For 1) I would just use a truth table.
For 3), can you just do this:
Let . Therefore .
Hence which is positive?
1) "Prove ." (This is very first workbook exercise for our Real (and Complex) analysis.)
I have more: 2) "Prove that if " 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 we must prove :
.But ,is equivalent to:
.................................................. .............................1
And if we put xεA=P,THEN (1) becomes:
, which is a law in propositional logic , called idempotent law,hence (1) is true ,hence .
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: .
ASSUME 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,~( ).Hence we have a contradiction:
and ~( ).
Now to prove : .
Again assume ,p and ~( ), which by De Morgan is equivalent to ~p v ~p,which is equivalent (by using another law of logic called material implication) to ~p.
But now since we have p and ~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 .
Now to prove: AUA =A.
But AUA=A is equivalent { } ,if we put again: xεA=P
Again 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