverse of (a, b) is (a, b) = (ab , 1b ):

(a, b) (a, b) = (a, b) (ab,

1

b)

= (ab

+ a 1b, b 1

b)

= (ab

+a

b,b

b) = (0, 1).

(d)

Determine whether is commutative.

Problem 2 [(d)] continued on next page. . . Page 3 of 5

Jonathan Ross MATH 3124 (Brown 8:00am): Problem Set 2 Problem 2

Define (x, y) = (2, 3) and (z, w) = (6,4

5). Then

(x, y) (z, w) = (2, 3) (6, 45

) = (6 + 2 45, 3 4

5) = (

38

5,

12

5), and

(z, w) (x, y) = (6, 45

) (2, 3) = (2 + 6 3, 45 3) = (20, 12

5).

Since (x, y) (z, w) 6= (z, w) (x, y), is not commutative.

Problem 3

Our checkerboard has only four squares, numbered 1, 2, 3, and 4. There is a single checker on the board,

and it has four possible moves:

V : Move vertically; that is, move from 1 to 3, or from 3 to 1, or from 2 to 4, or from 4 to 2.

H: Move horizontally; that is, move from 1 to 2 or vice versa, or from 3 to 4 or vice versa.

D: Move diagonally; that is, move from 2 to 3 or vice versa, or move from 1 to 4 or vice versa.

I: Stay put.

We may consider an operation on the set of these four moves, which consists of performing moves successively.

For example, if we move horizontally and then vertically, we end up with the same result as if we had moved

diagonally:

H V = DIf we perform two horizontal moves in succession, we end up where we started: H H = I. And so on. IfG = {V,H,D, I}, and is the operation we have just described, write the table of G. Granting associativity,explain why G, is a group.

I V H DI I V H D

V V I D H

H H D I V

D D H V I

First, we must define an identity element. It is fairly clear that the identity element is I, because I paired

with any move (I, V,H, or D) results in the move itself. This does not depend on whether I comes before

or after the move because the table of G is symmetric; therefore is commutative. Also, it is fairlyclear that each move is its own inverse. The corresponding move cells in the table of G result in no net

movement, symbolized by I, the identity element. Because is associative, G contains an identity element,and each element in G has an inverse, G, is a group.

Problem 4

Let X be a nonempty set and let be an associate operation on X.It is known that X contains an element J such that J b = b J = b for all b X.It is also known that for every c X, there exists c X such that c c = J . (Every element has a leftinverse.)

Give a careful proof of the following statement: For all c X, c c = J . [HINT: Let c X; think about(c).]

Problem 4 continued on next page. . . Page 4 of 5

Jonathan Ross MATH 3124 (Brown 8:00am): Problem Set 2 Problem 4 (continued)

Proof. By the second axiom, we know that X contains, by definition, an identity element J . Now, let

c X and let K = c c. Then

K K = (c c) (c c) = c (c c) c = c (J c) = c c = K,

and so

J = K K = K (K K) = (K K)K = J K = K = c c,which is what we wanted to show.

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