Mathematical Quickies
Maintained by Boaz Tsaban
Mathematical quickies are questions in the level of
mathematics college or university students.
These questions where emailed as an experiment to
the teaching assistants at BIU Math&CS dept.
Solutions, as well as
the names of the first to solve appear after the questions.
(All questions were also solved by me, but I do not count that.)
The quickies are sorted backwards, so that the first quickie you
see is the last one posted.
Quickie number 3
Proposed by Avital Oliver
Assume that A is a nilpotent matrix.
Prove that for each two nonzero matrices B and C,
ABC is not equal to B.
Quickie number 2
Proposed by Misha Sklarz
Prove that it is impossible to write the "mod" function using only
"elementary" algebraic operations: +, -, *, /. As a matter of fact, even
f(x)=x mod 2 cannot be written that way. Same goes for the min and max
functions.
Solution
Any such function is of the form f(x)/g(x) for polynomials f(x),g(x), and
therefore has only finitely many roots. But the mod function has infinitely
many roots.
Solved by:
Avi Mintz, Ohad Lypski.
Quickie number 1
Proposed by Eli Bagnoe
Let P be a point in the plane (RxR).
Prove that there exists a set D of (straight) lines
going through the point P in the plane, which satisfy:
(a) The angle between each two lines in D (measured in degrees, say) is
irrational.
(b) Any line through P which does not belong to D has a rational angle
to some line in D.
Solutions
Avital Oliver's solution.
Look at the group R/Z. Now, R/Z represents the possible angles (the angle x
\in R/Z represents 360x degrees). The question is now specified:
Now, it is trivial to build such a set, and actually to understand all these
possible sets.
Observe Q/Z, which is a subgroup of R/Z. Now, a set A \subset R/Z represents
a solution iff A is a set of coset representatives of Q/Z in R/Z (since Q/Z
is normal in R/Z, of course). Therefore, there are many such sets.
Another solution.
Immediate application of Zorn's lemma.
This solution is less "constructive", that is, both solutions
use the Axiom of Choice, but in the first one we see what actually happens.
In fact, the axiom of choice is essential here since such a construction
gives rise (as in Oliver's solution)
to a subset of the unit interval which is not Lebesgue measurable, and it is well
known that one cannot prove the existence of such a set without any form
of the axiom of choice.
Also solved by:
Misha Sklarz, Avi Mintz, Michael Shynar, Ohad Lipsky.
Links to other problem sites (under construction)
David
Molnar's Discrete Mathematics problem page
Boaz Tsaban
Last modified: Wed Sep 11 15:41:02 IDT 2002