Numeral set

Leandrowski

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Sort on true or false this sentence:

(A-B)∪(A∩B) = A

Ok, we know that: (A-B) = (x|x ∈ A ∧ x ∉ B)
we also know that: (A∩B) = (x|x ∈ A ∧ x ∈ B)
resulting in the sentence = (x|x ∈ A ∧ x ∉ B) ∪ (x|x ∈ A ∧ x ∈ B)

True or false? why? i need help.

OBS: i'm brazillian, sorry my english.
 

Dr.Peterson

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I'm not sure what you mean by "sort". You could mean to determine conditions under which the sentence is true, or to determine whether it is true for all A and B.

What you've done so far is valid, but not (yet) helpful. I might next try sketching a Venn diagram of the LHS, to guide my thinking.
 

Leandrowski

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I'm not sure what you mean by "sort". You could mean to determine conditions under which the sentence is true, or to determine whether it is true for all A and B.

What you've done so far is valid, but not (yet) helpful. I might next try sketching a Venn diagram of the LHS, to guide my thinking.
I meant that I want to know if the expression is true (results in A) or false (does not result in A), sorry. but regarding its method, I can understand through the diagram. Results in A (True), but my problem is in the expression part (written sentences) How is that (x | x ∈ A ∧ x ∉ B) ∪ (x | x ∈ A ∧ x ∈ B)
can prove that results in A?
 

Dr.Peterson

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Okay, you have determined by Venn diagram that the statement is true, but now you want to prove it symbolically.

A useful method is to first prove that any element of the LHS is in the RHS, and then prove that any element of the RHS is in the LHS.

I'll take the former: Suppose that x is in (A-B)∪(A∩B). Then x is either in A-B or in A∩B. In the first case, it is in A; in the second case it is in A. Therefore it is in A.

Now you do the other direction.

If you want a fully symbolic proof, you can start with your {x | x ∈ A ∧ x ∉ B} ∪ {x | x ∈ A ∧ x ∈ B} and continue: ... = {x | (x ∈ A ∧ x ∉ B) v (x ∈ A ∧ x ∈ B)} ...
 

pka

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Jomo

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What you did so far is fine. You just need to state your conclusion.
A is made of two disjoint parts. Basically A is partitioned in one part that includes B and the other part that does not include B.
Your next line should that (x|x ∈ A ∧ (x ∉ B ∪ x ∈ B)) = (x|x ∈ A ∧ U)=A
 

Leandrowski

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Okay, you have determined by Venn diagram that the statement is true, but now you want to prove it symbolically.

A useful method is to first prove that any element of the LHS is in the RHS, and then prove that any element of the RHS is in the LHS.

I'll take the former: Suppose that x is in (A-B)∪(A∩B). Then x is either in A-B or in A∩B. In the first case, it is in A; in the second case it is in A. Therefore it is in A.

Now you do the other direction.

If you want a fully symbolic proof, you can start with your {x | x ∈ A ∧ x ∉ B} ∪ {x | x ∈ A ∧ x ∈ B} and continue: ... = {x | (x ∈ A ∧ x ∉ B) v (x ∈ A ∧ x ∈ B)} ...
Ok, but...

(I)
(A-B)∪(A∩B)
[x ∈ A or x ∈ A]

x ∈ A --> 1° Case (A-B) (in A)
x ∈ A --> 2° Case (A∩B) (in A)

you determined that x belongs to A, so x is in A (A-B) and A (A∩B) so it shows that (A-B)∪(A∩B) ⊂ A ?


(II)
[[x ∈ A]]
(A-B)∪(A∩B)
[x ∈ B or x ∉ B]

x ∈ B --> 1° case (A∩B) (in B)
x ∉ B --> 2° case (A-B) (in -B)

you determined that x belongs to B or x does not belong to B (-B) in (A-B) in(-B) and in (A∩B) in(B) it shows that A ⊂ (A-B)∪(A∩B)? wht, i don't understand ;;(((

View attachment 16028
Look at the images. Unite the two. Is all of A included?
I understand that, but my problem is not in the diagrams, but in the symbology, thanks for the diagram illustration!

What you did so far is fine. You just need to state your conclusion.
A is made of two disjoint parts. Basically A is partitioned in one part that includes B and the other part that does not include B.
Your next line should that (x|x ∈ A ∧ (x ∉ B ∪ x ∈ B)) = (x|x ∈ A ∧ U)=A
that! symbologies !, but I didn't understand the part (x ∉ B ∪ x ∈ B) because that means they cancel each other out? for example two prepositions B and -B in a conjunction (OR) do they cancel each other out like this? why is only A? help me please.

I didn't understand what U means
 

Dr.Peterson

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(I)
(A-B)∪(A∩B)
[x ∈ A or x ∈ A]

x ∈ A --> 1° Case (A-B) (in A)
x ∈ A --> 2° Case (A∩B) (in A)

you determined that x belongs to A, so x is in A (A-B) and A (A∩B) so it shows that (A-B)∪(A∩B) ⊂ A ?
No, I didn't say "x belongs to A, so x is in A (A-B) and A (A∩B)"; I said that if x is in either (A-B) or (A∩B), then x is in A.

But, yes, my conclusion is that (A-B)∪(A∩B) ⊂ A.

(II)
[[x ∈ A]]
(A-B)∪(A∩B)
[x ∈ B or x ∉ B]

x ∈ B --> 1° case (A∩B) (in B)
x ∉ B --> 2° case (A-B) (in -B)

you determined that x belongs to B or x does not belong to B (-B) in (A-B) in(-B) and in (A∩B) in(B) it shows that A ⊂ (A-B)∪(A∩B)? wht, i don't understand ;;(((
Maybe I'm just not understanding your notation here (or above). What do you mean when you write "x is in A (A-B)" and, here, "x does not belong to B (-B) in (A-B) in(-B)"?

This is the case I didn't do for you, so I'm not sure what "you determined" refers to.

If x is in A, then either it is not in B (so it is in A-B), or it is in B (so it is in A∩B). Therefore it is in either A-B or A∩B. Therefore A ⊂ (A-B)∪(A∩B).
 

pka

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Prove \(\displaystyle (A\setminus B)\cup(A\cap B)\equiv A\)
Proof
\(\displaystyle \begin{gathered} \left( {A\backslash B} \right) \cup \left( {A \cap B} \right) \equiv \left( {A \cap {B^c}} \right) \cup \left( {A \cap B} \right) \\
\equiv A \cap \left( {{B^c} \cup B} \right) \\ \equiv A \cap U \\ \equiv A \\ \end{gathered} \)
 

Cubist

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I didn't understand what U means
U is the symbol for "universal set".

I didn't understand the part (x ∉ B ∪ x ∈ B) because that means they cancel each other out?
They don't cancel because this is a union - the result is "all elements in B" union with "all elements that are not in B". In other words you end up with every possible element, which is called the universal set U
 

Jomo

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Ok, but...

(I)
(A-B)∪(A∩B)
[x ∈ A or x ∈ A]

x ∈ A --> 1° Case (A-B) (in A)
x ∈ A --> 2° Case (A∩B) (in A)

you determined that x belongs to A, so x is in A (A-B) and A (A∩B) so it shows that (A-B)∪(A∩B) ⊂ A ?


(II)
[[x ∈ A]]
(A-B)∪(A∩B)
[x ∈ B or x ∉ B]

x ∈ B --> 1° case (A∩B) (in B)
x ∉ B --> 2° case (A-B) (in -B)

you determined that x belongs to B or x does not belong to B (-B) in (A-B) in(-B) and in (A∩B) in(B) it shows that A ⊂ (A-B)∪(A∩B)? wht, i don't understand ;;(((



I understand that, but my problem is not in the diagrams, but in the symbology, thanks for the diagram illustration!



that! symbologies !, but I didn't understand the part (x ∉ B ∪ x ∈ B) because that means they cancel each other out? for example two prepositions B and -B in a conjunction (OR) do they cancel each other out like this? why is only A? help me please.

I didn't understand what U means
No, B AND not B is the empty set (cancels out). However B OR not B is everything which is called the universal set U. A intersect U = A
 

pka

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Sort on true or false this sentence:
(A-B)∪(A∩B) = A
Let's all start over. Let's answer the question as asked.
There is nothing posted about a formal proof, just true or false.
One set is \(\displaystyle A\) the other \(\displaystyle (A\setminus B)\cup (A\cap B)\)
Are those sets the same? If \(\displaystyle x\in(A\setminus B)\cup (A\cap B)\)x\in A[/tex] in either case!
If \(\displaystyle x\in A\) then either \(\displaystyle x\in B\) or \(\displaystyle x\notin B\), so \(\displaystyle x\in(A\cap B)\text{ or }x\in(A\setminus B)\).
Thus the the statement is true.
 

Leandrowski

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No, B AND not B is the empty set (cancels out). However B OR not B is everything which is called the universal set U. A intersect U = A
thanks, with that I solved what I wanted in symbolic form, so:
(x ∈ B ∪ x ∉ B) results = U (Universe) -->
Screenshot_1.png
resulting in the sentence = (x|x ∈ A ∧ x ∉ B) ∪ (x|x ∈ A ∧ x ∈ B) = A
= (x|x ∈ A ^ (x ∈ B ∪ x ∉ B)) = A
= (x|x ∈ A ^ U) = A
Thanks for everbody.
 

Jomo

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5,407
thanks, with that I solved what I wanted in symbolic form, so:
(x ∈ B ∪ x ∉ B) results = U (Universe) -->
View attachment 16074
resulting in the sentence = (x|x ∈ A ∧ x ∉ B) ∪ (x|x ∈ A ∧ x ∈ B) = A
= (x|x ∈ A ^ (x ∈ B ∪ x ∉ B)) = A
= (x|x ∈ A ^ U) = A
Thanks for everbody.
I wouldn't write '=A' until the last line since only then do you truly know that the lhs equals A. If you insist on writing =A then at least put a question mark above the equal sign until you are sure about the equality.
 

Leandrowski

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I wouldn't write '=A' until the last line since only then do you truly know that the lhs equals A. If you insist on writing =A then at least put a question mark above the equal sign until you are sure about the equality.
I got it, thanks!
 
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