Operation on Sets Homework: B-A & C-A

  • Thread starter Thread starter ver_mathstats
  • Start date Start date
  • Tags Tags
    Sets
Click For Summary

Homework Help Overview

The discussion revolves around set operations involving the sets A = {1, 2, 3}, B = ℤ+, and C = [1, infinity). Participants are tasked with determining the results of the operations B - A and C - A.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the meaning of set subtraction, particularly what elements remain in B and C after removing those in A. There is uncertainty about whether B - A results in an empty set or a set of natural numbers greater than 3. Questions arise about how to express C - A correctly, with some suggesting it includes all real numbers greater than 3.

Discussion Status

Participants are actively engaging with the definitions and implications of set operations. Some guidance has been provided regarding the correct interpretation of the sets and the nature of the elements involved. There is ongoing clarification about the boundaries of the sets and the specific elements being excluded.

Contextual Notes

Participants express confusion regarding the definitions of the sets and the implications of the operations, particularly with respect to the inclusion of numbers less than 1 in C. There is a recognition of the need to accurately represent the sets in mathematical notation.

ver_mathstats
Messages
258
Reaction score
21

Homework Statement


Let A= {1, 2, 3}, B= ℤ+, C= [1, infinity)

That is C= {x∈ℝ:x≥1}

What is B - A and C - A?

Homework Equations

The Attempt at a Solution


I am unsure of how to go about answering this.

I know that B - A means what elements are in B that aren't in A. Would that make the answer B - A = {Z+>3}? Or is B - A an empty set since 1, 2, 3 are positive integers?

As for C - A, I am unsure of how to approach this one, would it be C - A = {x > 3}?

Would both just be greater than 3 since A = {1, 2, 3}?
 
Last edited:
Physics news on Phys.org
ver_mathstats said:

Homework Statement


Let A= {1, 2, 3}, B= ℤ+, C= [1, infinity)

That is C= {x∈ℝ:x≥1}

What is B - A and C - A?

Homework Equations

The Attempt at a Solution


I am unsure of how to go about answering this.

I know that B - A means what elements are in B that aren't in A. Would that make the answer B - A = {Z+>3}?
Correct, although I would either write it ##\mathbb{N}_{>3}## or ##\{\,4,5,6,\ldots\,\}##
Or is B - A an empty set since 1, 2, 3 are positive integers?
No. B-A are all natural numbers (B) which are not 1,2 or 3 (A). How do you get an empty set?
As for C - A, I am unsure of how to approach this one, would it be C - A = {x > 3}?
No, since e.g. 5/2 is still in C and therewith in C-A. Draw C, take scissors and cut out 1,2,3.
Would both just be greater than 3 since A = {1, 2, 3}?
No, you only take away the three points, not those in between.
 
fresh_42 said:
Correct, although I would either write it ##\mathbb{N}_{>3}## or ##\{\,4,5,6,\ldots\,\}##
No. B-A are all natural numbers (B) which are not 1,2 or 3 (A). How do you get an empty set?
No, since e.g. 5/2 is still in C and therewith in C-A. Draw C, take scissors and cut out 1,2,3.

No, you only take away the three points, not those in between.

I got confused about the empty set but I know now it is natural numbers greater than 3. As for C - A I understand what you are saying now, however I am struggling to write it out. C - A = {x ε ℝ+ : x ≠ 1, 2, 3}, would this be okay?

Thank you.
 
ver_mathstats said:
I got confused about the empty set but I know now it is natural numbers greater than 3. As for C - A I understand what you are saying now, however I am struggling to write it out. C - A = {x ε ℝ+ : x ≠ 1, 2, 3}, would this be okay?

Thank you.

Not quite, since C doesn't include any numbers < 1 and your set does. Restrict those and you're fine.
I've got this nagging feeling that there's something you're still missing about the definition of these sets and of the subtraction operation.
 
RPinPA said:
Not quite, since C doesn't include any numbers < 1 and your set does. Restrict those and you're fine.
I've got this nagging feeling that there's something you're still missing about the definition of these sets and of the subtraction operation.

Oh I thought that if I put the ℝ then + that would make it clear it was positive only but I will try to write it out differently. I am new to this and I think just the fact C is [1, infinity) through me off a little. But thank you for the help.
 
ver_mathstats said:
Oh I thought that if I put the ℝ then + that would make it clear it was positive only but I will try to write it out differently. I am new to this and I think just the fact C is [1, infinity) through me off a little. But thank you for the help.

It does mean that it is positive. For instance 0.5 is a positive number which is included in ##R^+##. It's a real number, and it's positive. But it's not in C, which does not have any numbers less than 1.
 
  • Like
Likes   Reactions: PeroK
RPinPA said:
It does mean that it is positive. For instance 0.5 is a positive number which is included in ##R^+##. It's a real number, and it's positive. But it's not in C, which does not have any numbers less than 1.

Oh right, I am very sorry for missing that, I get what's wrong now. Thank you so much.
 
ver_mathstats said:
I am struggling to write it out
You cannot simplify much here, so something like ##\neq 2,3## cannot be avoided. What you can do, is to modify the interval here:
What is ##C - \{\,1\,\}\,?##.
Then you can write ##\left( C- \{\,1\,\} \right) - \{\,2,3\,\}## or ##\{\,x \in \mathbb{R}\,|\,\ldots \,\}##.
 

Similar threads

Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K