math1004 help (1 Viewer)

[Libbster]

i know there is another math help thread, but i couldn't be bothered to resurrect it.

anyway, in math1004 we have an assignent and for once, i'm not leaving it til the last minute, but i'm having trouble with some questions. i would just like some pointers on maybe where to start or what to look at. question 4 is the one i can't seem to do, i'm just going around in circles

 

[acmilan]

You'll need to basically translate the info they give you into definitions

An function from X->Y is onto if for every y in Y, there is (at least) an x in X such that y = f(x)

So use the above on the fact that g o f is onto. Also remember that g o f = g(f(x)), so the input of g basically comes from the output of f

 

[Libbster]

i still don't get it

 

[KeypadSDM]

Part ii is easy:
A = {0,1}
B = {0,1}
C = {0,1}
f(x) = x
g(x) = 0

That's the simplest non-trivial answer I can think of.

 

[KeypadSDM]

Initial assumption gof is onto C

Second assumption: Then assume g is not onto C

Then there exists an element in C, y say, such that
For all x in B, g(x) != y

But from our initial assumption there exists some element, z say, in A such that:
g(f(z)) = y
Moreover, f(z) is an element of B, w = f(z) say, with the property:
g(w) = y
Hence our second assumption must be incorrect.

Thus g is onto - proof by contradiction.

 

[Libbster]

thank you so much, i could sort of prove that if g and f were onto, g dot f was onto but not the other way around i think i was/am having a retard day

 

[Libbster]

just a few more questions, promise

with 3i the way i did it was 10!/(2!2!2!2!2!)= 113400. (However this method could be wrong, most likely). So w/ part ii, do i just use the same method 9!/(2!2!2!2!1!), . i'm confuzzled. (i edited because i should actually read the question )

and w/ 1iii, do u think they want the no. of ways to select the president, secretary and treasurer as well as the committee, or just the no. of ways to select the committee as a whole (which would be 14C6, i think)?

thanks