1. ## Urgent

hey i know that this forumn sometimes frown upon people needing urgent help, but like I really need it I dont know how that helps my case but in all sincerity i need this help really quick so please help me out.

Imageshack - helpwz

or

I know you guys say to post up what you attempted and etc. before going around asking for help but i really have alot of problems to solve and my exam is going to quite sooon so if you could help me out these are the ones i really couldn't understand for the life of me

Alright I got the first one using De Morgan Law's please if you can verify that it is correct

B - C = BnC^complement <= A^complement

AnB = (AnB)^complement => C^complment
Bn(A^complement u B^complement) => BnC^complement
A^complement <= BnCcomplement

alright i got the solution to the next one and the one after basically
$
\begin{array}{l}
{\rm{Let y 2 f(C \backslash D) }}{\rm{. Then y = f(x) for some x 2 C \backslash D }}{\rm{. This implies that x 2 C and}} \\
{\rm{x 2 D }}{\rm{. We therefore conclude that y = f(x) is an element of both f(C) and f(D) , and}} \\
{\rm{hence in f(C) \backslash f(D)}}{\rm{. This proves that f(C \backslash D) f(C) \backslash f(D) }}{\rm{.}} \\
\end{array}
$

DONT DO J AND K I SOLVED THEM

2. I am somewhat confused by your post.

B - C = BnC^complement <= A^complement

AnB = (AnB)^complement => C^complment
Is this a solution to (j)? First, $A\cap B =(A^c\cup B^c)^c$, not $(A\cap B)^c$. I also don't see how the second line follows from the first or vice versa. And the last line:
A^complement <= BnCcomplement
is it what one needs to prove?

Then, if the typesetting was not messed up, you are talking about the set difference $C\setminus D$, which I don't see in any problem (in fact, set difference in the picture is denoted differently).

3. Problem: Show that $C=\mathbb{N}\times\{a,b\}$ is countable by constructing a bijection between $C$ and [tex]

Solution: Let $f:C\mapsto \mathbb{N}$ by $f\left(n,m\right)=\begin{cases} 2n & \mbox{if}\quad m=a \\ 2n+1 & \mbox{if}\quad m=b \end{cases}$. To see that this is injective assume that $f\left(n,m\right)=f\left(n,m\right)=f\left(\tilde{ n},\tilde{m}\right)$. Since, to be equal they have to both be even or odd it follows that $m=\tilde{m}$. Clearly then we have that either $2n=2\tilde{n}\implies n=\tilde{n}$ or $2n+1=2\tilde{n}+1\implies n=\tilde{n}$; either way we may conclude that $\left(n,m\right)=\left(\tilde{n},\tilde{m}\right)$. To see that it is surjective let $k\in\mathbb{N}$. If $k$ is odd we see that $\left(\tfrac{k-1}{2},b\right)\mapsto k$ and if $k$ is even we see that $\left(\tfrac{k}{2},a\right)\mapsto k$. The conclusion follows.

Note: It was implictly assumed that $a,b$ were distinct. If not, then the projection $\pi:C\mapsto \mathbb{N}$ given by $\pi\left(n,m\right)=n$ is a clear bijection.

Is that solution correct? You tell me.

I can help you with all of these if you just tell me which in particular are bothering you.

4. Originally Posted by treetheta
hey i know that this forumn sometimes frown upon people needing urgent help, but like I really need it I dont know how that helps my case but in all sincerity i need this help really quick so please help me out.

Imageshack - helpwz

or

I know you guys say to post up what you attempted and etc. before going around asking for help but i really have alot of problems to solve and my exam is going to quite sooon so if you could help me out these are the ones i really couldn't understand for the life of me

Alright I got the first one using De Morgan Law's please if you can verify that it is correct

B - C = BnC^complement <= A^complement

AnB = (AnB)^complement => C^complment
Bn(A^complement u B^complement) => BnC^complement
A^complement <= BnCcomplement

alright i got the solution to the next one and the one after basically
$
\begin{array}{l}
{\rm{Let y 2 f(C \backslash D) }}{\rm{. Then y = f(x) for some x 2 C \backslash D }}{\rm{. This implies that x 2 C and}} \\
{\rm{x 2 D }}{\rm{. We therefore conclude that y = f(x) is an element of both f(C) and f(D) , and}} \\
{\rm{hence in f(C) \backslash f(D)}}{\rm{. This proves that f(C \backslash D) f(C) \backslash f(D) }}{\rm{.}} \\
\end{array}
$

DONT DO J AND K I SOLVED THEM
l) Does it satisfy all of the field axioms?

n) Suppose that $g$ was not injective, then there exists some $x,y\in A$ such that $x\ne y$ but $g(x)=g(y)\implies f\left(g(x)\right)=f\left(g(y)\right)$ which contradicts $fg$'s injectivity. Let $y\in C$, then since $fg$ is surjective there exists some $x\in A$ such that $f\left(g(x)\right)=y$, but since $g:A\mapsto B$ we see then that $g(x)\in B$ and $f\left(g(x)\right)=y$ which shows $f$'s surjectivity.

o) If $n>2$ then clearly $\left(-1\right)^2\equiv 1^2\text{ mod }n$ but $-1\equiv n-1\text{ mod }n$

p) ${n\choose a}\cdot {{b+c}\choose c}=\frac{n!}{a! (n-a)!}\cdot\frac{(b+c)!}{c! b!}$ but $n=a+b+c\implies n-a=b+c$ so then ${n\choose a}\cdot {{b+c}\choose c}=\frac{n!\cdot(b+c)!}{a!b!c! (b+c)!}=\frac{n!}{a!b!c!}$

5. For o) another instructive example is furnished by perfect square $n$ and considering for example $0,\sqrt{n}$.

6. For (i) find your multiplicative identity. Then can you think of any element which doesn't have an inverse?...There are uncountably many of them, so they shouldn't be too hard to find .

EDIT: I would advise you editing the picture in your previous post to get rid of some of your tabs etc. We all now know what university you go to. Of course, you may not mind this, but you might...I would.

7. Originally Posted by Swlabr
For (i) find your multiplicative identity. Then can you think of any element which doesn't have an inverse?...There are uncountably many of them, so they shouldn't be too hard to find .

EDIT: I would advise you editing the picture in your previous post to get rid of some of your tabs etc. We all now know what university you go to. Of course, you may not mind this, but you might...I would.
(i) huh? That's pretty impressive that you can deduce the question from half a summation and some extraneous terms

Also, not everyone minds people knowing what school they go to. Drexel pride!