# Prove A^A=AvA=A

• Apr 5th 2009, 10:28 PM
courteous
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?(Speechless)))?

How many ways can such be proven (I think we've done it in discrete math(Thinking), but workbook is from analysis)?
• Apr 6th 2009, 03:58 AM
Showcase_22
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?
• Apr 6th 2009, 08:08 AM
xalk
Quote:

Originally Posted by courteous
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?(Speechless)))?

How many ways can such be proven (I think we've done it in discrete math(Thinking), 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