# rational - irrational question

Discussion in 'General Math' started by ~greg, Apr 17, 2008.

1. ### ~gregGuest

Consider the lattice of integral points (i,j) in (Z,Z) in (R,R),
and probe lines on the origin, shooting through the lattice.

Then it is easy to see that a particular probe line
will miss all the lattice points IFF its slope is irrational
(excluding the horizontal and vertical lines,
and the point at the origin, in all this.)

Another way to say it is that "almost all" probe lines
miss all the lattice points.

And it's also easy to see that if disks of a fixed finite radius
are constructed on each lattice point, then no probe line
will miss all of them.

And it's not hard to see that,
given any particular slope w, then disks of diminishing radii,
(each radius depending on the particular lattice point),
can be constructed so that this particular probe line
misses all the disks.

And then it's immediate that you can select any finite or countable set
of probe lines and construct disks of diminishing radii
such that none of these particular lines hit any of the disks.

~~

It seemed to me, at first, that there might be a way
to construct diminishing radii so that, once again,
almost-all probe lines miss all the disks.
But now I think that's not be the case.
But I don't know.

So my question is, -- am I right that --that's not the case?
Is there is really no way to construct disks on lattice points
such that almost-all probe lines miss all the disks?

~greg

~greg, Apr 17, 2008

2. ### William ElliotGuest

What is a probe line on the origin?
Do you mean a line through the origin?

The set of lattice points is ZzZ and the real plane
is RxR. (Z,Z) and (R,R) are ordered pairs of two
elements from P(R).
What does a particular probe line mean?

Do you mean, a given line will miss all the lattice
points Z^2, iff it's has an irrational slope?

That is false. Do you mean, a given line through the origin will
miss all the lattice points ZxZ iff it's has an irrational slope?

BTW, horizontal lines have rational slope
and vertical lines require a special case,
but rest assured, their slope isn't rational.
Again, are you using probe line to mean line through the origin? Hm,
that's so because every real number is arbitrarily close to a rational.
False, let w = 1/2.
Again false.
Let L = { L_j | j in N } be a countable set of lines through
the origin (0,0), of the real plane R^2. For all j in N, let
s_j be the slope of L_j. Construct L so that for all j in N,
s_j is irrational and

lim(j->oo) s_j = 1/2

No matter how small of a disk you put around (1,1)
infinitely many of L will pass through the disk.

Thus the proposition you stated to which I said "Again false"
is still false with the added hypothesis that all the slopes
are irrational.
No you're not right for sloppy statement of your propositions
and you're not right for the corrected version of the
proposition discussed above.
Only if the set of lines is finite and all have irrational slope.

----

William Elliot, Apr 18, 2008

3. ### ~gregGuest

~~~~~~~~~~~~~~~~~~~~~~

I should not have used the word "probe"
if was going to throw you off that much.

However, the problem didn't originate in mathematics.
Moreover, in origin, it was probably closer in spirit to projective geometry
(where the usual expression is "line on a point")
than to high school analytical geometry
(where I suppose they do speak of "line through a point").

It's been decades since I was in school,
and I'm not used to picayune criticism.

~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~

ZzZ ?

If you study math awhille longer
then you will encounter expressions such as
"by an abuse of notation ... - in order to avoid pedantry".

Also, I wouldn't be so sure as you are
that ZxZ is, strictly speaking, a "subset" of RxR.

But I know what you mean.

And I know (or presume that) you mean power-set by P(R),
although I have never seen that notation for it before.

Likewise, the shorthand notation
(A,B) =(by definition)= { (a,b) e S : a e A & b e B}

does occur. Whether you've ever seen it before or not.

And in any case you evidently didn't have any trouble
understanding what I meant.

~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~

All you're saying there
is obviously what I meant by - -
~~~~~~~~~~~~~~~~~~~~~~

Good. You got that.

But again, I started this whole thing off by saying

That limited the universe of discourse.

It implicitly defined what I mean by a "probe line" in everything that followed.
"probe line"
means
"a line on (or through) the origin".

~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~

Well, that WAS a real MISTAKE on my part.

I meant, obviously, "given any particular irrational slope w".

~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~

Again, I meant, obviously,
"any finite or countable set of probe lines of irrational slope".

However, since, (as you point out with w=1/2)
the assertion is blatantly false as stated,
I should think it ought to have occured to you
that I must have meant irrational slopes.

Now, assuming that you understand that now,
if, then, still, you think the assertion is false,
then you just aren't thinking hard enough.

The distance of the probe line of slope w
to the lattice point (i,j) is

R(w,i,j,)
= | w - j/i | ( i / sqr(1+w²) )

So, all you have to do is make each disk on each (i,j)
have a smaller radius than that.

Or better, just use the interior of the disks of radii Rij.

Then the probe-line w will miss all the disks
on all lattice points.

And then using R(W,i,j) = min { R(w,i,j) : w e W }
you get the same assertion for a given finite set W
of irrational (of course) slopes w.

Now, using proof by waving of hands,
I asserted (I think I said "it's easy" or "obvious")
that the same thing holds for countable sets W.

And then, using proof by leap of faith,
I thought, at first, that the same might hold
for particular uncountable sets W too,
although obvioulsy not for all.

But then I thought that might not be the case.

Which is the question I asked.
~~

The rest of your response here is stuck
on my failure to have explicitly stated
that I meant irrational w.

Thank you for trying to help anyway,

~greg.

~~~~~~~~~~~~~~~~~~~~~~~~~

~greg, Apr 18, 2008
4. ### Dan in NYGuest

[... snip]

I have been working on something similar to your comments but I couldn't be
sure because I didn't understand some of your fine points. The comments by
William Elliot and your answers helped quite a bit. I would like to see a
that reflect your answers. Would you like to post it -- or send it to me
via email?

I think my ideas are a special case of yours with an added twist. In my
work, the irrational slope is a trigonometric function and I distinguish
whether it is trancendental or not when the angle (in degrees) is rational.

In particular, I want to represent the sine of an angle as a rational number
or as an irrational number that can be written as a (finite) expression
using the square roots of integers or the square roots of expressions that
contain the square roots of integers. What can be said about the angles
that have such a sine? To generalize, I want to represent cosines and
tangents in the same way.

I have a list of expressions for various angles and I would like to extend
that list.

Dan in NY, Apr 18, 2008
5. ### ~gregGuest

~~~~~~~~~~~~~~~~~~~~~~~

It's been a very long time, Dan in NY,
but you are definitely talking about "field theory" and "Galois theory",
....and beyond.

There're a lot of books about these subjects, but almost all of them
beat around the bush, probably more than necessary. The older ones
beat around older bushes, and the newer ones beat around newer bushes.

As soon as I straighten out this apartment, I'll have a recommendation
for you. It's a very little book. But it's the most direct route I know of,
- going through radicals and field extensions rather than too much
old fashioned manipulation-algebra, or too much newer fashioned
abstract-algebra.

"What can be said about ..." ?

-- They are field extensions.

More than that, I'm not qualified to say.

~greg

~greg, Apr 18, 2008
6. ### ~gregGuest

I wrote >
What I meant by that is that there is a very useful definition
of the cartesian product, - which for two sets A, B is this: ...

AxB = { The set of functions f: T => AUB,
where T = {1,2},
having dom = T and cod=AUB
such that f(1) is in A and f(2) is in B }

The reason it's useful is because it's the definitinon that's most readily generalized.

However, if you use it, then, strictly speaking, ZxZ is not a subset of RxR.

( Because the elements of ZxZ have co-domains= ZUZ = Z,
whereas the elements of RxR have co-domains = RUR = R,
so that no function in ZxZ is, strictly speaking, in RxR. )
~~~~~~~~~~~~~~~~~~~

~greg, Apr 18, 2008
7. ### William ElliotGuest

Math is exacting, so get used to it.
AxB = { (a,b) | a in A, b in B }, the set of all ordered pairs,
with the first element from A and the second element from B,
is the cross product of A and B
If (n,m) in ZxZ, then n,m in Z, hence n,m in R and thus (n,m) in RxR.
The word probe threw me off, requring deciphering of what you meant and
your ommision lines through the origin and then again the ommision of
irrational slope, left you not stating what you meant. A person with less
patience or less ability, wouldn't understand.
It wasn't obvious. It had to be deduced.
It should occure to you to proof read to assure completenss and accuracy
is what you assert. Math is an exacting science.
It does not because the minimum may be zero.
I gave a detailed counter example that showed just than.
Math is not religion. That which is apparent for all finite beings, holds
not for infinite beings. For example for all n in N,
0 < 1/n
and
inf{ 1/n | n in N } = 0

If you don't understand infinums, consider inf like min.
Read my counter example, where I explicity stated that
all the slopes s_j were irrational.
I've given you a counter example. Now read it.

William Elliot, Apr 19, 2008
8. ### ~gregGuest

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Yeah, but math also requires imagination and intuition
--which latter predominate in informal gatherings of people
all of whom know what they're talking about.

In such settings the ability to fill in the details is taken for granted.
Time isn't wasted on it. That's only done later, in private, and only
when necessary, eg for publication. There is just too much else
in life to get to.

I had hoped that this newsgroup would be more like an informal
brain-storming senimar.

That seems to be what Dan in NY had hoped it would be too.
He said "I have been working on something similar to your comments
but I couldn't be sure because I didn't understand some of your fine points."
And although he wasn't at all clear about his own fine points,
I didn't have any trouble seeing that he's right about that,
- there is a similarity. Unfortunately I can't help him because
I know next to nothing about transcendentals.

You (William) may be taking this ng to be more of a place where
high school students go to get help with their homework.

Nevertheless, a failure to communicate is not a good thing.

The problem between us is that there was never a precise statement
of the problem I'm asked about. And I wasn't looking for it.
I was looking for imaginative ways to extend the problem.

Dan in NY's question was actually more like what I was looking for.

I started from the observation that a probe into the integer-lattice
will hit a lattice point iff its slope is irrational.
Unless the "lattice" consists of disks of fixed finite radius,
in which case any probe, of rational or irrational slope,
will hit some of them.

It then it occured to me to go outward from there,
into a consideration of a sort of intermediate case,
in which the radii diminish outward from the origin,
---but never actually become 0 (--which I should have said clearly.)
Then a probe of irrational slope would again miss all the disks,
- provided their radii diminished in a specific way.

Likewise, given ANY FINITE SET L of probes of irrational slopes,
then radii can be defined in such a way that all the probes in L
miss all the disks.

Then it occured to me to ask if ANY EXAMPLE of a COUNTABLE
OR UNCOUNTABLE SET L, of probes of irrational slopes
can be constructed such that each probe in L
misses all the disks, provided their radii diminish in an appropriate way.

And you gave me A COUNTER-EXAMPLE
that shows that THIS IS NOT TRUE
FOR ALL COUNTABLE SETS L of probes of irrational slopes.

But I wasn't looking for that.
I was looking for ANY EXAMPLE WHERE IT IS TRUE.

In your counter-example the slopes of the lines in L approach 1/2.
Which is rational. So the probe-line of slope 1/2 is not in L.
It's just a limit of the set L.

And you say
And that's true.

And more generally, if any limit-line of the set L has
a rational slope a/b, then no matter how small a disk
is put around the specific lattice point (b,a),
then infinitely many lines in L will pass through it.

(ie, - I understand your counter-example.)

The upshot is that the countable or uncountable example
of a set L of probe lines of irrational slope
such as I am looking for
must neither contain (obviously)
- NOR APPROACH -
any probe line of rational slope.

The set of slopes W, of the lines in a set L such as I am looking for
WOULD HAVE TO BE A SET OF IRRATIONAL NUMBERS
SUCH THAT ALL ITS LIMIT POINTS ARE ALSO IRRATIONAL,
(whether or not any or all of them actually belong to W).

Which is why I called this thread a "rational - irrational question".

Because I don't know the answer to the question:
-- are there sets of irrational numbers
-- all of whose cluster points are also irrational?

And that's as far as I got with it
when I first asked the question.

Even if there does exist such a set, I am not 100% certain
that it could completely get around the gist of your
counter-example. But now I think it could. If it exists.

If such a set exists, then it gets around your specific
counter-example of a rational limit slope a/b implying
the existence of a specific lattice point (b,a) whose disk
would have to have a radius equal to zero.

But, while probes of irrational slope w, which are limits
of the set L, do have probes of rational slope a/b approaching
w arbitrarily closely, nevertheless the lattice point (b,a) in this case
is not fixed. As a/b->w, the disk on (b,a) does have to vanish.
But that's already true of all the probes of irrational slope in L anyway!

So, that line of thinking does not automatically
supply any dis-proof of the possible existence
of the kind of set I'm looking for.
~

I am sorry if I sounded testy before.

You hav've actually been of great help
in getting me to clarify my question.

Thank you!

~greg

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AxB = { (a,b) | a in A, b in B }, the set of all ordered pairs,
with the first element from A and the second element from B,
is the cross product of A and B
If (n,m) in ZxZ, then n,m in Z, hence n,m in R and thus (n,m) in RxR.
The word probe threw me off, requring deciphering of what you meant and
your ommision lines through the origin and then again the ommision of
irrational slope, left you not stating what you meant. A person with less
patience or less ability, wouldn't understand.
It wasn't obvious. It had to be deduced.
It should occure to you to proof read to assure completenss and accuracy
is what you assert. Math is an exacting science.
It does not because the minimum may be zero.
I gave a detailed counter example that showed just than.
Math is not religion. That which is apparent for all finite beings, holds
not for infinite beings. For example for all n in N,
0 < 1/n
and
inf{ 1/n | n in N } = 0

If you don't understand infinums, consider inf like min.
Read my counter example, where I explicity stated that
all the slopes s_j were irrational.
I've given you a counter example. Now read it.

~greg, Apr 19, 2008
9. ### ~gregGuest

"~greg" >
Obviously there are!

eg

{ pi + (n+2)/2n : n = 1..oo } => pi+1/2

~greg, Apr 19, 2008
10. ### William ElliotGuest

Such as posting to a newsgroup. take the time to proof before posting.
Even twice perhaps. It the long run it'll save you time and us time and
also perhaps you some embarrassment.
Your brain is storming. You want me to off the cuff, oh yea yea, I see.
Not only that but also, off the top of my head hear say and fairy tales or
do you want some actual facts and math?

If you're in too much of a hurry to do the math, then storm your brains
in a geometric art group.
That wasn't what you stated. Now that you have changed the inquiry more
thought, none of that quick splash dash thought, but real thoughtful
thought that may take more time and bother than instant gratification
media tutored consumers are want to take.
Good insight.
Hey hey, I can hear you from here. YOU DON"T HAVE TO SHOUT, that's
consider rude and lacking in netiquette.
{ pi }, { pi, e }, { pi, e, sqr 2 },

{ n.pi | n in N }, { pi + 1/n | n in N }

{ pi + 1/n + 1/m | n,m in N }

{ m.pi + 1/n + 1/k | n,m,k in N }

Thus you're left pondering the question if there's an uncountable
set of reals whose closure is disjoint from the rationals, Q.

That is an interesting question. A weaker interesting question is:
does an uncountable set of reals have uncountably many limit points?

According to my notes, such a set has uncountably many limit
points in the set itself. The other question I have no answer.
By convention, there isn't any.
Were you to do a rewright, posting in a newsgroup with a wider audience,
some having much greater learning than I, you should first clarify
A probe is a line of the real plane through the origin.

William Elliot, Apr 19, 2008
11. ### ~gregGuest

Now, given any set W of irrational numbers in (0,oo)
all of whose limit points are finite and irrational.

And consider any lattice point (i, j).

|- The distance from the "probe line" through the origin, with slope w,
to the point (i, j), is = | w - j/i | ( i / sqr(1+w²) )

So, define Aij = inf { | w - j/i | : w e W } .
This is > 0.

And define Bij = inf { i/ sqr(1+w²) : w e W } = i / sqr(1 + (supW)²) .
This is also > 0.

And define radius Rij = Aij · Bij,
This, too, is > 0,

And consider the disk Dij (the interior) of radius Rij
on the lattice point (i, j).

and consider the set L
= { "probe lines" on the origin having slopes w : w e W }

~~

Then I assert that every line in L misses all the disks Dij.
And I have answered my question.

~~~~

If this be error and upon me proved,
I never writ, nor no man ever loved

- Shakespeare

~greg, Apr 19, 2008
12. ### William ElliotGuest

So this set is bounded. Can it be uncountable?
Where's w coming from? What does the question mark mean?
What does the question mark mean?
What does the question mark mean?
What do the question marks mean? Can W be uncountable?

William Elliot, Apr 19, 2008
13. ### ~gregGuest

"William Elliot" >

It's been 40 years since I took topology, but that sounds right.

And it means there is no uncountable set of slopes and radii
such as I've been looking for.

(But I think I've proved that there is for the countable case
(in another post).)

Thak's again.

~greg, Apr 19, 2008
14. ### ~gregGuest

"William Elliot" >
Most likely it just means that we have our ng readers
set up to use different fonts.

The character is the standard ASCII for superscipt 2.
It's supposed to be the square of (supW).

~greg, Apr 19, 2008
15. ### OdysseusGuest

Not standard ASCII, but a superset thereof: ISO Latin-1 or, as indicated
superior-two character in your original posting as a double closing
quotation mark.) William is posting in ASCII, which doesn't have a slot
for the character.

Odysseus, Apr 20, 2008
16. ### William ElliotGuest

In other words, use n^2 for n squared, to be readable by all.
Some other question marks that weren't explainable by being the "squared"
symbol '^2'. What were they? Again, to be readable by all, become
acquainted with plane ascii text notation conventions used in these
groups.

William Elliot, Apr 20, 2008
17. ### William ElliotGuest

Me too. It makes for a nice retirement hobby, does it not?
Go to the news group sci.math and read the thread with subject
Q /\ cl A

where there is both a proof and a construction, that there can be
an uncountable A for which Q /\ cl A is empty.

William Elliot, Apr 20, 2008
18. ### ~gregGuest

"William Elliot" >
Thank you very very much.
That is exactly what I was stumbling around looking for.
And a very nice statement you made of it!

"quasi"s answer should not have surprised me the way it did at first.
After all, there really are an awful lot of irrational numbers,
- many many more than there are rationals. However, they
aren't talked about so much in the kinds of music and geometric art groups
that I've been hanging out in. I realize now that those other groups
are all bad influences, and I must disown, reject, and denounce them.

I do remember now the virtues of clarity precision and conciseness.
And I guess I have you to thank for that too.
~~

A few years ago I saw somewhere a FAQs about how to write math
in news groups. I have to find that again, if you know what I'm talking about.

~greg

~greg, Apr 20, 2008
19. ### se16Guest

It depends on what you mean by "almost-all".

If you mean all lines through the origin expect a set of measure zero,
then you are correct as it is impossible. Any single disk will cut
out a set of lines through the origin with positive measure (for any
sensible measure).

If you mean a set of lines through the origin with arbitrary large
measure, then you are wrong. Here is a simple example:

Let the radius of the disk at (i,j) be K/((|i|+|j|)*2^(|i|+|j|)) for
some constant K. Then for suitably small values of K (e.g. <1/4)
there will be lines through the origin which miss all the disks round
lattice points and this set of lines will have positive measure.
Decreasing K further increases the measure of disk-missing lines
through the origin towards, but not to, the total measure of lines
through the origin.

In effect the sets of lines are "nowhere dense sets with positive
measure"

se16, Apr 22, 2008