bio | website | math.berkeley.edu/~sramesh |
---|---|---|
location | Berkeley, CA | |
age | 29 | |
visits | member for | 4 years, 10 months |
seen | 2 days ago | |
stats | profile views | 3,635 |
I was a graduate student in the Logic program at Berkeley, broadly interested in categorical logic and foundations of mathematics, as well as in applications of category theory to the semantics of programming languages. I work for Google now.
Jun 9 |
comment |
What are good non-English languages for mathematicians to know?
Knowing some programming language well is probably useful. But C, specifically, needn't be it... |
Jun 1 |
comment |
Is complement of LL(k) grammar context free?
en.wikipedia.org/wiki/LL_parser |
May 23 |
awarded | Enlightened |
May 23 |
awarded | Nice Answer |
May 2 |
comment |
existence of a field that has a non surjective ring homomorphism
I usually leave only comments rather than answers, to bypass the silly reputation system. But for once, I thought, let me post an actual answer. And what's the result? A bunch of responses about somebody else's answer-posted-as-comment. :) |
May 2 |
comment |
existence of a field that has a non surjective ring homomorphism
Ah, good point! I should have thought of that. I kept implicitly thinking $\mathbb{R}$ was algebraically closed... |
May 2 |
revised |
existence of a field that has a non surjective ring homomorphism
D'oh! |
May 2 |
answered | existence of a field that has a non surjective ring homomorphism |
Apr 9 |
comment |
Why the underlying function of a monomorphism may not be an injection
What more are you looking for, beyond "A generalization may not always behave exactly the same as the thing it generalized, in all respects"? For what it's worth, in the quite common case of a concrete category in which the underlying set functor is representable, (i.e., in which there is a free object on one element), all monomorphisms will be injections. In some sense, the failure of monomorphisms to be injections more generally is just the failure of the underlying set functor to always be representable. |
Feb 19 |
comment |
Nontrivial question about fibonacci numbers?
Tilings using 1x1 and 1x2 tiles? Bah! It's a direct observation that the "right parents" of each diagonal comprise the previous diagonal, while the "left parents" comprise the twice-previous diagonal. |
Feb 12 |
comment |
Covering the Rationals — A Paradox?
So the problem is mainly your second bullet point, but it does not involve a new uncountable infinity (and what would it mean to be an uncountable infinity smaller than $\aleph_0$?). You are simply wrong in supposing that the number of gaps will be less than the number of rationals; you have no means of constructing a partial surjection from the latter to the former. |
Feb 12 |
comment |
Covering the Rationals — A Paradox?
Indeed, instead of using mini-intervals, we might imagine removing single points: removing one rational at a time from [0, 1], we end up with n + 1 many connected components left after the first n many rationals have been removed. But after removing every rational, we are not left with a countable collection of connected components; instead, we are left with an uncountable collection of single points (the irrationals). |
Feb 12 |
comment |
Covering the Rationals — A Paradox?
It is not true that the number of gaps must be countable. Your argument is simply "The number of gaps when the first n mini-intervals have been placed is <= n + 1; therefore, the number of gaps when all the mini-intervals have been placed is countable". But this argument is fallacious. |
Feb 9 |
awarded | Yearling |
Jan 13 |
comment |
Direct proof of irrationality?
Slight clarification: POSITIVE integers are products of powers of primes in which the exponents are natural numbers, and POSITIVE rational numbers are products of powers of primes in which the exponents can be any integer. |
Jan 13 |
comment |
Any example of a non-strong monad?
This interpretation will produce a non-strong monad on a CCC, if you add in all the rest of the rules of the simply-typed lambda calculus with pairs. Indeed, this produces the free monad on a CCC. |
Jan 11 |
comment |
Any example of a non-strong monad?
Oh, whoops, nevermind A); apparently, it was corrected 5 minutes before I said it. |
Jan 11 |
comment |
Any example of a non-strong monad?
More to the point, perhaps, depending on how you think about monads in functional programming: for a non-strong monad, one does not have a function bind : m a -> (a -> m b) -> m b in the programming language. Rather, there will be a method of taking any function of type a -> m b and turning it into a function of type m a -> m b, but this method is not carried out by any higher-order function. |
Jan 11 |
comment |
Any example of a non-strong monad?
In functional programming, what this amounts to is that you will have to look outside the monads/functors for which one has a function map : (a -> b) -> (m a -> m b). There will in fact be a way (external to the programming language) to take any function of type a -> b and turn it into the corresponding function of type m a -> m b, but there will not be any higher-order function (internal to the programming language) which does this for you. [This is the meaning, in this context, of Finn's statement that strong monads are those which respect the internal hom's enrichment] |
Jan 11 |
comment |
Any example of a non-strong monad?
A) How can the initial monoidal category equipped with a monad be empty? Surely, it must have an identity object for the monoidal structure. B) This is nice, as a free example with monoidal structure, but the question-asker seemed to want specifically cartesian product structure. |