# S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
 It is currently Wed, 19 Jun 2013 04:12:58 UTC

 All times are UTC [ DST ]

 Page 1 of 2 [ 19 posts ] Go to page 1, 2  Next
 Print view Previous topic | Next topic
Author Message
 Post subject: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 07:06:30 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182

Im doing numbers 1,2,3. Im confused on the notation.

For #1.) and #2.) To prove its injective or one-to-one, what do i need to show?

for example in 1.) to show its injective i wanted to show f(a,b)=f(c,d) which then a=c. But this is wrong.
Should it be , suppose f(a,y)=f(c,y)?

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 09:16:25 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12169
Location: Austin, TX
DgrayMan wrote:

Im doing numbers 1,2,3. Im confused on the notation.

For #1.) and #2.) To prove its injective or one-to-one, what do i need to show?

for example in 1.) to show its injective i wanted to show f(a,b)=f(c,d) which then a=c. But this is wrong.
Should it be , suppose f(a,y)=f(c,y)?

The first function is not 1-1 unless B only has one element, so you cannot prove it is injective because it's more or less guaranteed NOT to be.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 09:21:01 UTC
 Moderator

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6066
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
The first function is not 1-1 unless B only has one element

... or: if A or B is empty.

_________________

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 15:48:34 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12169
Location: Austin, TX
outermeasure wrote:
The first function is not 1-1 unless B only has one element

... or: if A or B is empty.

Ah yes, this pathology with the universal quantifier.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 16:31:15 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
So for #2). to prove its injective could i say if suppose f(a,b)=f(b,a) then (b,a)=(a,b)?

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 17:39:46 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
And #3 i'm at a loss as show if it's a bijection

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 19:00:06 UTC
 S.O.S. Oldtimer

Joined: Sat, 16 Aug 2008 04:47:19 UTC
Posts: 208
#3 cannot be a bijection. Consider such that , then but ....assuming the cardinality of is greater than 1.

And for #2 to show something is injective, you can usually follow the definition.
If , then which means and , so .

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 19:58:57 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12169
Location: Austin, TX
DgrayMan wrote:
So for #2). to prove its injective could i say if suppose f(a,b)=f(b,a) then (b,a)=(a,b)?

For #2 just find a left-inverse and that shows your function is injective.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 22:46:20 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
Got it for #2.)

For #3) ito show its not injective i made f(a)=f(c), with f(a)=(a,b) & f(c)=(c,b). since (a,b)=(c,b) it shows b is repeated so its not 1-1 . But my teacher said i need to check whether a=c?

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 23 Feb 2012 23:19:46 UTC
 S.O.S. Oldtimer

Joined: Sat, 16 Aug 2008 04:47:19 UTC
Posts: 208
I really think you should just follow the definition of injectivity. Which means do #3 similar to way I told you about number #2 because ultimately you want to show, by definition, that if , then which is what your teacher is telling you.

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Tue, 28 Feb 2012 22:26:26 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
For surjective. z in AxB, then z=(z,b)=f(z). So every f(x) for f(x)=(a,b)?

Maybe im having trouble visualizing what exactly. A->AxB looks like

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Tue, 28 Feb 2012 23:19:49 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12169
Location: Austin, TX
DgrayMan wrote:
For surjective. z in AxB, then z=(z,b)=f(z). So every f(x) for f(x)=(a,b)?

Maybe im having trouble visualizing what exactly. A->AxB looks like

Huh? Surely . In any case, it should be dead obvious that the map from cannot be surjective unless B has only one element.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Wed, 29 Feb 2012 21:37:25 UTC
 S.O.S. Oldtimer

Joined: Sat, 16 Aug 2008 04:47:19 UTC
Posts: 208
DgrayMan wrote:
For surjective. z in AxB, then z=(z,b)=f(z). So every f(x) for f(x)=(a,b)?

Maybe im having trouble visualizing what exactly. A->AxB looks like

I don't know if you are having problems visualizing what looks like since it's explicitly given to you by the mapping where and is always the same element.

I'll give an example to see if it helps.
Let and and assume that , then .

Even with this example you should see why is not surjective with these particular sets and .

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 1 Mar 2012 00:34:39 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
I think im starting to get it. So for example AxB contains for instance (2,8), which is not in the function f, so since it's not the image of anything then f is not surjective

Top

 Post subject: Re: Functions on Arbitrary Sets and GroupsPosted: Thu, 1 Mar 2012 00:41:05 UTC
 S.O.S. Oldtimer

Joined: Sat, 16 Aug 2008 04:47:19 UTC
Posts: 208
Yes. That typically is how to show something is not surjective. Find one element in the range that is not hit by any element in the domain or as you put it...not in the image.

Of course, I put this more mathematically in my first post.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 2 [ 19 posts ] Go to page 1, 2  Next

 All times are UTC [ DST ]

#### Who is online

Users browsing this forum: No registered users

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forum

Search for:
 Jump to:  Select a forum ------------------ High School and College Mathematics    Algebra    Geometry and Trigonometry    Calculus    Matrix Algebra    Differential Equations    Probability and Statistics    Proposed Problems Applications    Physics, Chemistry, Engineering, etc.    Computer Science    Math for Business and Economics Advanced Mathematics    Foundations    Algebra and Number Theory    Analysis and Topology    Applied Mathematics    Other Topics in Advanced Mathematics Other Topics    Administrator Announcements    Comments and Suggestions for S.O.S. Math    Posting Math Formulas with LaTeX    Miscellaneous