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

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In summary: 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 \,\}##Okay I think I got it now, thank you for your help!In summary, B - A = {x ε ℤ+ : x > 3} and C - A = {x ε ℝ : x >
  • #1
ver_mathstats
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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}?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
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  • #7
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.
 
  • #8
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 \,\}##.
 

1. What is the purpose of "Operation on Sets Homework: B-A & C-A"?

The purpose of this homework is to practice performing operations on sets, specifically finding the difference between two sets (B-A and C-A).

2. How do you find the difference between two sets?

To find the difference between two sets, you need to subtract the elements of the second set from the elements of the first set. This will give you a new set with only the elements that are present in the first set but not in the second set.

3. What are some common mistakes when performing operations on sets?

Some common mistakes when performing operations on sets include forgetting to include all elements in the set, mixing up the order of subtraction, and not properly representing the result as a set.

4. Can you give an example of "Operation on Sets Homework: B-A & C-A"?

For example, if set B = {1, 2, 3, 4} and set A = {2, 4}, then B-A = {1, 3} and C-A = {3, 5}. This means that the elements 2 and 4 are removed from set B, and the elements 2 and 4 are removed from set C, resulting in two new sets with different elements.

5. How can "Operation on Sets Homework: B-A & C-A" be applied in real life?

Operations on sets can be applied in many real-life scenarios, such as organizing data, analyzing survey results, and solving problems in mathematics and computer science. For example, finding the difference between two sets can help identify unique elements or patterns in a data set.

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