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Thread: Proving A is a subset of B...

  1. #1
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    Proving A is a subset of B...

    This is my first attempt at a proof like this (and even proofs in general) and my textbook gives no direct example for me to compare mine with. Can you examine this and tell me if I have proved what I intended to?



    Thank you all.
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  2. #2
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    Quote Originally Posted by mjlaz View Post
    This is my first attempt at a proof like this (and even proofs in general) and my textbook gives no direct example for me to compare mine with. Can you examine this and tell me if I have proved what I intended to?



    Thank you all.
    You're more or less right, but there's no need to pick an $\displaystyle x \in B$. Remember, that if it is the case that $\displaystyle A \subseteq B$, then for every $\displaystyle x \in A$, we have that $\displaystyle x \in B$. So, pick an $\displaystyle x \in A$ - then, we have that $\displaystyle x = 4c$, but we can rewrite this to $\displaystyle x = 2(2c)$, from which we conclude that $\displaystyle x \in B$, which satisfies our definition.

    I've just put all your ideas into a shorter proof. If this is indeed your first proof, it's an excellent attempt
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  3. #3
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    Thank you, friend!

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  4. #4
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    Quote Originally Posted by mjlaz View Post
    This is my first attempt at a proof like this (and even proofs in general) and my textbook gives no direct example for me to compare mine with. Can you examine this and tell me if I have proved what I intended to?



    Thank you all.
    Remember a proof is a series of logical conclusions, therefor the following proof shows that:


    xεA===>( x is an integer divided by 4) ====> (there exists an integer c such that x=4c) ====> (there exists an integer c such that x =2(2c))====> ( there exists an integer d such that x= 2d) =====>(x is an integer divided by 2)====> xεB.

    Hence $\displaystyle A\subseteq B$


    Where the double arrow means : logically implies
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