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Math Help - In over my head - proofs questions.

  1. #1
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    In over my head - proofs questions.

    Today was only the third lecture of my intro to higher math class, yet the first homework is due in two days and we're required to prove a number of things without even being first instructed on what qualifies as a "proof", or given any idea how to approach them. Unfortunately at the moment there is literally no one I can go to with these questions, so I'm hoping you can help.

    Question 1: We've been told that: F \Rightarrow T = T. Can someone explain why this is the case? Is it just convention?

    Question 2: Let P be the statement " x^2+2=11 for all real numbers x such that x^3+32=5". This statement appears to be true, as the real solution for " x^3+32=5" also solves " x^2+2=11" But is it required that the "real number x" be a solution to both problems? If I were to change this to: Let P be the statement " x^2+2=11 for all real numbers x such that x^3+32=6", does the overall statement become false? It seems like a silly question, but how can something that is broken down into a logic dependence also have a second dependence on individual solution sets? For example: Let R be the statement " x^3+32=5 for all real numbers x such that x^2+32=0" Since there are no real solutions for x^2+32=0, the statement gets reduced to F \Rightarrow T = T or F \Rightarrow F = T. But without an x to plug into x^3+32=5, how can it exist in any state, true or false? Isn't it just unknown?

    Question 3: Let A and B be non empty sets. Prove that: A \times (B \cup C) = (A \times B) \cup (A \times C)
    This is my pitiful attempt at some kind of "proof":
    Let A = \{a\}, B = \{b\}, and C = \{c\}.
    B \cup C = \{b,c\}
    A \times (B \cup C) = \{(a,b),(a,c)\}
     A \times B = \{(a,b)\}
     A \times C = \{(a,c)\}
     (A \times B) \cup (A \times C) = \{(a,b),(a,c)\} = A \times (B \cup C)
    Unfortunately that seems a lot more like a "demonstration" than a proof. I have no idea how to approach this.

    Question 4: Let A and B be sets. Prove that A \times \emptyset = B \times \emptyset = \emptyset I understand that a Cartesian product requires the creation of an ordered pair, and the pair  \{(a,\emptyset)\} is not an ordered pair. As far as structuring a proof to say that, I'm lost.

    I apologize for the sloppiness of this post, and for the barrage of questions. If anyone can offer any help or insight, I would very VERY much appreciate it. Thanks in advance.
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  2. #2
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    Question 1
    One of the most often quoted statements in a logic class is: ďA false statement implies any statement and true statement is implied by any statement.Ē
    In other words, we are interested only in the case where the antecedent is true then the consequence must also be true. If the antecedent is false we donít care so label it true.
    Look at the truth tables for p\to q~\&~ \neg p \vee q. Those are equivalent statements.
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  3. #3
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    Qestion 3: No! You're giving an example, bot a proof.

    You want to show that each side is a subset of the other side. In general, to show X\subseteq Y, you begin with x\in X, and then argue that x\in Y.

    A typical element of A\times (B\cup C) has the form (x,y) where x\in A and y\in B\cup C. Can you now argue that (x,y)\in (A\times B)\cup (A\times C) ?

    Don't forget to do the other direction too (you may actually be able to do both directions simultaneously in this particular problem using "if and only ifs", but I would do them seperately at first for practice).


    When you really understand Question 1, and the method I just showed you in Question 3, then Question 4 is quite simple.
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    Question 4
    If A\times B\not=\emptyset then \exists (a,b) \in A \times B.
    That means a\in A~\&~b\in B or A\not=\emptyset~\&~ B\not=\emptyset .
    Use the contrapositive.
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  5. #5
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    Quote Originally Posted by ChrisEffinSmith View Post
    Question 1: We've been told that: F \Rightarrow T = T. Can someone explain why this is the case? Is it just convention?
    Yes. Implication is a function of two arguments (each of the arguments is T or F). The result is F, by definition, if the first argument is T and the second one is F. Here this is not the case, so the result is T, by definition.

    Question 2: Let P be the statement " x^2+2=11 for all real numbers x such that x^3+32=5". This statement appears to be true, as the real solution for " x^3+32=5" also solves " x^2+2=11" But is it required that the "real number x" be a solution to both problems?
    No. P can be written symbolically as \forall x\,(x^3+32=5\Rightarrow x^2+2=11), i.e., P is true iff every real number x makes the implication x^3+32=5\Rightarrow x^2+2=11 true. If there were a conjunction ("and") instead of an implication, then P would require that every x is a solution to both equations. As it is, only those x that satisfy the first equations have to satisfy the second one.

    If I were to change this to: Let P be the statement " x^2+2=11 for all real numbers x such that x^3+32=6", does the overall statement become false?
    Yes. For x=-\sqrt[3]{26}, x^3+32=6 is true, but x^2+2=11 is false. This makes the whole implication false, and the whole "for all" statement is also false because of this one counterexample.

    Let R be the statement x^3+32=5 for all real numbers x such that x^2+32=0" Since there are no real solutions for x^2+32=0, the statement gets reduced to F \Rightarrow T = T or F \Rightarrow F = T. But without an x to plug into x^3+32=5, how can it exist in any state, true or false? Isn't it just unknown?
    For each particular x, the conclusion x^3+32=5 is either true or false (namely, it is true iff x = -3). However, its truth value does not matter since the implication is true for all x due to the false premise.
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