Myra wrote:I'm just going to pretend this a family table and talk to Greg while you guys are chatting.
Hi, Myra.
I don't do pretend or make-beleive.
I happen to know for a fact
that you are a very bright student who worships the ground I walk on.
Therefore I am happy to teach you new things.
(And if you meet me after class, you can learn even more!
More math I mean. There is never enough math. )
~~
I probably didn't get very far with my question when it first occurred to me as a kid.
But when I recalled it the other day it took me less than a tenth of second to solve.
In other words it's really trivial. If you know certain things.
Otherwise it isn't trivial.
(It's the same with everything in math
It either utterly defeats you. Or else it is trivially obvious. )
And I have some experience trying to communicate these particular set-theoretic ideas
to intelligent enough people, - who simply flat-out refused to believe them!
~~
Let the lattice points be the integral points in the plane--
-- ie, all the points with integer coordinates like
(1,1), (1,2), ..., (2,1), (2,2), (2,3)... etc ...
And let the probe line be any line through the origin,
so that its slope is == y/x,
where (x,y) is any point on the line.
Then the probe line will hit a lattice point
if and only if
its slope is
rational.
(Because if the slope is rational,
then it is equal to some j/i, where i and j are integers,
which implys that the lattice point (i,j) is on the line.
Whereas if a lattice point (i,j) is on the line,
then the line's slope is == j/i, - which is rational.)
Therefore a probe line misses all the lattice points
if and only if its slope is
ir-rational.
Now, probe-line slopes go through all the real numbers between 0 and infinity
(-- the angle with respect to the x-axis going from 0 to 90 degrees.
But it is best to exclude the extremes cases, as you already noticed.)
And it is a fact that
"almost all" the real numbers in that set are
irrational .
Which then implies that
almost all the probe lines
miss all the lattice points.
~~
"Almost all" is a technical expression.
To say that "almost all A are B"
means that the set of As that are not Bs is "countable"
"Countable" is another technical expression
A set is
countable iff it is either "finite", or else "countably infinite".
And you know what "finite" means.
Whereas a "countably infinite" set is simply one
that can be put into into one-to-one correspondence with the integers, 1,2,3, ....
~~
The fact that
almost all real numbers are irrational
is shown by means of a little "zig-zag" argument due to Cantor.
The rational numbers can easily be put into an explicit one-to-one correspondence with the integers.
And it's just as easy to show that the irrational numbers can't.
I say "easily".
But, like I said, I've known smart enough people who for some
reason just can't get past this step! And I don't know why.
But I have concluded that it's because set-theory is a form of
love.
So some people get it. But others just don't.
~~
Finally, another fact is that the rational numbers are "dense" in the reals.
"Dense" meaning, in this case, that for any real number e > 0, however small,
and for any real number r, we can find integers i and j such that |r - j/i| < e.
In other words, given any irrational number, we can find
a rational number arbitrarily close to it.
Which implies, pretty quickly, that if the lattice points aren't true points,
but are rather little disks with a fixed finite radius, then every probe line
will intersect an infinity of them.
(On the other hand, if the radii diminish fast enough
towards infinity, then you might be back at almost all probe lines
missing all lattice points again.)
~~~~~~~~~~~~~~~
Myra wrote:is it counter-intuitive to think that they are good?
One would think it ought to have been for Anne Frank!
You do know the story of Anne Frank, don't you?
