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42m
comment Is category theory ambiguous? or it just is the case for beginners?
The question should never be at the end. The question should be at the beginning. Tell us what your question is, without all the preamble. Lots of people read these questions, and want to know quickly whether they can help you. What happens when they can't figure out the question? They either stop reading, and you fail to get help from people who might be able to, or they plow through all the nonsense only to find at the end that they cannot help you. Ask your question first. Help people help you. Your first two paragraphs are inessential.
54m
comment Is category theory ambiguous? or it just is the case for beginners?
If you could get to the question rather than judgements about the writing. State the thing that is confusing you, and why it is confusing you. For example: "He seems to be too wordy and honestly speaking, I could not realize/rewrite this principle, nor the notions [such as the "language of category theory" (here with objects contained), or "dual statement" of a statement and the way we construct it] in a rigorous formal way. However, intuitively speaking or by using diagrams, I could understand what he is claiming." That give us no idea what you are confused about. Be specific.
11h
comment Completing the sequence, is answer $98$ or $99$?
Note that in all cases, $a-b=1$, so you could have just said $a^2-2$. \ Or it could be $a*b+b-1$, then the answer is $98$ again. But these sorts of problems are ill-defined at best - there are often multiple possible answers.
11h
comment Showing that $\mu$ is a measure when continuous from above
Wait, you want to prove a measure is a measure?
11h
comment Showing that $\mu$ is a measure when continuous from above
What is a "measurable"?
11h
revised Showing that $\mu$ is a measure when continuous from above
edited title
12h
comment Suppose $a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.
It's somewhat sloppy, language-wise, but it is basically correct.
12h
answered On the relationship between $\phi(n)$ and $\sigma( n)$
12h
comment Suppose $a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1+a_n)$ converges if and only if $\sum_{n=1}^\infty a_n<\infty$.
Hint: You haven't used yet that $\log(1+a_n)>\frac{a_n}{1+a_n}$. Show that if $\sum \frac{a_n}{1+a_n}$ converges then $\sum a_n$ converges.
12h
comment Prove that 10101…10101 is NOT a prime.
Specifically, $10^{2017}-1$ is divisible by $9$ and $10^{2017}+1$ is divisible by $11$. So this factorization can be written as $(111\cdots 1)\cdot (909090\dots91)$.
13h
comment Prove that 10101…10101 is NOT a prime.
@ChadShin 4034 in the exponent.
13h
comment What is the the integral of $\sqrt{x^a + b}$?
There's almost certainly no close-form answer for general $a$.
15h
comment Is there a short proof of the existence of $a$ so that $a$ is a primitive root for infinitely many primes $p$?
My comment wasn't intended for you, but as a response to a now-deleted comment from another user who thought you were asking about Artin's full conjecture.
16h
comment Is there a short proof of the existence of $a$ so that $a$ is a primitive root for infinitely many primes $p$?
For example, that Wikipedia link says: "For example, it follows from the theorem of Heath-Brown that one or more of 3, 5, and 7 is a primitive root modulo p for infinitely many p." @ajotatxe
16h
comment Is there a short proof of the existence of $a$ so that $a$ is a primitive root for infinitely many primes $p$?
Actually, this isn't Artin. @ajotatxe This question is asking if there exists an $a$ which is a primitive root of unity. The partial results for Artin prove that there are.. But the existence of one $a$ of this sort is much weaker than Artin.
18h
comment Prove that if $\sum a_n$ converges, then $na_n \to 0$.
How do you get $\leq 2a_{n+1}$? It's not possible for $2(a_{n+1}+\text{ positive values})\leq 2a_{n+1}$.
18h
comment Can sum of rationals be irrational?
The key thing to note is that an infinite sum is not just "a sum." It is a limit of a bunch of sums. Lots of things you expect from finite sums don't work in infinite sums - re-arrange an infinite sum, for example, and you get different results.
19h
comment Can sum of rationals be irrational?
Induction doesn't apply to infinite sums. You know by induction that any finite sum of rationals is a rational.
20h
revised Proof Verification: If $x$ is a nonnegative real number, then $\big[\sqrt{[x]}\big] = \big[\sqrt{x}\big]$
added 472 characters in body
21h
revised Proof Verification: If $x$ is a nonnegative real number, then $\big[\sqrt{[x]}\big] = \big[\sqrt{x}\big]$
added 296 characters in body