# question on sets

• Sep 27th 2008, 08:50 AM
zerodigit
question on sets
Consider the sets A={x | x=2n, n is natural integer} & B = {y | if y is a natural
integer and is divisible by 2}. Both the sets A and B seem to signify the same
thing. Do you agree? Justify your answer in the light of set definition.

• Sep 27th 2008, 09:08 AM
Plato
$p \in A \Rightarrow \left( {\exists j \in \mathbb{N}} \right)\left[ {p = 2j} \right] \Rightarrow \left[ {2|p} \right] \Rightarrow p \in B$

$q \in B \Rightarrow \left[ {2|q} \right] \Rightarrow \left( {\exists k \in \mathbb{N}} \right)\left[ {q = 2k} \right] \Rightarrow q \in A$
• Sep 28th 2008, 05:46 AM
zerodigit
could you explain it to me a little

I could not understand the signs
• Sep 28th 2008, 06:02 AM
Plato
Quote:

Originally Posted by Plato
$p \in A \Rightarrow \left( {\exists j \in \mathbb{N}} \right)\left[ {p = 2j} \right] \Rightarrow \left[ {2|p} \right] \Rightarrow p \in B$ [/tex]

If p is in A then there is a natural number, j, and p=2j; then 2 divides p so p belongs to B.

Quote:

Originally Posted by Plato
$q \in B \Rightarrow \left[ {2|q} \right] \Rightarrow \left( {\exists k \in \mathbb{N}} \right)\left[ {q = 2k} \right] \Rightarrow q \in A$

If q is in B then because 2 divides q there is a natural number, k, such that q=2k; then q must belong to A.