What is the meaning of a double implication in aproof .
This is a question for those that they have difficulties in following double implications proofs,which is a very short way of presenting some of the proofs
If you're referring to
A is true B is true
that is read "A is true if and only if B is true".
It's the same as saying A implies B and B implies A.
It's also the same as saying statements A and B are equivalent.
There is some additional commentary in remark 1.4 here:
http://www.bobostrategy.com/PartIChapter1.jpg
which is a sample page of my book
(see Amazon.com: Topology and the Language of Mathematics: Chris Cunliffe: Books)
which contains a fair amount of material that would be helpful with other questions like this.
Thank you bobostrategy,but i open this thread for people that they have difficulties in following double implications proofs.
I have written couple of proofs using double implications and i have discovered that some members cannot follow the proofs because i get no respond to my proofs.
Finally i will be interested to read your book and, if possible, to communicate with you and discuss logical aspects in mathematics
I truly disagree with that assessment.
I think that your overly formal logical proofs confuse the issue.
No working mathematician uses formal logic in proofs.
As I have said to you in another post, many mathematicians loathe logicians.
Sad to say, the proofs that you offer will not past muster in most mathematics classes. It is not because they are incorrect, they are not, but they are over-talk.
What is wrong with double implications? Double implications is a mathematical tool ,although of logical origin as you very well know, that mathematicians use:
1) To write short compact proofs
2)To discover proofs.
For example we want to prove that: 0x = 0 for all x real Nos. An inexperience student would not know ,sometimes ,where to start from.
However having an idea of the concept and use in proofs of double implication ,he could put it into practice in the following way.
Lets say 0x=x is correct ..
0x=0 η=θ ( implies and is implied) 0x +x = 0+x. Because if we add x to both sides of 0x=0 we get 0x +x =0+x ,or if we subtract x from both sides of 0x +x = 0+x we get 0x=0.
The law allowing the above is : adding or subtracting a No to both sides of an equation the equation does not change.
So far we have that: ..0x=0 ηθ 0x +x = 0+x 1
By looking now on the right hand side of (1),we observe that 0x+x it is equal according to distributive property of Real Nos to x( 0 +1),which if we substitute into (1) ,it becomes: x(0 +1) = 0+x,
HENCE : 0x + x = 0+x ===θ(implies) x(0+1) = 0+x ..2
But ,since again x(0 +1) = 0x + x,if we substitute that into (2),it becomes: 0x+x=0+x,
THUS .......x(0+1)=0=x ====θ(implies 0x+x=0+x .3.
And ..0x+x=0+x η==θ x(0+1)= 0+x 4
And .(1) becomes: 0x=0 η==θ 0x+x=0+x η=θ x(0+1) = 0+x 5.
By similar reasoning and using the facts that ,0+1=1 and 0+x=x we conclude:
x(0+1) = 0+x η===θ 1x= x WHICH is an axiom in Real Nos 6
And (5) becomes : 0x=0η==θ 0x+x=0+x η==θ x(0+1)=0+x η=θ 1x=x ..7.
Now since the last step of (7) is a well known axiom and therefore always true .WE CAN SAY BASED ON THEOREMS IN LOGIC that :
..0x=0 is true for all Real x ..
So not only we have proved the above identity with minimum brain energy but also have discovered a proof of single implications.
Therefore we can reason:
1x=x ==θ x(1+0) =0+x ===θ x+0x = 0+x ===θ 0x=0
.Which are the back wards steps in (7).
Another high school example is the following:
Ix+YI =< IxI + IyI i.e the absolute value of the sum is less or equal to the sum of the absolute values,where x, y are real Nos.
A double implication procedure will be as follows:
Ix+YI =< IxI + IyI η===θ Ix+yI^2 =< (IxI + IyI)^2 η===θ x^2 + y^2 + 2xy =< x^2 + y^2 +2IxIIyI η====θ xy =< IxIIyI A WELL known fact.
FOR the above proof note IAI^2 =A^2.
Coming now to proofs concerning set identities,which are very often appear in this subforum ,an example maybe will show the benefits of a double implication application.
Let us take one of the distributive identities in sets:
AU(B C) =( AUB) (AUC).........1
.......Assume again (1) to be true and:
AU(B C) =( AUB) (AUC) [by equality of sets]<========>{ xε[AU(B C)] <======> xε[(AUB) (AUC)]}[by definition of intersection and union of sets] <======> {xεΑ v( xεB ^ xεC) <======> (xεA v xεB) ^ ( xεA v xεC)}.
But if in the equivalence: xεΑ v( xεB ^ xεC) <======> (xεA v xεB) ^ ( xεA v xεC) ,we put ...xεA=p, xεB=q,and xεC=r the above becomes:
pv(q^r) <====> (pvq) ^ (pvr) which is one of the distributive properties in propositional calculus.
So double implication reveals that for the set identity to be true the distributive property of propositional logic must hold. A fact which is definitely true.
And conclusively the set identity is true.
Double implication can be used in slightly different way in the above problem.
Let,
..............xε[AU(B C)] [by definition of union and intersection of sets] <=======>xεΑ v( xεB ^ xεC)[by distributive property in logic]<=====>[(xεA v xεB) ^ ( xεA v xεC)][again by definition of union and intersection of sets] <=====>xε[(AUB) (AUC)].
Again from logic we can infer :
xε[AU(B C)]<======>xε[(AUB) (AUC)].
Thus AU(B C) =( AUB) (AUC).
.
"Perhaps a few examples will put things in their right place.
From an analysis book i quote the following proof:
It is easy to show that when a limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that
lim f(z) =l and lim f(z)=m as z----> a (z goes to a).
Then,for any positive number ε,there are positive numbers r,δ such that
I f(z)-lI< ε whenever 0<Iz-aI< r
and
If(z)-mI < ε whenever 0<Iz-aI < δ.
So if 0< Iz-aI< θ ,where θ denotes the smaller of the two Nos r and δ,we find that
Im-l I = I(f(z)-l)-(f(z)-m)I =< I f(z)-l I + If(z)-mI < ε+ε =2ε.
But Im-lI is a nonnegative constant, and ε can be chosen arbitrarily small.
Hence
l-m =0 , or l=m.
And i am asking you and captain black is that proof correct??
where for absolute value of z i use IzI"