| Assigned |
Target date | Read |
Exercises |
|---|---|---|---|
| 9/5 |
9/12 |
Ch. 0, pp.
3-17 |
0.4, 10, 12, 14, 17 |
| 9/12 |
9/19 |
Ch. 0, pp.
8-22 |
0.8, 13, 15, 18, 19 |
| 9/19 |
9/26 |
Ch. 0, pp.
8-22 |
0.22, 24, 25 |
| 9/26 |
10/3 |
Ch. 0, pp. 17-22 |
0.16, 26, 28, 49 |
| 10/3 |
10/10 |
Ch. 0, pp. 17-22 |
0.48, 50, 51 1. Find the error in the following argument. Let ~ be a relation on a set S. Suppose ~ is summetric and transitive. We claim that ~ is an equivalence relation. Since ~ is already symmetric and transitive, all we need to prove is that ~ is also reflexive. Let x be an element of S. If x~y then y~x by symmetry. Now x~y and y~x imply x~x by transitivity. Hence ~ is indeed reflexive and an equivalence relation. 2. Find a relation ~ on a set S which is symmetric and transitive, but not reflexive. |
| 10/10 |
10/17 |
Ch. 1 |
0.53 1. Let f:A->B and g:B->C be functions. Prove that if gf is onto, then g is onto. 2. Let S be a finite set and f:S->S a function. Prove that f is one-to-one if and only if it is onto. 3. Let S=natural numbers. Find a function S->S which is (a) one-to-one but not onto, (b) onto but not one-to-one. |
| 10/17 |
10/24 |
Ch. 1 |
Let f:S->T be a function. Prove that 1. f is one-to-one if and only if there exists g:T->S such that gf=idS, 2. f is onto if and only if there exists g:T->S such that fg=idT. |
| 10/24 |
10/31 |
Ch. 2 |
1.5-8, 14, 22 |
| 10/31 |
11/7 |
Ch. 3 |
2.4, 6, 14, 16, 18, 32 |
| 11/7 |
11/14 |
Ch. 3 |
3.6, 11, 23, 28, 44 Find a mistake in the statement in exercise 3.26. (Hint: are the elements of Zn really integers?) If you don't see a mistake, prove that the statement is false. |
| 11/14 |
11/21 |
Ch. 3 |
3.10, 12, 14, 18, 27 |
| 11/21 |
11/28 |
Ch. 4 |
3.15, 16, 20, 24, 32, 52 |
| 11/28 |
12/5 |
Ch. 4 |
4.14, 18, 22, 36, 42, 54 |