
#1
Mar2109, 03:47 PM

P: 109

Hello,
Trying to go through Tom Apostle text on Calculus. There is an exercise about proving the associative law for sets: So, (A U B) U C = A U (B U C) So, if we assume x to be an element in set in left hand side, than we can say x belongs at least to either A, B or C which in turn means that x is also an element in set in right hand side and then we can say that the LHS and RHS are subsets of each other... Is this a valid proof? I am never sure with these. It is really tricky to prove such ideas that we take for granted in every day life! Anyway, I would be really grateful for any help you can give this old man. /Luca 



#2
Mar2109, 04:04 PM

P: 322

Hi pamparana,
What it comes down to is that "or" (logical disjunction) is associative, so ((x in A or x in B) or (x in C)) is the same as (x in A or (x in B or x in C)). Either you take this for granted, or you check that that the truth tables for ((p or q) or r) and (p or (q or r)) are the same. 



#3
Mar2109, 08:31 PM

HW Helper
P: 1,344

[tex] \begin{align*} x \in (A \cup B) \cup C & \leftrightarrow x \in (A \cup B) \text{ or } x \in C \\ & \leftrightarrow x \in A \text{ or } x \in B \text{ or } x \in C \\ & \leftrightarrow x \in A \text{ or } x \in (B \cup C) \\ & \leftrightarrow x \in A \cup (B \cup C) \end{align*} [/tex] I've use [tex] \leftrightarrow [/tex] to represent the phrase "if and only if" (I couldn't get the usual double arrow to work, sorry). Hope this helps. 


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