3,569 reputation
936
bio website bu.edu/sed/about-us/faculty/…
location Boston University, SED
age 28
visits member for 2 years, 3 months
seen 6 hours ago

Profile for Benjamin Dickman on Stack Exchange: MESE, MSE, and MO

Postdoctoral Fellow, Mathematics Education, Boston University SED.

Ph.D., Mathematics Education, Columbia University. 2014. NSFGRFP.

Dissertation: Conceptions of Creativity in Elementary School Mathematical Problem Posing. (Relevant MESE post here.)

M.Phil., Mathematics Education, Columbia University. 2014.

Fulbright Grant, Mathematics/Mathematics Education, Nanjing Normal University. 2008-2009.

Research Topic: High School Mathematics Teacher Training in China.

B.A., Mathematics, Amherst College. 2008. Honors Program.

Undergraduate Thesis: On a Theorem of Dwork. (p-adic proof of the rationality part of the Weil Conjectures; relevant MSE post here.)


Have you solved any of my puzzles? If so, send me an email!
Email: bdickman[at]bu。edu


Nov
14
revised Possible values of $\gcd(a+b, a\times b)$
Cleaned up presentation now that an answer has been given.
Nov
13
comment Possible values of $\gcd(a+b, a\times b)$
This is great; thanks!
Nov
13
accepted Possible values of $\gcd(a+b, a\times b)$
Nov
13
revised Possible values of $\gcd(a+b, a\times b)$
Corrected after answer!
Nov
13
revised Possible values of $\gcd(a+b, a\times b)$
rolled back to a previous revision
Nov
13
revised Possible values of $\gcd(a+b, a\times b)$
Conjecture 1 removed thanks to D. Fischer; too unwieldy too leave up as a vestige.
Nov
13
comment Possible values of $\gcd(a+b, a\times b)$
@DanielFischer Ah, I had miscalculated; I had thought $a = 6$ and $b = 30$ worked, but I see now that $\gcd(+, \times) = 36$ (and not, as I had thought, $6$). So you have found the clear error that I worried existed in the latter conjecture; thank you. Do you see how to resolve the initial question?
Nov
13
comment Possible values of $\gcd(a+b, a\times b)$
@DanielFischer Okay; working out an example. Let $N = 3$; let $a = 3 \cdot 2$ and $b = 3 \cdot 5$. Then $\gcd(2, 5) = 1$ (and $\gcd(6, 15) = 3$ as desired). Now $\gcd(7, 21) = 7$. I have the sense that I'm overlooking something obvious, but I do not see the issue with the above. For $N = 3$, there are certainly $a,b$ such that $\gcd(+, \times) = 3, 6, 9$. Are you indicating other values can be taken on, too? (More generally: Should the answer to the main question be clear? I do not see it...)
Nov
13
asked Possible values of $\gcd(a+b, a\times b)$
Oct
29
revised Metrics and the Kuratowski closure axioms
Request for alternative proofs and a corresponding tag
Oct
26
comment Metrics and the Kuratowski closure axioms
@user87690 Thanks for the comments and answer; it is quite nice to see a new way of looking at a problem, and not only the new ways of thinking that come with it, but also the new vocabulary.
Oct
26
accepted Metrics and the Kuratowski closure axioms
Oct
25
comment Metrics and the Kuratowski closure axioms
This looks great; I will have to read up on how to use one-point compactification arguments (specifically in the context of discrete spaces). You have definitely satisfied the request for parsimony! I will give the question a bit more time before accepting in case there are alternate answers; I have posted my own approaches to (ii) and (iii) as well.
Oct
25
answered Metrics and the Kuratowski closure axioms
Oct
25
asked Metrics and the Kuratowski closure axioms
Sep
30
awarded  Explainer
Aug
31
revised Frechet differentiable implies reflexive?
xpost to MO link
Aug
31
answered Frechet differentiable implies reflexive?
Aug
28
revised How to divide two inequalities
Typo in title corrected
Aug
28
answered How to divide two inequalities