Urgent

**treetheta** · December 17th 2009, 01:48 PM

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

**emakarov** · December 17th 2009, 03:48 PM

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).

**Drexel28** · December 18th 2009, 01:24 AM

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.

**Drexel28** · December 19th 2009, 12:44 AM

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 ${n\choose a}\cdot {{b+c}\choose c}=\frac{n!\cdot(b+c)!}{a!b!c! (b+c)!}=\frac{n!}{a!b!c!}$

**Drexel28** · December 19th 2009, 12:55 AM

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

**Swlabr** · December 19th 2009, 02:37 AM

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.

**Drexel28** · December 19th 2009, 02:42 AM

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!

Thread: Urgent

LinkBack

Thread Tools

Display

Urgent

Similar Math Help Forum Discussions

Urgent!!!help with attempted questions - urgent

urgent help please!!

I need urgent urgent hel with this parabola question

urgent help

Help me again, urgent

Search Tags