Disturbing math facts that you didn't ask for
by greenturtle3141, Dec 19, 2021, 8:34 PM
Reading Difficulty: 0-5/5
Some classics:
2 Hint
3 Comment
Some others:
1 Reason
2 Comment
4 Reason
5 Reason
and
both work.
6 Proof
7 Hint
7 Another Hint
Feel free to share others in the comments.
Some classics:
of
is equal to
of
.
- It's possible to drill a hole into a cube such that you can pass a bigger cube through the hole.
- Take a map of the world. Crumble it into a ball and throw it outside. There exists a point on the map that lies directly above the real-world location in the world that it corresponds to.
- The harmonic series diverges. But it will converge if you remove terms that contain a
as a digit in the denominator.
2 Hint
Rotate the cube so it looks like a regular hexagon from above.
3 Comment
This follows from the Brouwer's Fixed Point Theorem. Essentially, if you let
be the point in the world that's directly under the point on the map corresponding to the point
, then
is a continuous map from the world to the world. Some compactness argument lets you apply Brouwer's to deduce that
has a fixed point.




Some others:
- The last digit of
is not
.
for all
.
- Consider a unit cube in
. It can be anywhere and with any orientation. No matter what, the length of its projection unto the
-axis will be equal to the area of its projection unto the
-plane.
- There exists a relation that is symmetric and transitive, but not reflexive.
- How many open covers are there of
in the empty topology? Answer
2 - Suppose
is a monic polynomial with integer coefficients whose roots are real numbers in
. Then
.
- Is it true that every chain of subsets of
is countable? Chain Definition
A chain of sets is a family of setsAnswersuch that for any
, either
or
. For example,
is a chain of subsets of
.
False. There exists an uncountable such chain.
1 Reason
It's negative.
2 Comment
Of course, this requires accepting that
.

4 Reason
Consider the empty relation
on
, i.e.
.



5 Reason


6 Proof
The source for this absolutely bonkers solution is not known.
Let
be the roots of
. The magic is the following:
Claim. [Black magic] For all real
, we have
with equality iff
.
Proof. Note that
It is now not hard to finish verifying the inequality and finding its equality cases. 
With the claim, the proof is straightforward: note that by the claim applied to each
,
On the other hand, since
has no integer root, we know
.
Thus equality holds everywhere. In particular, equality holds in the claim for each
, so
for all
. The end is simple: there need to be an equal amount of each for
to have integer coefficients, so
.
Let


Claim. [Black magic] For all real
![$x \in [0, 3]$](http://latex.artofproblemsolving.com/7/7/a/77ac5e3365959039da8431fe01d976f9d70a5b21.png)
![\[ |x(x-1)(x-2)(x-3)| \le 1, \]](http://latex.artofproblemsolving.com/1/9/0/190ad09c14bd9974119d576b739b213b492fd5ba.png)

Proof. Note that
![\[ x(x-1)(x-2)(x-3) = (x^2-3x)(x^2-3x+2) = (x^2-3x+1)^2 - 1. \]](http://latex.artofproblemsolving.com/6/f/1/6f1511bb1ca13cbc4bdba1e0f15b6612458b9f06.png)

With the claim, the proof is straightforward: note that by the claim applied to each

![\[ |Q(0)Q(1)Q(2)Q(3)| = \prod_{i=1}^n |x_i(x_i-1)(x_i-2)(x_i-3)| \le 1. \]](http://latex.artofproblemsolving.com/e/a/a/eaa00d4bd4b9f355cce6746773e1d1490dfaf35d.png)


Thus equality holds everywhere. In particular, equality holds in the claim for each





7 Hint
Instead of
, consider
.


7 Another Hint
Dedekind Cuts. That is, consider sets of the form
for
.


Feel free to share others in the comments.
This post has been edited 4 times. Last edited by greenturtle3141, Dec 19, 2021, 8:45 PM