Subsets of two-dimensional space

  • Thread starter Dafe
  • Start date
  • #1
145
0

Homework Statement



a) construct a subset of two-dimensional space closed under vector addition and even subtraction, but not under scalar multiplication.

b) construct a subset of two-dimensional space (other than two opposite quadrants) closed under scalar multiplication but not under vector addition.


The Attempt at a Solution



a) A line going through a point on the y-axis. Not a line through the origin though. I figure since the line does not have a null-vector in it, it is not closed under scalar multiplication.

b) I only know of two opposite quadrants :/

Anyone? Thanks!
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
964

Homework Statement



a) construct a subset of two-dimensional space closed under vector addition and even subtraction, but not under scalar multiplication.

b) construct a subset of two-dimensional space (other than two opposite quadrants) closed under scalar multiplication but not under vector addition.


The Attempt at a Solution



a) A line going through a point on the y-axis. Not a line through the origin though. I figure since the line does not have a null-vector in it, it is not closed under scalar multiplication.
Un fortunately, if it does not go through the origin, that is, does not contain 0, it is not closed under "vector addition and even subtraction" because v-v= 0.

b) I only know of two opposite quadrants :/

Anyone? Thanks!
 
  • #3
145
0
Maybe I've missunderstood what even subtraction is. Does that mean that I subtract a vector by the same vector? I thought it meant that it was closed under addition and also under subtraction..

If even subtraction really is v-v, then the answer to a) is a line from the origin into the first quadrant, right?

Thank you
 
  • #4
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,623
630
It means if u,v are in your subset, u+v AND u-v are in the subset. One particular example is for all v in the subset, v-v = 0 is in the subset also
 
  • #5
145
0
Damn, then I'm out of quesses.. Please help? :)
 
  • #6
145
0
I just went back to this problem and still can't find an answer.
Could someone please help?
 
  • #7
179
0
the answer to a) is a line from the origin into the first quadrant, right?

Then for any v on this line, 0 - v must also be on it, but it's not because your line is only in the first quadrant.

a) How about vectors with integer norm?

b) Two non-parallel lines through the origin?
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,847
964
You are looking for sets of numbers that are closed under addition and subtraction but not under multiplication by real numbers.

Try {(n,n)} where n is an integer.
 
  • #9
179
0
a) How about vectors with integer norm?

Now I know that this is wrong; e.g. (1, 0) - (0, 1) does not have integer norm. See what HallsofIvy said.
 
  • #10
145
0
HallsofIvy: does {(n,n)} mean vectors in R^2 like (1,1), (2,2), (-4,-4), etc.?
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
41,847
964
I wrote what it meant: "Try {(n,n)} where n is an integer."
Surely you know what an integer is?
 
  • #12
145
0
Yes, I do know what they are.
To me {(n,n)} is a line through the origin, 45 degrees on the x-axis, which is a subspace closed under addition and scalar multiplication.

You do not have to reply to these stupid questions of mine HallsofIvy, so there is no need to be an ***.

Thank you.
 
  • #13
HallsofIvy
Science Advisor
Homework Helper
41,847
964
Do you understand that insults will get you kicked of this board permanently?
You were the one who asked if "Try {(n,n)} where n is an integer" meant "(1,1), (2,2), (-4,-4), etc." so my question was perfectly reasonable.


In any case, do you now see what the answer to your problem is?
 
  • #14
179
0
{(n, n) : n is an integer} is not a line. If you connected the points for some reason, then it would be a line.
 
  • #15
145
0
So it is closed by addition and subtraction but not by multiplication by real numbers because then they might not be integers any more?

I apologize for being an ***. It sometimes happens without me knowing it :p

Thank you for your patience.
 
  • #16
HallsofIvy
Science Advisor
Homework Helper
41,847
964
Yes, that is correct. (n, n)+ (m, m)= (n+ m, n+ m) and (n, n)- (m, m)= (n-m, n-m), both of the form (k, k) for k integer. But (1/2)(1, 1)= (1/2, 1/2) which is NOT.
 
  • #17
145
0
I am obviously quite bad at this and am not able to figure out the second problem.
(subset closed under multiplication but not under addition)

Am I allowed to say that it is closed under multiplication by integers or something like that?

Thank you.
 
  • #18
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,623
630
For part (b) you know you need a non-zero point (call it A)... then as it's closed under scalar multiplication, you need the line through the origin and A. This is a subspace, so we can't use this. So pick another point B that's not on the line. Again, you need the line passing through the origin and B. Do you need any more points?
 
  • #19
145
0
Two points in the 1st quadrant and two points in the 3. quadrant. I draw lines through the origin between the points in opposite quadrants. Then I take linear combinations of the vectors on the two lines, so I get a sector in the first quadrant, and a mirror image of the same sector in the 3. quadrant. Then it is closed under multiplication but not under addition.
Is this right?

I guess this would work for all "mirrored sectors".. Am I totally wrong?

Thank you!
 

Related Threads on Subsets of two-dimensional space

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
2K
Replies
18
Views
6K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
892
  • Last Post
Replies
1
Views
866
Top