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Math Help - Prove A^A=AvA=A

  1. #1
    Member courteous's Avatar
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    Question Prove A^A=AvA=A

    1) "Prove A\cap{A}=A\cup{A}=A." (This is very first workbook exercise for our Real (and Complex) analysis.)

    I have more: 2) "Prove that if a>b\Rightarrow -a<-b" and prove that 3) "square of every negative number is a positive number".

    I presume that the more elementary it is, the harder it is (like Principia proving that 1+1=2 (after how many pages?))?

    How many ways can such be proven (I think we've done it in discrete math, but workbook is from analysis)?
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    Super Member Showcase_22's Avatar
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    For 1) I would just use a truth table.

    For 3), can you just do this:

    Let k \in (0, +\infty). Therefore -k \in (- \infty,0).

    Hence (-k)^2=(-1)^2k^2=k^2 which is positive?
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    Quote Originally Posted by courteous View Post
    1) "Prove A\cap{A}=A\cup{A}=A." (This is very first workbook exercise for our Real (and Complex) analysis.)

    I have more: 2) "Prove that if a>b\Rightarrow -a<-b" and prove that 3) "square of every negative number is a positive number".

    I presume that the more elementary it is, the harder it is (like Principia proving that 1+1=2 (after how many pages?))?

    How many ways can such be proven (I think we've done it in discrete math, but workbook is from analysis)?
    To prove  A\cap A=A we must prove :

    x\in[A\cap A]\Longleftrightarrow x\in A.But x\in[A\cap A]\Longleftrightarrow x\in A ,is equivalent to:

    x\in A\wedge x\in A\Longleftrightarrow x\in A.................................................. .............................1

    And if we put xεA=P,THEN (1) becomes:

     p\wedge p = p, which is a law in propositional logic , called idempotent law,hence (1) is true ,hence  A\cap A=A.

    However if one wishes to prove the above law using other laws of propositional logic this can be done in the following way:

    1st we prove: p\wedge p\Longrightarrow p.

    ASSUME p\wedge p and ~p(= not p) then ~p implies ~p v ~p(BY a law in logic called disjunction introduction).
    But ~p v~p by using De Morgan ( another law in logic) is equivalent to,~( p\wedge p).Hence we have a contradiction:

    p\wedge p and ~( p\wedge p).

    Now to prove : p\Longrightarrow p\wedge p.

    Again assume ,p and ~( p\wedge p), which by De Morgan is equivalent to ~p v ~p,which is equivalent (by using another law of logic called material implication) to p\Longrightarrow~p.

    But now since we have p and  p\Longrightarrow~p,by using M.Ponens( another law in logic) we can infer ~p.Hence we have:

    ~p and p, a contradiction ,therefor we can say p\Longrightarrow p\wedge p.

    Now to prove: AUA =A.

    But AUA=A is equivalent { x\in (A\cup A)\Longleftrightarrow x\in A} \Longleftrightarrow [(x\in A\vee x\in A)\Longleftrightarrow x\in A]\Longleftrightarrow[(p\vee p)\Longleftrightarrow p],if we put again: xεA=P
    Again p\vee p\Longleftrightarrow p is a law of logic called again idempotent.

    Now to prove : a>b ===> -a<-b.

    If we multiply both sides of the inequality by ,-1 the inequality changes direction since -1<0.But we can prove the above implication in another way.

    Add to both sides of the inequality -b and then -a ,and we get successively :a-b>0 and a-a-b>-a ====> -a<-b.



    To prove : x<0 ====> x^2>0,multiply x<0 by x<0 ,and since x is less than zero we have x^2>0.

    And finally to prove : 1+1=2.

    Define the successor of zero denoted by, 0' equal to 1 i.e 0'=1 and 1'=2.

    AND by using the Peano axioms :

    1) for all x ,( x+0 =x)
    2) for all x,y [ x+ y'= (x+y)'],we have:

    1+1=1 +0' =( 1+0)' = 1' =2
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