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comment Can anyone help me answering this ? :(
Find a good title for this question and show us your thoughts.
Jul
23
comment Is there a way to show the following equality without induction
Nah, apparently it is what the questioner wants, as she or he accepted xpaul’s answer. Not sure how the site’s policy is on duplicate answers, I’d just leave it until someone complains, possibly.
Jul
23
comment Is there a way to show the following equality without induction
@SimonS Ok, that seems like a legit interpretation. One should nevertheless state that induction is still needed to prove the result when collapsing the telescope sum.
Jul
23
comment Is there a way to show the following equality without induction
You’re like the fifth person to answer that.
Jul
23
comment Is there a way to show the following equality without induction
By the use of “$\cdots$”, you are implicitly carrying out induction.
Jul
23
comment Is there a way to show the following equality without induction
I’m actually not even sure if it is possible to do that without induction, because finite sums are defined inductively. Maybe if you can give a non-inductive definition of finite sums …?
Jul
23
comment Is there a way to show the following equality without induction
But “$\ldots$” is hiding induction.
Jul
23
comment Is there a way to show the following equality without induction
To all the hints: Please note that using telescope sums just postpones the use of induction to showing that $\sum_{k=1}^n (a_{k+1} - a_k) = a_{n+1} - a_1$ (unless, of course, you have a non-induction way of showing that). I think it would be nice to have an answer which doesn’t suggest hiding the induction, but really has an idea of avoiding it.
Jul
23
comment Why there is no sign of logic symbols in mathematical texts?
Actually, you are joking right? I was falling for it, right?
Jul
23
comment Why there is no sign of logic symbols in mathematical texts?
@MKR The “logical” version of you saying you washed your hand doesn’t make any sense to me – even when knowing it should express you having washed your hands. Saying “$∃t_1 \colon f(t_1) → ∞$” doesn’t make any sense. Also I think that should rather read “$∃t_2 > t_1$” instead of “$∃t_2 ≥ 0$” (If $t_2$ should be the time when you wash your hands). Also there’s more and less to washing hands that getting rid of bacteria. Bottomline: The second version is definitely not more elegant than the first one, in fact it’s absolutely awful.
Jul
19
comment Showing $\dim T_0 ℝ^n = n$ using a derivation definition for the tangent space.
Found out: If one only considers continuous derivations, it follows from Stone–Weierstraß without any further analysis.
Jul
14
comment In how many ways can a natural number be written as a sum of $2$ natural numbers?
I don’t think it’s a good idea to answer this question without letting the questioner think about it a bit more.
Jun
29
comment A good book on humankind’s understanding of primes?
Thank you for your recommendation. The book seems to be on freshman level, though, and only covers the Riemann hypothesis as you already stated. Anyway, thanks for bringing it up!
May
31
comment Show that a map $f:X\to Y$ is onto iff $f(f^{-1}(C))=C$ for all subsets $C\subseteq Y$.
@Andrew You mean “$f(x) ∈ C$”. Other than that, the preimage of $C$ under $f$ is defined as $f^{-1}(C) = \{x ∈ X;~f(x) ∈ C\}$ and can be defined for any map $f$ – it doesn’t need to be surjective. (Maybe you thought of $f^{-1}(C)$ as the image of an inverse $f^{-1}$ of $f$ which would only exist if $f$ was already bijective. But that’s not how the symbol “$f^{-1}$” is used in this case.)
May
29
comment Why is the cartesian product so categorically robust?
I don’t know all of these categories, but – at least as far as I know – for most of them the corresponding forgetful functor has a left adjoint, thus is continuous. Morally speaking, I’d guess that so many of these forgetful functors have left adjoints because mathematicians (for historic reasons alone) tend to build stuff from sets and that already suggests a forgetful functor as well as an left adjoint for it.
May
28
comment How to not feel bad for doing math?
@AsafKaragila Why not? It’s a real concern to some and I guess it can be a real mood killer if you’re considering spending 3–5 years working hard on one particular problem/question. Even though I don’t pursue a career as a mathematician, I can relate to this concern. Why not take it seriously?
May
28
comment How to not feel bad for doing math?
@Did I added a comma and read the sentence as “It’s the only job I can think of, which doesn’t contribute to bettering the world somehow.”.
May
23
comment How to insert Gothic letters in Word?
That’s not a math question, though.
May
21
comment Proving that the ball is converx
Invoke the definition of convexity and the triangle inequality.
May
15
comment If a field extension contains a cyclotomic extension is it solvable?
Recall that a solvable group requires a subnormal series in which all factors are abelian. What can you really say about $\mathrm{Gal}(L/M)$? As an extreme case, take any non-solvable extension $L/K$ and choose $M = K$, which trivially is a cyclotomic extension.