- #1
Lajka
- 68
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Hello,
I've just recently stumbled upon this forum, in search for an answer for my little dilemma, so I hope someone can help me. This is the question:
Given that [tex]dx[/tex] is an infinitesimal interval on set [tex]R[/tex], does it mean that it has infinitly many points in itself as well?
If I understood correctly, its length does approaches zero (but never seems to quite get there), however, it's still an interval by its nature, and all intervals have the cardinal number c.
The reason I ask this is because a friend told me that there is a theorem, or a consequence of some theorem, that an infinitesimal region can contain at most one real number, and this pretty much rocked my world.
Here's some reasoning on my behalf:
1) Like I sad, it's an INTERVAL. If it merely contains a one real number, what else does it contain then? Seems not much of an "interval" to me...
2) This again is more of a question, really. If i take dx, and divide it by two, I should have gained two new (even) smaller intervals. I don't know if I have a right to do this. But if I do, then each one of them can contain at most one real number, so the original interval can now containt at most two real numbers. You can imagine what happens if I continue dividing these smaller intervals. So, it's a paradox. Something tells me this is logic is flawed, so I would appreciate if someone could point out where I made a mistake, thanks.But then again, even I sometimes, intuitively, believe that dx can't be "normal" interval.
Let me give an example of this:
If I look at the concept of integration, but not in the ordinary "area under the curve" sense, but in the sense that reflects the Fundamental Theorem:
[tex]\int^{b}_{a} f'(x)dx = f(b) - f(a)[/tex]
i.e. I imagine that I'm summing up a bunch of [tex]dy's (dy=f'(x)dx)[/tex], e.g. [tex]dy_{1}+dy_{2}+...[/tex], and in that process the upper boundaries of the previous [tex]dy[/tex] and lower boundaries of the next one cancel each other. Following this process, and going from [tex]x=a[/tex] to [tex]x=b[/tex], we finally get some finite distance: [tex]f(b)-f(a)[/tex]
What I'm trying to say, is that these intervals, these [tex]dy's[/tex], they don't overlap. But now I wonder. Why wouldn't they? If I imagine [tex]dx's[/tex] as "normal" intervals, I would actually expect that they do. I am not sure if I made myself clear enough, so I drew a little picture:
http://imagebin.ca/img/CFuUFxr.png
The lower part in the picture illustrates the way I reason the process of integration, and the upper part my new dilemma.
Again, if my logic is flawed, I would really appreciate if someone could point out why.
So, that's it. If I imagine a [tex]dx[/tex] as an infinitesimal distance, but still with infinitely many points, I have this problem. Then again, if I don't, I have dilemmas that I explained in the beginning. A bloody nightmare.
Any help is appreciated. Also, please bear in mind that I study to be an electrical engineer, so, although I like to think that I'm not a newbie in mathematics, I''m sure I can't compare with you guys. So be gentle, please. :D
Much obliged,
Lajka
I've just recently stumbled upon this forum, in search for an answer for my little dilemma, so I hope someone can help me. This is the question:
Given that [tex]dx[/tex] is an infinitesimal interval on set [tex]R[/tex], does it mean that it has infinitly many points in itself as well?
If I understood correctly, its length does approaches zero (but never seems to quite get there), however, it's still an interval by its nature, and all intervals have the cardinal number c.
The reason I ask this is because a friend told me that there is a theorem, or a consequence of some theorem, that an infinitesimal region can contain at most one real number, and this pretty much rocked my world.
Here's some reasoning on my behalf:
1) Like I sad, it's an INTERVAL. If it merely contains a one real number, what else does it contain then? Seems not much of an "interval" to me...
2) This again is more of a question, really. If i take dx, and divide it by two, I should have gained two new (even) smaller intervals. I don't know if I have a right to do this. But if I do, then each one of them can contain at most one real number, so the original interval can now containt at most two real numbers. You can imagine what happens if I continue dividing these smaller intervals. So, it's a paradox. Something tells me this is logic is flawed, so I would appreciate if someone could point out where I made a mistake, thanks.But then again, even I sometimes, intuitively, believe that dx can't be "normal" interval.
Let me give an example of this:
If I look at the concept of integration, but not in the ordinary "area under the curve" sense, but in the sense that reflects the Fundamental Theorem:
[tex]\int^{b}_{a} f'(x)dx = f(b) - f(a)[/tex]
i.e. I imagine that I'm summing up a bunch of [tex]dy's (dy=f'(x)dx)[/tex], e.g. [tex]dy_{1}+dy_{2}+...[/tex], and in that process the upper boundaries of the previous [tex]dy[/tex] and lower boundaries of the next one cancel each other. Following this process, and going from [tex]x=a[/tex] to [tex]x=b[/tex], we finally get some finite distance: [tex]f(b)-f(a)[/tex]
What I'm trying to say, is that these intervals, these [tex]dy's[/tex], they don't overlap. But now I wonder. Why wouldn't they? If I imagine [tex]dx's[/tex] as "normal" intervals, I would actually expect that they do. I am not sure if I made myself clear enough, so I drew a little picture:
http://imagebin.ca/img/CFuUFxr.png
The lower part in the picture illustrates the way I reason the process of integration, and the upper part my new dilemma.
Again, if my logic is flawed, I would really appreciate if someone could point out why.
So, that's it. If I imagine a [tex]dx[/tex] as an infinitesimal distance, but still with infinitely many points, I have this problem. Then again, if I don't, I have dilemmas that I explained in the beginning. A bloody nightmare.
Any help is appreciated. Also, please bear in mind that I study to be an electrical engineer, so, although I like to think that I'm not a newbie in mathematics, I''m sure I can't compare with you guys. So be gentle, please. :D
Much obliged,
Lajka
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