Reputation
45,228
Next tag badge:
373/400 score
110/80 answers
Badges
4 59 130
Newest
 terminology
Impact
~576k people reached

1h
comment Why does mathematics fits so well with reality?
I'm not entirely sure your question is sufficiently pin-pointed (you seem to talk of applicability and the logic as somehow interchangeable), but regardless, it would be quite astonishing if using any system (symbolic or whatever) which is capable of complex manipulations, one would not arrive at something useful in the real world after centuries upon centuries of continuous refinement and the weeding out of the useless stuff in the system. Thus, it is not surprising at all that mathematics is useful. It is in fact inevitable.
Apr
17
comment Need help with proof of existence of $\sqrt{2}$
if $q\ge 2$, then $qq\ge 2q\ge 2\cdot 2=4>2$.
Apr
17
comment Need help with proof of existence of $\sqrt{2}$
and what precisely $f(x)=\sqrt x$ is then? You are trying to argue $\sqrt 2$ exists by arguing that $\sqrt x$ exists for which $x$ exactly?
Apr
17
comment Need help with proof of existence of $\sqrt{2}$
You seem to already assume quit a lot about $\sqrt 2$....
Apr
16
comment What are some illustrative examples that demonstrate how $\succ$ can differ in behavior from $>$ and/or $\geq$?
note that in the reference the well-below relation is the classical one from domain theory, i.e., it talks about directed sets. Flagg's notion is slightly more general.
Apr
15
comment Looking for info on power set functor
and I suggest again: you may be interested in topos theory.
Apr
14
comment Are clopen sets Borel?
you state yourself that open sets are Borel. So certainly clopen sets are Borel.
Apr
12
comment Proving the set identity $(X \cup Y) = X + Y - (X \cap Y)$
You need to be precise about how you model sets with repetition, and then how you define the set operations, and the meaning of equality. Before you do that, the question is too vague (though certainly there is a grain of truth to it).
Apr
12
comment if f:[0,1]$\to$ $\Bbb R$ is continuous and has only rational [respectively, irrational] values, must f be constant? Prove your assertion.
hint: intermediate value theorem + density = qed.
Apr
11
comment ISO information on powerset functor
You may be interested in topos theory. But your question is so vague it's hard to answer.
Apr
11
comment Deriving the value of $\pi$ from a dart board
You can run a one-line R code that will simulate this approach. You'll find that you need about 100,000 darts to get $\pi$ correct to about 2 decimal places. This method is quite inefficient, though cool.
Apr
10
comment Can we get categories like $\mathbf{TopGrp}$ as some kind of a pullback?
sorry, I was on auto-pilot. For abelian groups this works because finite products agree with finite coproducts, but now change from abelian groups to groups and it stops working since the coproduct is much larger than the product.
Apr
10
comment Why is delta used to describe difference between two entities
$\Delta$ifference.
Apr
10
comment Can we get categories like $\mathbf{TopGrp}$ as some kind of a pullback?
@goblin consider $F:Set\to Ab$, the free abelian group functor, left adjoint to $U$. Now consider both categories as multicategories via the cartesian product. Then, denoting $*$ any singleton set,$Set(*, *; UA)\cong UA$ for any abelian group $A$ while $Ab(F*, F*;A)$ is considerably different, so there isn't even a bijection, let alone a natural isomorphism. This happens for many other free construction, so this definition, while it logically makes sense, suffers from a severe lack of examples.
Apr
10
comment Proof of infinite primes
@MurtuzaVadharia you have been given a value for $n$ such that neither is prime. What else do you want?
Apr
10
comment Proof of infinite primes
see the comments to your question.
Apr
10
comment Proof of infinite primes
you are mistaken.
Apr
10
comment Proof of infinite primes
@MurtuzaVadharia that would only prove there are infinitely many primes if you can show that any number of the form 6n+1 or 6n-1 is prime. This is of course not the claim.
Apr
9
comment Can this be shown: $\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\dots}}} = \sqrt a$?
As a first step, try to define the sequence rigorously, i.e., give the explicit recursive definition. Then you should be able to show the sequence is monotonic, bounded, and then figure out which equation does the limit satisfy.
Apr
8
comment functors with a morphism lifting property
@ZhenLin yes ! :) the unique lifting property is precisely a discrete fibrations (so thanks already). What is the lifting is not unique?