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??

 

So what is the negation of this? Simple: The series is bounded.

 

[math]\exists H\in \mathbb R^+:\,\forall n\in \mathbb N \ |S_{n}|<H[/math]

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

You made a universal statement. A simple counterexample *is* proof that your claim is false.

What is proof??

Knowing that a sequence doesn't diverge to infinity just doesn't tell you anything, it might converge to a limit or it might oscillate within certain bounds.

Sorry,

When i asked to substantiate that i mean to prove your claim.

An example does not constitute , a proof.

Knowing that a sequence doesn't diverge to infinity just doesn't tell you anything, it might converge to a limit or it might oscillate within certain bounds.

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 ??

No. Just imagine a sequence alternating between the values 1 and 0 as a counter example.

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??

What is the definition used in that book for upper bound? If it is something along the lines of given an ordered structure [math](E,\le)[/math] and a set [math]A\subset E[/math], then an element [math]c\in E[/math] is an upper bound of [math]A[/math] if [math]\{x \in E:x > c\} = \Phi[/math], then it should be obvious that every element of [math]E[/math] is an upper (and lower) bound of the empty set. If the set [math]E[/math] is itself unbounded (e.g., the reals), then there is no least upper bound (greatest lower bound).

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.

 

 

 

 

 

Does the statement that "nothing" is greater than (or smaller than) something make any sense?

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??

Oh, x's and y's the other way around, whatever.

 

Equivalence does sort of carry redundancy with it, when the equivalence is trivial.

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

triclino:

For 4, just take 1 and set X=sqrt(x) and Y=sqrt(y). Similarly for 5, set X=sqrt(x) and Y=0. All you've done is give specific values to the variables, so it is the same lemma.

 

D H:

That's sufficient to prove that a positive real has no more than one positive root. Proving that it has at least one would in fact be much easier: I think I just ran into a brick wall by trying to do both at once, thanks though.

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

Quite seriously ajb and DH know what they are talking about and it would be a good idea to y'know, listen to him.

 

But while we're here:

 

9 and 10 are both special cases of [imath]x(a+b)=xa+xb[/imath], which is one of the field axoims anyway, so they are clearly true (which means that you wouldn't bother going through it rigoursly unless you were made to for the sake of an exercise).

 

6 is nearly how you'd define sqrt in the first place.

 

4 and 5 are made redudant by 1.

 

1,2,3,7 and 8 might be worth proving once but really it doesn't matter if you don't.

In which way 4 and 5 are redundant by 1

Merged post follows:

Have a look at the synthetic approach to real numbers.

 

On Wiki.

On PlanetMath.

 

I expect, with some work you could prove all you need from these axioms.

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

That derivation is fine. What is your objection?

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)

. a must be non-zero because 1/0 is undefined.

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 ??