Discrete math problem checks and questions.

[Faxon]

I have no idea what 1[sub:3cdctw48]N[/sub:3cdctw48](n) = n, means. If you could clear it up so that I can find out if it's one-to-one and if it maps N onto N that would be great.

This next one I think I have right, but I'm not sure if I know why:

If f maps N into N, and f(n) = n+1 show that f does not map N onto N. I think that f does not map N onto N because it doesn't account for 0. And in order to get 0, n=-1 and since -1 is not a N but a Z is the reason why.

Also if g(n) = max{0,n-1} it maps N onto N because it accounts for all N but I'm not sure how to formally represent my answer. I can give examples like: g(0)= 0 and g(123413)=123412.

For the next problems prove that functions are their own inverses:
For F(x) = 1/x I can simply switch F(x) out for y and then change y = 1/x for x = 1/y and solve for y. That would be the easy non-book way. So here is the only example the book gives me:

g(m,n) = (-n,-m)
To see that g maps onto Z onto Z X Z, consider (p,q) in Z X Z. We need to find (m,n) in Z X Z so that g(m,n) = (p,q). Thus we need (-n,-m) = (p,q), and this tells us that n should be -p and m should be -q. In other words, given (p,q), and in Z X Z we see that (-q,-p) is an element in Z X Z such that g(-q,-p) = (p,q). Thus g maps Z X Z onto Z X Z. Hence g[sup:3cdctw48]-1[/sup:3cdctw48](p,q) = (-q,-p) for all (p,q) in Z X Z.

I'm not seeing simply switch F(x) out for y and then change y = 1/x for x = 1/y and solve for y, in that garbage above.
So, I'm assuming that's not the "official" way to do that problem. So, I'm assuming I'm doing it wrong.

Also, I have no idea on how to prove, The function empty set: P(S) ---> P(S) defined by, (empty set)(A) = A[sup:3cdctw48]c[/sup:3cdctw48] [sub:3cdctw48]*[/sub:3cdctw48].

If you could explain it to me that would be great. Note, I don't have the symbol for empty set, (the 0 with a slash through it), so I just typed, empty set, in it's place.

 

[pka]

Also, I have no idea on how to prove, The function empty set: P(S) ---> P(S) defined by, (empty set)(A) = A[sup:12slwf2j]c[/sup:12slwf2j] [sub:12slwf2j]*[/sub:12slwf2j].

If you could explain it to me that would be great. Note, I don't have the symbol for empty set, (the 0 with a slash through it), so I just typed, empty set, in it's place.

I think that the symbol is \(\displaystyle \phi (A) = A^c\) and that is the Greek letter phi, \(\displaystyle \phi\). That symbol is often used as a mapping symbol.

The mapping \(\displaystyle \phi :\,P(S) \mapsto P(S),\;\phi (A) = A^c\) maps a subset of S to its complement. Because each subset of S has a unique complement, it is easy to see that mapping is one-to-one and onto.

 

[daon]

Normally to show a function g is an inverse of a function f you must show that g is both a left and right inverse. i.e. \(\displaystyle (g \circ f)=(f \circ g) = 1\), where \(\displaystyle 1\) is the identity function \(\displaystyle 1(x)=x\) for all x. To show F is its own inverse just see if \(\displaystyle (F \circ F)(x)=x\).

Also, if A is some set, \(\displaystyle 1_A\) usually is the notation for the identity function on the set A (in my experience anyway). So in your case \(\displaystyle 1_\mathbb{N}\) is the idenity function on the naturals. Identity functions are always bijective.