mathmath333
  • mathmath333
Set Theory
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \normalsize \text{A ∩ B = A ∩ C need not imply B = C.} \quad \normalsize \text{ Explain through an example.}\hspace{.33em}\\~\\ \end{align}}\)
Kainui
  • Kainui
A = {1} B = {1, 2} C = {1, 3}
ikram002p
  • ikram002p
huh kai was so fast :P

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More answers

Kainui
  • Kainui
Hahaha there could have been a simpler example: A = {} B = {1} C = {2}
mathmath333
  • mathmath333
thnx
Kainui
  • Kainui
I think that last example I gave sorta shows that it's almost like this is a lot like multiplying by zero. a*b=a*c This doesn't mean b=c, since a=0 is possible.
mathmath333
  • mathmath333
Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.
ikram002p
  • ikram002p
in general any case with \[ A\cap B \neq \varnothing \] does not imply A=C
ikram002p
  • ikram002p
or \(C\cap B \neq \varnothing \)
mathmath333
  • mathmath333
i need an example like the previous one
Kainui
  • Kainui
ikram needs more owl bucks sry
ikram002p
  • ikram002p
A={1,2,3} B={1,2,3} C={1,2,3,4}
ikram002p
  • ikram002p
:O @Kainui i dont lol
Kainui
  • Kainui
Hahaha jk :P
mathmath333
  • mathmath333
is that the example for this que Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.
ikram002p
  • ikram002p
you wanna prove this ?
mathmath333
  • mathmath333
yes with example
ikram002p
  • ikram002p
ok example A={1,2,3} B={1.4} C={1,2.3}
mathmath333
  • mathmath333
but C is not there in the question
ikram002p
  • ikram002p
sorry i ment to say x :P
ikram002p
  • ikram002p
now to prove they are two side of the prove 1- show A is a subset of B 2- show B is a subset of A
ikram002p
  • ikram002p
let \(c\in A \cup X\) so \( c\in A ~or~c\in X \) if c in A then \( c\in B\cup X \) and \( c\in B\) (since c is not in x) thus \(A\subset B\) the other way is alike :P sorry im lazy to type
thomas5267
  • thomas5267
Is it possible to do something like this? \[ A\cup X=B\cup X\land A\cap X=B\cap X=\emptyset\implies A\cup X\setminus X=B\cup X\setminus X\implies A=B \]

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