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1h
comment How to prove the limit exists for function of two variables?
The first reasoning is correct, well done. The second (Squeeze) has an easily fixed flaw at points $(0,y)$.
9h
comment Elementary Substitution in Solving Equations - Why it works
When we square, or take the sine, or make some other not one to one transformation, verification that we have not introduced an extraneous root becomes necessary.
11h
awarded  computability
11h
answered If a professor has 7 students and they have to at least do 2 assignments each…
11h
revised The existential theory is undecidable
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13h
answered The existential theory is undecidable
16h
comment Proving $10\cdot n=0$ for all $n\in\mathbb{Z}$ with $n\geq 0$ using strong induction
The step from $0$ to $1$.
16h
revised Calculus: Finding Arc Length--Squaring the Derivative Where did the -1/2 come from?
added 21 characters in body
16h
revised Calculus: Finding Arc Length--Squaring the Derivative Where did the -1/2 come from?
added 268 characters in body
16h
answered Calculus: Finding Arc Length--Squaring the Derivative Where did the -1/2 come from?
17h
comment The existential theory is undecidable
Because of my background, this stuff is easy for me, a standard reduction to Hilbert's 10th. I had noticed it but have no time for a while. Am bookmarking, and if no one else has given a full answer by the time I get to it, I will write out an answer.
22h
comment Sum of remainders of $2^n$
We can start the cycle for $125$ anywhere we like, such as $1$. But modulo $1000$ things are different, because $0,1,2$ are special modulo $8$. Starting at any $k\ge 3$ will work modulo $1000$.
22h
comment Sum of remainders of $2^n$
We can start anywhere. Traditional is $k=1$.
22h
comment Sum of remainders of $2^n$
Length $6=\varphi(7)$.
23h
comment Sum of remainders of $2^n$
Every introductory number theory book has it. You can also search, say starting with Wikipedia. Comments are not a useful place to describe the notion to someone who has not met it.
23h
comment Sum of remainders of $2^n$
$150$ does not have a primitive root.
23h
comment Sum of remainders of $2^n$
The powers later on repeat. Since $2$ is a primitive root of $125$, the powers $2^3$ to $2^{102}$ are distinct modulo $125$, and hence modulo $1000$. With the powers of $2$ method, we are summing the geometric series $2^3$ to $2^{102}$. By the usual formula the sum is $S=\frac{2^3(2^{100}-1)}{2-1}$. Since $2^{100}\equiv 1\pmod{125}$, $S$ is divisible by $1000$, so has $0$ remainder.
23h
comment Does a mathematical construct exists which explains all theories?
Set theories, such as ZFC, come fairly close to being comprehensive enough that most mathematical objects can be "constructed" within them.
1d
comment Proof that the sum does not depend on enumeration …
The argument only works for $I$ a countable set, we have used in a fundamental way the definition of "sum of a sequence" as the limit of the partial sums.
1d
comment Sum of remainders of $2^n$
For every $a$ relatively prime to $125$, the congruence $2^n \equiv a\pmod{125}$ has a solution, and therefore for every $b$ divisible by $8$ but not by $5$, the congruence $2^n\equiv b\pmod{1000}$ has a solution. For other $b$ there is no solution. Now note that $b$ is divisible by $8$ but not by $5$ iff $1000-b$ is divisible by $8$ but not by $5$. Now $b$ and $1000-b$ cancel modulo $1000$. It follows that the sum of the numbers between $1$ and $999$ that are congruent to some $2^n$ ($n\ge 3$) is divisible by $1000$. (Basically, a pairing argument.)