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15h
comment Is the null set $\emptyset$ a real subset of any set?
What language is your textbook in? Surely not English? As Hagen says, the correct term is proper subset.
1d
comment Help Understanding Gradients
That's better!$ $
1d
comment Help Understanding Gradients
This function is continuous at $(0,0)$.
1d
comment Is there a difference between discount and VAT calculations?
The second one, for £0.80.
2d
comment Riddle: Minimum time to cross the bridge?
But you're guessing! There are two probably's in there.
Apr
11
comment Infinite Union of Sets with Content Zero
$\cup_{x\in\mathbb R}\{x\} = \mathbb R$. This answers one of your interleaved questions.
Apr
11
comment Interesting horserace counting problem
This is sequence A000670 in the On-line Encyclopedia of Integer Sequences. There its elements are called the Fubini numbers. I don't think there is a closed form for this.
Apr
10
comment What is the negation of the statement “Every odd nmber is divisible by 2”.
@MichaelChirico: That's not right. The negation of $P \implies Q$ is not $\lnot P \implies \lnot Q$.
Apr
10
comment $a^2\equiv1 \pmod n$ iff $a\equiv\pm\,1\pmod p$ for all $p\mid n$
@Rellek: The question specifically mentions "primes dividing $n$". So $n$ itself obviously doesn't have to be prime.
Apr
10
comment $a^2\equiv1 \pmod n$ iff $a\equiv\pm\,1\pmod p$ for all $p\mid n$
It's not clear which direction you are trying to prove here, especially as your conclusion is $p\equiv\pm 1 \mod p$. In any case, the conjecture is false $-$ see my comment to the question.
Apr
10
comment $a^2\equiv1 \pmod n$ iff $a\equiv\pm\,1\pmod p$ for all $p\mid n$
This is wrong. Take $n=9$ and $a=4$.
Apr
10
revised $a^2\equiv1 \pmod n$ iff $a\equiv\pm\,1\pmod p$ for all $p\mid n$
deleted 6 characters in body
Apr
8
comment If $a_7 = 120$ then find $a_8$.
@tone: In answer to your question "What is the general process?", I would say that I have never come across such a problem before!
Apr
8
comment If $a_7 = 120$ then find $a_8$.
@achillehui: Yes, of course. My bad. But then the sequence is not increasing. (I have edited my comment to incorporate your correction.)
Apr
8
comment If $a_7 = 120$ then find $a_8$.
@achillehui: I agree with your 194 ($a_1=8,a_2=10$). But I think 193 is wrong: $193,120,73,47,26,21,5$ gives $a_1=16$. So 194 is the unique solution in strictly positive integers.
Apr
8
comment If $a_7 = 120$ then find $a_8$.
@tone: By "numbers", do you mean "positive integers"? If so, then you should say so.
Apr
8
comment If $a_7 = 120$ then find $a_8$.
@PrasunBiswas: That is not the problem here. The problem is hitting 120 exactly.
Apr
8
comment Any subfield of $\mathbb{C}$ must contain every rational number
As far as I can see, you haven't used the fact that your $q$ is rational. So how can the proof be correct?
Apr
8
comment Accumulation point(s) of $\mathbb{R} \setminus \mathbb{N}$ in $\mathbb{R}$
I think points of $\mathbb N$ should also be included, according to this definition from the on-line Encyclopedia of Mathematics.
Apr
7
comment How do we identify twin primes .
"As we know each prime number...": Each prime number greater than three.