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I've been having some trouble with this question while preparing for college algebra. Can anybody give an explanation of the proofs for the following question step by step so that I can comprehend? (I'm only $16$ for reference)

Let $S$ be a non-empty set, $A(x)$ and $B(x)$ can be open sentences evaluated on $S$.

a) Prove or disprove for all elements in set $S$, $A(x) \implies B(x)$, then there exists an element in set $S$ such that $B(x)$

b) If $S$ is the empty set, the prove the converse is true.

Stewart
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    If you are trying to prepare for college algebra this is not what you should be studying. This is abstract algebra and is something you would normally not take until after at least two semesters of calculus. – Squirtle May 28 '14 at 02:27
  • @Squirtle Calculus isn't required for basic abstract algebra..., but of course, you're right. On a formal level, you usually do not take abstract algebra before calculus. –  May 28 '14 at 02:28
  • Even though the OP may be preparing for collega algebra, the actual question is not about abstract algebra, but about set theory or logic. – Magdiragdag May 28 '14 at 02:29
  • Stewart, what have you tried? –  May 28 '14 at 02:29
  • Ugh, way to state the obvious. You say this as if I didn't know, regardless.... the fact remains that this is usually how it is done (regardless if it's optimal for the learning experience). In any case, the OP said that they are preparing for college algebra and clearly this question isn't that subject. – Squirtle May 28 '14 at 02:30
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    Okay, call it whatever you want.... I'd be completely shocked if we're teaching this in "college algebra". My point being, has the OP considered the possibility that this study is a waste of their time if they just want to do college algebra (trig functions, trig identifies, basic properties of a function, what a function IS, etc) – Squirtle May 28 '14 at 02:31
  • My logic for a) is that since S is non-empty, and A(x) implies B(x) but it does not necessarily mean that there is an element in S such that B(x) unless A(x) was equal to B(x) – Stewart May 28 '14 at 02:37

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Your logic for (a) in the comments is mostly correct. The only minor mistake is: even if $A(x)$ and $B(x)$ are the same, this still might not mean that any element of $S$ satisfies $B(x)$. For example:

  • if $S$ is the set of even integers,
  • $A(x)$ is true if $x$ is odd, and
  • $B(x)$ is true if $x$ is odd (so $A(x)$ and $B(x)$ are the same),
  • then $A(x)\Longrightarrow B(x)$ is true,
  • but not a single element in $S$ satisfies $B(x)$.

Part (b) is sort of a "trick question", but here is a statement of the converse with the $S$ removed, which might help.

If there exists an element in $\varnothing$ such that $B(x)$, then it follows that $A(x)\Longrightarrow B(x)$.

(Note that we were able to remove the $S$ because we have a different name that we could it instead, i.e. $\varnothing$, the empty set)

Eric Stucky
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  • I'm still not sure why it would follow that A(x) implies B(x) if there exists an element in the empty set such that B(x). I understand why it would be the other way around but having B(x) shouldn't necessarily mean a implies b right? Thanks! – Stewart May 28 '14 at 03:49
  • The only hint I can give you is: Read the sentence very carefully. You will notice something strange. – Eric Stucky May 28 '14 at 05:02