question on sets

**zerodigit** · September 27th 2008, 08:50 AM

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.

Please help me as i could not solve the question

**Plato** · September 27th 2008, 09:08 AM

$p \in A \Rightarrow \left( {\exists j \in \mathbb{N}} \right)\left[ {p = 2j} \right] \Rightarrow \left[ {2|p} \right] \Rightarrow p \in B$

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

**zerodigit** · September 28th 2008, 05:46 AM

could you explain it to me a little

I could not understand the signs

**Plato** · September 28th 2008, 06:02 AM

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.

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.

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