
Re: Defining "f o g" with Functions
Posted:
Oct 15, 2012 3:17 AM



f={(a,b)} implies f(a)=b f o g (x)=f(g(x)) f o g =(1,6), (2,4),(3,7),(4,4),(5,3) since g maps 1 to 2, which f maps to 6, and g maps 2 to 3, which f maps to 4 etc. g is not onetoone since both 2 and 4 are mapped to 3. f is onetoone since no two ordered pairs have the same number as either their first or second element. A function has an inverse if and only if it is onetoone. f^1={(6,2),(7,1),(4,3),(3,5),(5,4)} which is the ordered pairs with the first and second elements swapped.
________________________________ From: Edward <discussions@mathforum.org> To: discretemath@mathforum.org Sent: Sunday, 14 October 2012, 18:32 Subject: Defining "f o g" with Functions So I've been learning about functions. Ideas such as onetoone, onto, bijections....and so on. Things have been going smoothly and I've been going on my own through the text trying to do all of the exercises to prove to myself I understand the concepts on my own but I came into a problem where I don't know where to begin. I'm not really familiar with "g o f" and "f o g". Can somebody show me how to figure this out:
Let S = {1,2,3,4,5} and let T = {3,4,5,6,7}. Define function f : S > T and g : S > S as follows: f = {(2,6), (1,7), (3,4), (5,3), (4,5) and g = {(2,3), (1,2), (4,3), (5,5), (3,1)}.
(a.) Find "f o g" or explain why "f o g" is not defined, Repeat for "g o f", "f o f" and "g o g". (b.) Which (if any) of f, g are onetoone? Which (if any) are onto? Explain (c.) Find f^1 if it exists. If not explain why. (d.) Find g^1 if it exists. If not explain why.
*** ^ represents exponent ***
Anybody who understands these concepts better than I do?

