Otherwise...

George,Y · 2007-01-26 16:37:29

Do you agree with the following statement:

There exists an amount c that throughout (0,c) some status(You can image it as a bulb) is "on",
Otherwise there should be an amount d satisfying throughout (0,d) the status is "off".

?

Or put it another way, do you think the two statements are negtive to each other?

luca-deltodesco · 2007-01-26 19:12:58

i dont understand the change to using d instead of c, but supposing ive understood you right otherwise, id think so yes.

to put it the way ive understand it.

there exists c, such that for all (0,c) some status is on

then the negation would be

for all c, there does not exist (0,c) such that some status is on:

i.e. for all c, some status of (0,c) is off.

which seems to be saying to me, your second statement.

George,Y · 2007-01-26 22:43:36

No, that comes the tricky part.

Logically and Typically when we negate "all" we just need one example "not".

But the second statement says "all", too.

And it is my error not having expressed the meaning thoroughly:

Actually both c and d are arbitary.

So when you list, for instance, the point 2/3c that the status is contrary, the new c could be c/2, this time (0,c) satisfy the first statement. More precisely, c has infinite choices(potential infinity, a far more perplexing topic). So, it seems only another "all" could negate the first statement. But how can we get it solidly?

Ricky · 2007-01-27 04:29:13

Yes, I would agree that they are negations of each other.

Logically and Typically when we negate "all" we just need one example "not".
But the second statement says "all", too.

But you aren't negating "all". You are negating an "exists":

There exists an amount c that throughout (0,c) some status(You can image it as a bulb) is "on",

Or in symbolic form, with f being a binary function (on and off):

Negation:

Now we see in that negation that if there does exist such a c, then there must exist a minimum c with this property in (0, c). This is equivalent with the statement that there exists a d such that f(a) = 0 for all a in (0, d).

George,Y · 2007-01-27 23:58:55

I am sorry that E-inverse and "any" symbol in your two Latex statements do not make sense to me. I cannot understand why you are sure about the negation stands.

George,Y · 2007-01-28 00:25:37

However I tried another approach

Starting from the latter statement.

There exist a d, all right. Let's analyze c, c is either less than d or larger than d if they are not equal.

If c is less than d, when the latter statement holds, a such c would make bulb off throughout (0,c), strictly negative to bulb on throughout (0,c);

If they are equal, the same.

If c is larger than d, when the latter statement holds, a such c fail to satisfy the circumstance that the bulb is on throughout (0,c). This is a very critical situation, because at this point we may be confused about whether it is a negation or not. But noting Ricky's brilliant reminder, the first statement is not about the status of the interval but about whether a c satisfying the certain condition exists. Surely no c exists.

Thus when the second statement is true, no c is satisfying the condition that throughout (0,c), the bulb is on. Thus inorder to have a c satisfying the condition, we need the second statement false, right? ......Oh, I am confused...

Ricky · 2007-01-28 07:05:02

I am sorry that E-inverse and "any" symbol in your two Latex statements do not make sense to me. I cannot understand why you are sure about the negation stands.

First off, what do you mean by "do not make sense"? Are you unfamiliar with my notation above, or are you familiar and you just don't see it making sense?

Second, what do you mean "negation stands"? It is a statement which may be true or false, depending on the circumstances. It is not always true, nor is it always false. But, if the first statement is not true, then the negation must be true.

George, can we lose the "bulb" stuff and replace it with a function which is either 0 or 1? It really shouldn't make a difference overall, but I believe it will be easier to talk about. But if you really wish to hold on to it, I have no problems. Like I said, shouldn't make a difference.

If c is less than d, when the latter statement holds, a such c would make bulb off throughout (0,c), strictly negative to bulb on throughout (0,c);

The way I read your question, I thought it was implied that d < c. If this is not implied, then your negation is invalid. In my negation, it states that d is in (0, c), and thus it must be that d < c.

But noting Ricky's brilliant reminder, the first statement is not about the status of the interval but about whether a c satisfying the certain condition exists. Surely no c exists.

I'm sorry, but how do you reach the conclusion that no such c exists? I take it you mean under all circumstances.

Thus inorder to have a c satisfying the condition, we need the second statement false, right? ......Oh, I am confused...

Heh, and it's wearing off on me. In order to have c satisfy the first condition (that (0, c) is all on), it must be that the negation is false.

George,Y · 2007-01-28 13:22:58

No, I think we have different definations on the term "negation".

I treat negation as "otherwise situation", if A is a situation, then
_
A is its negation, meaning "all the other situations".

And this is definately not simply using all the negatives.

For example:
All the birds can fly.
Your negation would be:
All the birds cannot fly.
Or
None the birds cannot fly.

But what I am concerned about is otherwise, what?

It should be, however:

Otherwise, some bird (at least one bird) can fly.

I'm certain this negation concludes all the other situations.

Ricky · 2007-01-28 15:01:56

No, I think we have different definations on the term "negation".
I treat negation as "otherwise situation", if A is a situation, then
_
A is its negation, meaning "all the other situations".

If A is a statement, then the negation of A, also known as "not A" is all the situations where A is false.

For example:
All the birds can fly.
Your negation would be:
All the birds cannot fly.
Or
None the birds cannot fly.

No, that is incorrect. The negation of "All the birds can fly" is "There exists a bird which can not fly." In symbolic notation, this is:

Negation:

In fact, in general, all one must do is switch the quantifiers (there exists and for all), then negate the "formula", and one may arrive at the negation of any statement.

George,Y · 2007-01-28 19:36:37

There exists a bird which can not fly- Uh, that is just what I meant, sorry for missing one "not"

Now I understand what you have said in Post 4:

To get the negation, you first negate the quantity of (All or at least one) c.

Then you should negate the qualifier, the description. We could simply say that exist c, not clause 2 and clause 3 in your first statement. As I wrote "Surely no c exists"(All c are not the c that).

But this "not" is another statement. You further simplified this statement. Not <All alphas in the interval (0,c) satisfy f(alpha)=1> means <at least one alpha in the interval doesn't satisfy f(alpha)=1>. However, your statement is more complex: you used d instead of alpha, and simplified <doesn't satisfy f(alpha)=1> to <satisfy f(alpha)=0>

In my point of view, you did two steps of negation technique, and they are one outside, one inside. Further, you made use of the negation of f(alpha)=0 as f(alpha)=1.

Apparently, the negation of f(alpha)=0 is not f(alpha)=0, but if we assume f(alpha) exists, (the existential assumption), we can then develop this negation to f(alpha)<>0. Since f can only be 0 or 1, the true negation of f(any)=0 or f(any)<>0 is that f(the any)=1. And I call this the derivative of the negation that can be deducted from the negation.

So actually we need to check is whether my second statement is the derivative of the negation of the first, or the derivative of your second statement.

But I proved is that the negation can be deducted from the second statement. So this proof is useless. It only states that IF the second statement is true, the first is false, and hence the negation of the first is true.

More on Negations:
Negations of one statement are basically Not+the statement, although it may take on a more complex form.
A negation must have two properties regarding what it negates:
1) The negation must contradict what it negates. That is, if one true, the other must be false.
For example, I played video games yesterday 16:00-17:00 contradicts I studied in the liberary in the mean time.
2) The negation must complement what it negates. That is , if one is false, the other must be true.
Playing videogames surely fail to have this property. I did anything but studying in the liberary can meet this property.

So the negation could be: I did anything but studying in the liberary. Or I didn't studied in the liberary.
Or It's not the case that I studied in the liberary. (Not that I studied in the liberary)As I have said, negations may have many forms.

George,Y · 2007-01-28 19:44:29

2 id Not a negation, but it isn't because the reason you've offered.

I proved if 2 is true, 1 is false, that is 2 contradicts 1. However, 2 is not complementry to 1, meaning there may be other situations that 2 is false( outside of 2), and 1 is false.

By simply picking up 1,1/2,1/4,1/8... as the points that make the bulb off, or your f=0

Ricky · 2007-01-29 03:55:20

But this "not" is another statement. You further simplified this statement. Not <All alphas in the interval (0,c) satisfy f(alpha)=1> means <at least one alpha in the interval doesn't satisfy f(alpha)=1>. However, your statement is more complex: you used d instead of alpha, and simplified <doesn't satisfy f(alpha)=1> to <satisfy f(alpha)=0>

I could have just used alpha. In fact, I did at first, but you were using d to be such an element, and so I went back and edited by post.

Also, f <> 0 is equivalent to f = 1, as my function has a range of only 0 and 1.

Apparently, the negation of f(alpha)=0 is not f(alpha)=0, but if we assume f(alpha) exists, (the existential assumption), we can then develop this negation to f(alpha)<>0. Since f can only be 0 or 1, the true negation of f(any)=0 or f(any)<>0 is that f(the any)=1. And I call this the derivative of the negation that can be deducted from the negation.

There isn't really an existential assumption in there however. Your bulb must be either "on" or "off". I defined my function to parallel this. Thus, even though I never explicitly stated it, my function is defined on all of the reals. So it's not an assumption, but rather a definition.

2 id Not a negation, but it isn't because the reason you've offered.
I proved if 2 is true, 1 is false, that is 2 contradicts 1. However, 2 is not complementry to 1, meaning there may be other situations that 2 is false( outside of 2), and 1 is false.

I'm sorry, "2 id"? What are 2 and 1?

By simply picking up 1,1/2,1/4,1/8... as the points that make the bulb off, or your f=0

You're exactly right. Open sets got the best of me:

I wrote:

Now we see in that negation that if there does exist such a c, then there must exist a minimum c with this property in (0, c). This is equivalent with the statement that there exists a d such that f(a) = 0 for all a in (0, d).

I stated the existence of a minimum. However, it may be the case that such a minimum lies on a limit point of the set, which in this case would be 0. As 0 is not in (0, k) for any real k, the minimum is not in the set, and thus, (0, d) does not exist.

However, I believe the fix is simple:

Now we see in that negation that if there does exist such a c, then there must exist a minimum c with this property in (0, c). This is equivalent with the statement that there exists a d such that f(a) = 0 for all a in [0, d).

The problem was not in my negation, but rather my translation of that negation into your equivalent statement.

George,Y · 2007-01-31 02:06:18

I'm sorry, "2 id"(is)? What are 2 and 1?
Statement 1 and Statement 2 in post 1.

Now we see in that negation that if there does exist such a c, then there must exist a minimum c with this property in (0, c). This is equivalent with the statement that there exists a d such that f(a) = 0 for all a in [0, d).
-sorry I don't understand you with "this property" (f=1 or f=0), and the left-closed interval[0,d). However, I believe it has something to do with Cantor Sets. Nevertheless, we talk about virtually the same thing.

George,Y · 2007-01-31 02:07:23

Thank you Ricky and Luca

Math Is Fun Forum

#1 2007-01-26 16:37:29

Otherwise...

#2 2007-01-26 19:12:58

Re: Otherwise...

#3 2007-01-26 22:43:36

Re: Otherwise...

#4 2007-01-27 04:29:13

Re: Otherwise...

#5 2007-01-27 23:58:55

Re: Otherwise...

#6 2007-01-28 00:25:37

Re: Otherwise...

#7 2007-01-28 07:05:02

Re: Otherwise...

#8 2007-01-28 13:22:58

Re: Otherwise...

#9 2007-01-28 15:01:56

Re: Otherwise...

#10 2007-01-28 19:36:37

Re: Otherwise...

#11 2007-01-28 19:44:29

Re: Otherwise...

#12 2007-01-29 03:55:20

Re: Otherwise...

#13 2007-01-31 02:06:18

Re: Otherwise...

#14 2007-01-31 02:07:23

Re: Otherwise...

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