Thread:

1. Functions and properties

Hi guys, im am so frustrated right now. I suck at discrete math. I dont understand it at all, the proffesor talks to fast. Im on the vergre of tears. Plus theres no tutoring for this class, and i really need help. right now i have two problems

1. suppose g: A--->B and f: B---> C are both one to one, prove f (o) g what does the little o mean?) i have no idea how to start this.

2. if E is an equivalance equation on A , then prove or give an counter example, E (o) E is an equivalnce on A. Im not even sure what this is asking. i dont really get equivalance equations or sets and relations really, all i know is E are transitive, reflexive and symmetric.

2. Originally Posted by Wardub
1. suppose g: A--->B and f: B---> C are both one to one, prove f (o) g $what does the little o mean?$) i have no idea how to start this.
This notation suggests the succcessive operation of the two functions.

Map this through 'g' to the range of 'g'.
This becomes the Domain of some subset of 'f'.
Map the new value through 'f'.

3. but there is no value of f or g, what do i do when theres nothing to begin with?

4. prove f (o) g
Is this REALLY the problem statement? Doesn't seem like much to go on.

5. I assume that #1 is, if $f:A \mapsto B\quad \& \quad g:B \mapsto C$ are both one-to-one then so is $g \circ f$.
The proof is quite easy:
$\begin{array}{rcl}<br /> g \circ f(p) & = & g \circ f(q) \\ <br /> f(p) & = & f(q),\; \mbox{g is 1-1} \\ <br /> p & = & q,\quad \mbox{f is 1-1} \\ <br /> \end{array}$

For #2. If $R\quad \& \quad S$ are relations on a set that are reflexive, symmetric and transitive then $R \circ S$ is reflexive, symmetric and transitive.
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