Slog Week 12
This week’s material covering multiple infinities is
different from the understanding I had before. I did not use the concept of
having infinities being mapped out on a one to one scale before. I did not even
realize that rational numbers could be even compared to natural numbers. In my
mind the set of rational numbers and the set of natural numbers has such an immense
gap they could not possibly be mapped out. As for the induction, as I mentioned
in a previous Slog, this course is actually quite similar to MAT137. I’ve
already had a experience in highschool with induction actually as well as in
MAT137. I feel like it’s somewhat repetitive however, I do not mind it.
I like this week’s problem and seeing that I have not done a
problem solving slog yet I will seize this opertunity and do it.
Understanding the
problem:
To my understanding the question is asking me to draw a rectangle
(squares count as rectangles technically) with squares inscribed into them which
have the sides m and n such that they represent the number of squares inscribed
on each side. Now given sides m and n, I am tasked to be able to predict the
number of squares which are crossed when a line is drawn from one corner to the
opposite corner.
Planning a Solution:
When I first looked at this problem, I thought of listing
the various situation which can happen and then thought of ways which the
problem could be solved. Then I made a description of what would happen in each
case.
Carrying Out the
Plan:
To start off I first drew several practice examples and tried
how the relation of similar triangles behaved in various cases. Then I made
several observations about how the triangles react when given different lengths
then I came convinced myself of the properties which I have observed and tested
them and reason them out to convince myself they work, and then I listed all
the properties out.
For the sake of ease whenever we are given numbers I am
going to refer to the longer side as m, and the shorter side as n. Also the
rectangle will be rotated such that the long side is horizontal such that side
m is the length of the bottom or top and side n is the side of the right or
left.
Property 1:
Property 1:
If side m is can be divided by n then the number of boxes is
m
Property 2:
If the if both m and n are even then they will meet at the
mid-corner of the triangle before crossing over.
Property 3:
If m is odd and n is even then the middle square will always
cross twice
Property 4:
If m and n have common factors then they will make a series
of smaller rectangles within the bigger rectangle. The smaller rectangles will
have the dimensions of what’s left after m and n have their greatest common factor
factored out. The number of boxes will equal the number of
Property 5:
If m and n are both odd then the middle square is only
crossed once.
Case 1:
M = N then squares = m
Case 2:
M and N have a greatest common factor, then treat m and n as
a series of smaller rectangles with the number being the greatest common
factor. And the smaller rectangles have side’s m and n such that they are
divided by the greatest common factor.
Case 3:
1)
M is even and n is odd
Boundary crosses = celling ( m/ n ) * n
2)
M is even and n is even
(Does not exist because of Case 2 Greatest
common factor)
3)
M is odd and n is even
Remove middle block so then m = (m-1) n = n
treat it as an m = even and n = even case and factor the greatest common factor
out. And plus 2 to the final result.
4)
M is odd and n is odd
Remove middle block and apply the M is even
and n is odd case and plus 2 to the final result.
Reflection:
I treated this as a program but then I’m not even sure if
this works for all cases. I did not come up with an elegant way of solving this
and instead used a very brute force method which I’m not even sure works with
all the cases.