MHB Proving $A=B$ from Sets $A,B,C$

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Sets
Click For Summary
Given sets A, B, and C, the conditions A ∪ C = B ∪ C and A ∩ C = B ∩ C are established. From the first condition, it can be inferred that the elements of A and B must be equal when elements of C are removed. The second condition indicates that any common elements between A and C are also present in B and C. Therefore, by eliminating the influence of set C, it can be concluded that A must equal B. This demonstrates that under the provided conditions, A = B is proven.
Albert1
Messages
1,221
Reaction score
0
Three sets $A,B,C$ given:

(1)$A\bigcup C=B\bigcup C$

and

(2)$A\bigcap C=B \bigcap C$

Prove: $A=B$
 
Mathematics news on Phys.org
Albert said:
Three sets $A,B,C$ given:

(1)$A\bigcup C=B\bigcup C$ and (2)$A\bigcap C=B \bigcap C$
Prove: $A=B$

Let's prove that $A\subset B$ $$x\in A\Rightarrow x\in A\cup C=B\cup C\Rightarrow \left \{ \begin{matrix} x\in B\\\vee\\x\in C\end{matrix}\right.$$ $$\Rightarrow \left \{ \begin{matrix} x\in B\\\vee\\x\in A\cap C=B\cap C\end{matrix}\right.\Rightarrow \left \{ \begin{matrix} x\in B\\\vee\\x\in B\end{matrix}\right.\Rightarrow x\in B.$$ In the same way, we could prove that $B\subset A.$
 
Fernando Revilla said:
Let's prove that $A\subset B$ $$x\in A\Rightarrow x\in A\cup C=B\cup C\Rightarrow \left \{ \begin{matrix} x\in B\\\vee\\x\in C\end{matrix}\right.$$ $$\Rightarrow \left \{ \begin{matrix} x\in B\\\vee\\x\in A\cap C=B\cap C\end{matrix}\right.\Rightarrow \left \{ \begin{matrix} x\in B\\\vee\\x\in B\end{matrix}\right.\Rightarrow x\in B.$$ In the same way, we could prove that $B\subset A.$
Thanks! very good !
sol-1
$A=A\cup(A\cap C)
=A\cup(B\cap C)
=(A\cup B)\cap (A\cup C)$
$=(B\cup A)\cap (B\cup C)
=B\cup (A\cap C)$
$=B\cup(B\cap C)=B$
sol-2
$A=A\cap(A\cup C)
=A\cap(B\cup C)
=(A\cap B)\cup (A\cap C)$
$=(B\cap A)\cup (B\cap C)
=B\cap (A\cup C)$
$=B\cap(B\cup C)=B$
 
Last edited:
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K