[BilboGrabbins]

@BilboGrabbins

Hi there!
🙋

I have a Query, Especially for You.
Interested?

A_______B

A simple line.
Start point A.
End point B.

I suppose i can keep dividing it into Halves.

But considering Planck's Length, Can i Divide it until Infinity?

If i Know the Line's point of Origin(A) & am Aware of the Line's point of Cessation(B)...Does that sound like Infinity?

Or if i didn't know the Origin, n was clueless bout the End.
Infinity!

Ps - Thanks E_S for the suggestions.
I've Realized my capacity to understand & learn is Not infinite.
Hence i do not bother myself, & also spare Others of goin thru the troubles of explaining me things which i Firmly Believe i shall never understand.
(Reason i Request for short & brief answers, i Hate it when Teachers waste their Precious Time on Futile Things)
: )

No, to divide through the Planck length infinitely many times would be equivelent to a singularity. That is itself equivelent to a breakdown in physics.

[Eternal Student]

Hi.

I've Realized my capacity to understand & learn is Not infinite.

   Neither is mine.  I was especially "thick" as a child and couldn't read, write or do arithmetic until I was about 10 years old.  Fortunately the UK doesn't hold people back a year when they don't pass exams, so I did get out of primary school.

No, to divide through the Planck length infinitely many times would be equivalent to a singularity. That is itself equivelent to a breakdown in physics.

    This is about the limitations of our ability to probe below the planck length not the actual nature of space.

...The Planck length is expected to be the shortest measurable distance, since any attempt to investigate the possible existence of shorter distances, by performing higher-energy collisions, would inevitably result in black hole production....         [from Wikipedia]
    It doesn't follow that space isn't continuous, just that our current experiments to probe below that length are expected to create a black hole and change the nature of space in the region.

Best Wishes.

[BilboGrabbins]

It's sort of the point here. Physics breaks down below this length.

[hamdani yusuf (OP)]

...The Planck length is expected to be the shortest measurable distance, since any attempt to investigate the possible existence of shorter distances, by performing higher-energy collisions, would inevitably result in black hole production....         [from Wikipedia]

What prevents us from expanding a 1 meter rod to 1 meter plus half Planck's length, eg. by heating it up?

[Eternal Student]

Hi.

What prevents us from expanding a 1 meter rod to 1 meter plus half Planck's length, eg. by heating it up?

   Are you asking me?  BilboGrabbins was the one proposing that space cannot be divided into units smaller than the Planck length.
   There's nothing I know of that would prevent space from being continuous.  Indeed, the standard formulation of General Relativity requires that space is continuous.
   However, when you come to measure a rod there are practical limits in our ability to measure things that accurately.   It also assumes the rod is some idealised body and not a quantum mechanical system.  Treating the particles that make up the rod as quantum mechanical objects would put a theoretical limit on the ability to localise the ends of the rod. 
    If I recall correctly, it was this sort of limitation on the ability to localise a particle that lead to the first proposals that space (and time) may not be continuous since it almost becomes irrelevant:  You can't localise a particle to one point in space unless it's momentum* → ∞, there is some uncertainty in it's position and therefore there isn't any great difference if you just split space up into discrete intervals rather than assuming it is continuous.  However, I prefer to remain open to both possibilities.  Space might be continuous or it might be discrete.
    From a purely mathematical point of view.  We often take derivatives or integrals with respect to some co-ordinate.  This Calculus does assume the co-ordinate will be a continuous variable.

Best Wishes.

LATE EDITING:   * See comment below by Hamdani Yusuf.  "uncertainty in momentum".

[hamdani yusuf (OP)]

You can't localise a particle to one point in space unless it's momentum → ∞

It's not the momentum itself that becomes infinite. It's the uncertainty of momentum.

[hamdani yusuf (OP)]

https://twitter.com/pickover/status/1541962088926560256?t=ml0TBOGlhgCZIKDaLXserg&s=03
There are many hypercomplex numbers known by modern math. Do they have the same size as real number?

[Bored chemist]

It's sort of the point here. Physics breaks down below this length.

I wonder if Bilbo thinks that physics breaks down for objects smaller than the Planck mass which is about 21 µg
On a day-to-day basis 21µg is a pretty small mass, but it's not impossibly small.
It's about the mass of a million bacteria.

(And I'm still waiting for him to explain what angles I'm not allowed to orient my coffee table.)

[Bored chemist]

https://twitter.com/pickover/status/1541962088926560256?t=ml0TBOGlhgCZIKDaLXserg&s=03
There are many hypercomplex numbers known by modern math. Do they have the same size as real number?

Apparently, the complex numbers have the same cardinality as the reals and I'm tempted to assume that the hypercomplex ones do too.
https://quizlet.com/explanations/questions/show-that-c-the-set-of-complex-numbers-has-the-same-cardinality-as-r-the-set-of-real-numbers-cf3e343a-9b5b-4f78-8a0c-0dcfe9ee5b15

[alancalverd]

Reverting to the question a bit, thanks to Heisenberg physics may demonstrate a limit to any experiment involving matter, but mathematics does not involve matter and therefore does not pose limits on what can be imagined or discussed.

[hamdani yusuf (OP)]

Apparently, the complex numbers have the same cardinality as the reals and I'm tempted to assume that the hypercomplex ones do too.

There are as many real numbers in a set of complex numbers where imaginary part is 0. Like wise, there are as many real numbers in a set of complex numbers where imaginary part is 1. You can change the imaginary part with any magnitude. Each of them contains a whole set of real numbers. How can we say that they have one to one relationship?

[Bored chemist]

Apparently, the complex numbers have the same cardinality as the reals and I'm tempted to assume that the hypercomplex ones do too.

There are as many real numbers in a set of complex numbers where imaginary parts is 0. Like wise, there are as many real numbers in a set of complex numbers where imaginary parts is 1. You can change the imaginary part with any magnitude. Each of them contains a whole set of real numbers. How can we say that they have one to one relationship?

Infinite sets are weird.
https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel
There are as many even numbers as there are integers (because every even number is twice an integer).
There are as many odd numbers as there are integers (because every even number is twice an integer with 1 added to the product).

There are similarly as many multiples of 3 as there are integers.
Ditto multiples of 4, 5 ... etc.
So the integers are the union of an infinite number of infinite sets sets.
But the cardinality stays the same.

So why would there be a problem with the complex numbers having the same cardinality as the reals?

[hamdani yusuf (OP)]

So why would there be a problem with the complex numbers having the same cardinality as the reals?

Why would real numbers have different cardinality than rational numbers?
Where do algebraic numbers belong to?

[alancalverd]

And the answer to the question is "as many as you like, plus at least as many again". To take the most obvious case, if you specify any integer N, all those from 0 to N are presumed to exist, as do all those up to N² or any other integer power.

[Eternal Student]

Hi.

Apparently, the complex numbers have the same cardinality as the reals and I'm tempted to assume that the hypercomplex ones do too.

   Yes, that's correct.
The usual proof uses interleaving of decimal digits to generate a unique real number.  Here's an extract that explains it with minimal distraction:

To show this (a bijection exists between C and R )  you need to show every real number maps to a unique complex number and that every complex number maps to a unique real number. The first part is easy -- every real number already is a unique complex number. So let's concentrate on the second part, mapping the complex numbers to the reals.

Each complex number a+bi, a and b real, may be mapped to a unique real number as follows: Expand a and b as decimals, taking care not to end either in repeating 9s for uniqueness. Then you can interleave the digits of the two decimal expansions so, e.g, the even numbered digits are from a and the odd from b. Start at the decimal points and work out in both directions. Call the result of this interleaving c.
So you now have a real number c  from which you can recover a and b, and thus the original complex number a+bi.

Since you did this mapping both ways, by the Cantor-Bernstein theorem (or common sense) the cardinality of the complex numbers and real numbers are the same.

[Quote taken from:  https://www.quora.com/Do-the-complex-numbers-have-a-greater-cardinality-than-the-real-numbers ]

The proof requires something like the Schröder-Cantor-Bernstein theorem which means that you do require some of the higher axioms of set theory   (e.g. the Axiom of Choice or axioms of logic such as the principle of excluded middles).   Wikipedia has some discussion:  https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem#Prerequisites .    To say that a simpler way,  in some systems of mathematics you would not be certain that C and R have the same cardinality.

    Anyway,  with  |R X R| = | R |   it would be simple to generalise and show  | R^k | = |R|  for any natural number k.     Since the hypercomplex numbers are elements of a finite dimensional algebra over R,  we have that any algebra of hypercomplex numbers would have the same cardinality.

There are as many real numbers in a set of complex numbers where imaginary parts is 0. Like wise, there are as many real numbers in a set of complex numbers where imaginary parts is 1. You can change the imaginary part with any magnitude. Each of them contains a whole set of real numbers. How can we say that they have one to one relationship?

    Indeed it is weird.   It follows from a set of abstract axioms.    "Intuitionists" would not accept many of the results concerned with cardinality of infinite sets.   As already mentioned, there are systems of mathematics (or axiom schemes) that would not conclude  |R| = |C|.   If I recall correctly, you ( @hamdani yusuf  ) have an interest in Philosophy.   You might understand the Wikipedia description of Mathematical Intuitionism  better than I do:   https://en.wikipedia.org/wiki/Intuitionism.

Best Wishes.

LATE EDITING:  Sorry for the overlap.  Several new posts have appered since I started writing this.  I'll read those in a moment.

[hamdani yusuf (OP)]

There isn't a Mathematics section in this forum.  It seemed that the OP was asking something about Mathematics without any reference to some application in Physics.

To be clear, I want to discuss pure maths here, unrestricted by physical application. What's important here is the consistency of definitions and relationships among every used concepts in a theory.

[Origin]

To be clear,

Looks like another never ending, merry-go-round thread by Hamdani to put on ignore.  Bye and have fun.

[Bored chemist]

I didn't read the whole thread, but perhaps you could quote the place where somebody asserted that this would be problematic.

Here

How can we say that they have one to one relationship?

[hamdani yusuf (OP)]

Since you did this mapping both ways, by the Cantor-Bernstein theorem (or common sense) the cardinality of the complex numbers and real numbers are the same.

We can represent any rational numbers by combining two integers as numerator and denominator. 
Some irrational numbers can be stated as a rational number powered by another rational number.  Let's call them power numbers. How many more expansion procedures like that are required to cover the whole real numbers?

[Eternal Student]

Hi.

We can represent any rational numbers by combining two integers as numerator and denominator.
Some irrational numbers can be stated as a rational number powered by another rational number.  Let's call them power numbers. How many more expansion procedures like that are required to cover the whole real numbers?

   I honestly don't know.    It depends what you consider to be an "expansion procedure like that".

For example, would you allow taking limits?     If   {X_n}  is a sequence of numbers you already admit in your set,  then  why not also include the Real number X whenever the sequence X_n → X.   That seems like a reasonable expansion procedure.    Anyway, that expansion procedure gets you straight from Q  to  R   in one step.

   The old fashioned view is that when trying to build the Real Numbers getting from N to Q is just algebra.  You can indeed go a bit further and include some irrational numbers in a field extension of Q if you tried.    However, crossing that last mile to build the complete, Archimedian, ordered field we call the Reals steps outside the territory called "Algebra" and into "Real Analysis"  (i.e. where things like limits need to be used).   It's a bit arbitrary to draw a line somewhere and say this bit and no further is algebra, while that bit over there is analysis but that's what is generally done.   I'm mentioning this because you might be considering "an expansion like that" to be some procedure involving only algebra and not analysis:   If you denied yourself access to the techniques of Real Analysis then conventional wisdom is that you'll never get to R.   

[However, I don't know if anyone was so worried about not crossing that arbitrary line that they really tried hard.  I do know that they wouldn't have been able to prove they had got to R - which is mildly amusing.   This is because they must demonstrate the field is complete and "completeness" is a property firmly held in Real Analysis territory, so they can't even use those ideas let alone perform some test for it].

Best Wishes.