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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 Isomorphism under integer multiplication
@lulu Multiplication isn’t in the cards for $G_1$ either, since all elements in it (except $1$) are zero divisors modulo $18$ and therefore not multiplicatively invertible.
Jul
23
comment Isomorphism under integer multiplication
@A.P. (1) Why would they need to be abelian, (2) the task is to find an isomorphism, that is to explicitly write one down, I presume.
Jul
23
comment Isomorphism under integer multiplication
You have written down sets $G_1$, $G_2$ and implied they ought to be groups. But they are not so in an obvious manner. What do you consider to be the neutral elements and the group operations for them?
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
20
revised Showing $\dim T_0 ℝ^n = n$ using a derivation definition for the tangent space.
update, typo fixed
Jul
19
accepted Are there noncontinuous derivations $C^1(X) → ℝ$?
Jul
19
revised Are there noncontinuous derivations $C^1(X) → ℝ$?
edited body
Jul
19
revised Are there noncontinuous derivations $C^1(X) → ℝ$?
added 212 characters in body
Jul
19
revised Are there noncontinuous derivations $C^1(X) → ℝ$?
added 212 characters in body
Jul
19
asked Are there noncontinuous derivations $C^1(X) → ℝ$?
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.