triclino

I am sorry i should have been more clear . By Aristotelian logic ,i mean the two value logic i.e true or false. So the problem becomes a problem ,of whether we have results in physics with a value more than two ( true or false ),or not.

is there any place in physics where , the Aristotelian logic does not apply??

Let us stick to sequences The question now is : Does the sequence converge

Instead of showing your knowledge about quantifiers by using silly examples ,why don't you prove by using quantifiers and their laws ,what is the statement resulting from the negation of the following statement: [math]\forall\epsilon[\epsilon>0\Longrightarrow\exists N(n\in N\wedge\forall n(n\geq N\Longrightarrow S_{n}\geq\epsilon))][/math] Which is the definition of a sequence diverging to[math]+\infty[/math] And thus answering my opening post

What is proof??

. Sorry, When i asked to substantiate that i mean to prove your claim. An example does not constitute , a proof.

How can you substantiate that ??

So let us make one point very clear. A sequence or perhaps a series that does not diverge to [math]+\infty[/math] does not converge to a limit. Agree ??

You mean, that if the sequence does not diverge to[math]+\infty[/math] then vibrates ( alternates) between two Nos ??

Suppose that the sequence [math] S_{n}[/math] diverges to [math]+\infty[/math]. Now if [math]S_{n}[/math] does not diverge to [math]+\infty[/math] , does that imply that [math]S_{n}[/math] goes to a limit s

I presented a professor with the following proof: Prove that the empty set is closed. Proof : By definition : [math]\emptyset[/math] is closed <=> cl[math]\emptyset\subseteq\emptyset[/math] <=> ([math]x\in[/math] cl [math]\emptyset[/math]=>[math]x\in\emptyset[/math]) cl [math]\emptyset[/math] is the closure of the empty set,and B(x,r) is a ball of radius r round x But ,by definition again [math]x\in[/math]cl [math]\emptyset[/math] <=> for all r>0 ,B(x,r)[math]\cap\emptyset\neq\emptyset[/math]....................................................................1 But , B(x,r)[math]\cap\emptyset =\emptyset[/math] => (B(x,r)[math]\cap\emptyset =\emptyset[/math] or [math]x\in\emptyset[/math]) <=> (B(x,r)[math]\cap\emptyset\neq\emptyset[/math] =>[math]x\in\emptyset[/math]) And using (1) we get : [math]x\in\emptyset[/math] Thus ,we have proved: ([math]x\in[/math] cl [math]\emptyset[/math]=>[math]x\in\emptyset[/math]) And the empty set is closed. The professor did not accept the proof as correct . Do you agree with him??

So, in your definition ,let A = (0,1) ,an open interval and E = R ,the real Nos. According to your definition 2 is not an upper bound of (0,1) ,because {xεR : x>2} is not empty. But according to the definition of the book ,page 186, 2 is an upper bound. The definition is: " if A is a set of real Nos then y is an upper bound of A if and only if y is areal No and for every x in A ,[math]x\leq y[/math]"

But Patrick Suppes in his book "AXIOMATIC SET THEORY" on page 186 ,ask us to give a proof that the empty set has no least upper bound,meaning that there is a proof that the empty set has no least upper bound.

No. But the correct statement is: for all x : if x belongs to "nothing" then this x is smaller or greater than something. And now we wander whether this conditional is true or false. According to wikibedia it is "vacuously" true,meaning that ,since x belongs to 'nothing " is false the conditional is true,irrespectively whether x is smaller or greater than something. However assuming that the conditional is not true then its denial must be true. But its denial is the following: There exists an x belonging to "nothing" and this x is greater or smaller to something, ending in a contradiction ,since there dos not exist an x belonging to "nothing". Hence "nothing" is bounded. Now the question is, if "nothing" has a supremum (infemum) or not

I was hoping some one will produce a solid proof that the formula: there exists aεR such that,for all xεΦ ,[math] x\leq a[/math] , is true and not the vacuously true nonsenses of the wikipedia

is the empty set bounded from above? has it got a supremum (l.u.b)? Whether it has or not how can we prove that??

If you put x=0 and y =sqrt(y) we have: [math] 0\geq 0[/math] and [math]\sqrt y\geq 0\Longrightarrow(0\leq\sqrt y\Longleftrightarrow 0\leq(\sqrt y)^2)[/math] From which formula you can never prove : [math]y\geq 0\Longrightarrow\sqrt y\geq 0[/math] unless you assume: 1) [math] y\geq 0\Longrightarrow\sqrt y\geq 0[/math],a formula that you want to prove ,and: 2) [math] y\geq 0\Longrightarrow(\sqrt y)^2 = y[/math] Also in proving the equivalence between (1) and (4) it is not a trivial matter because we have other theorems involved in the proof and the proof it is not just a substitution

For (5): if you put x=sqrt(x) and y=0 in (1) you have: [math]\sqrt x\geq 0[/math] and [math]0\geq 0\Longrightarrow (\sqrt x\leq 0\Longleftrightarrow (\sqrt x)^2\leq 0)[/math] Where you have : [math] \sqrt x\geq 0[/math] and [math]\sqrt x\leq 0[/math] and you cannot get (5) in any way. As for formula (4) it is equivalent to (1) ,hence not redundant

. In which way 4 and 5 are redundant by 1 Merged post follows: Consecutive posts merged In taking the first implication of the proof i.e [math] 0\leq a\leq 1\Longrightarrow 0\leq 1-a^2\leq 1[/math] what would be the theorems that are involved and how are they involved

I propose for the above proof the following set of theorems: 1) for all ,x,y : ( [math]x\geq 0[/math] and [math]y\geq 0\Longrightarrow ( x\leq y\Longleftrightarrow x^2\leq y^2)) [/math] or in words: for all ,x,y : if x and y are greater or equal to zero ,then x is lees or equal to y iff x to the square is less or equal to y to the square. 2) for all ,x,y,z : [math]x\leq y\leq z\Longleftrightarrow -x\geq -y\geq -z[/math] 3) for all ,x,y,z,w :[math] x\leq y\leq z\Longleftrightarrow x+w\leq y+w\leq z+w[/math] 4) for all ,x,y : [math] x\geq 0[/math] and [math]y\geq 0\Longrightarrow( x\leq y\Longleftrightarrow\sqrt x\leq\sqrt y)[/math] 5) for all x : [math] x\geq 0\Longrightarrow \sqrt x\geq 0[/math] 6) for all x : [math] x\geq 0\Longrightarrow(\sqrt x)^2 = x[/math] 7) [math] 1\geq 0[/math] 8) for all x,y : [math]x\geq 0[/math] and [math]y\geq 0\Longrightarrow\sqrt x.\sqrt y = \sqrt xy[/math] 9)for all ,x,y : [math] ( x+y)^2 = a^2 +b^2+2ab[/math] 10) for all ,x,y : [math] (x+y)(x-y) = x^2-y^2[/math] And finally the definition of sqrt. Can any one propose a different or shorter list

I was wandering what are the theorems or definitions used in the proof

for the following problem: For all ,a : if [math]0\leq a\leq 1[/math], then [math]\sqrt{1+\sqrt{1-a^2}} +\sqrt{1-\sqrt{1-a^2}} =\sqrt{2+2a}[/math].....................................................................................A the following proof is suggested: [math]0\leq a\leq 1\Longrightarrow 0\leq 1-a^2\leq 1\Longleftrightarrow 0\leq \sqrt{1-a^2}\leq 1[/math][math]\Longrightarrow(\sqrt{1+\sqrt{1-a^2}}\geq 0\wedge \sqrt{1-\sqrt{1-a^2}}\geq 0)\Longrightarrow[/math] [math](\sqrt{1+\sqrt{1-a^2}}+\sqrt{1-\sqrt{1-a^2}})^2[/math] = [math]1+\sqrt{1-a^2}+1-\sqrt{1-a^2}+2\sqrt{(1+\sqrt{1-a^2})(1-\sqrt{1-a^2})}=2+2|a|=2+2a[/math] And since [math] 2+2a\geq 0[/math] (A) is valid

To prove : for all natural Nos ,n : [math]n^2 [/math] is even[math]\Longleftrightarrow[/math] n is even.....................A The following proof was suggested: for all ,nεN : n=2k or n= 2k + 1 ......κεN But ,[math] n^2 = (2k + 1)^2 =4k^2 + 4k +1 = 2(2k^2 +2k) + 1[/math] Hence : [math]n^2[/math] is not even or n is even....................1 Also: [math] n^2 =(2k)^2 = 2(2k^2)[/math] Hence: [math]n^2[/math] is even or n is not even......................2 From (1) and (2) we conclude: (A)

In that case the theorem to prove is: if [math]a\neq 0[/math] and (1/a)>0,then a>0 ,and not: if (1/a)>0,then a>0

Can we prove : if (1/a)>0 ,then a>0 ??