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1d
revised Is $2^{\aleph_0} = \aleph_1$?
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1d
answered Is $2^{\aleph_0} = \aleph_1$?
1d
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
Not sure I understand all you say. But if you want to prove a total amicable tuple cannot exist for $m>1$, be sure to use the "totality" in a crucial way. Because plenty of amicable tuples are known. If for example you could prove that for every non-trivial amicable tuple there exist a "new" integer which is not in the tuple, and whose $\sigma$ value is also $n$, it would suffice. Is there a way to construct such a witness for non-totality?
1d
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
Since neither of the two late "answers" really gives the answer, I shall award the bounty to this one.
1d
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
Similarly, for an amicable triple like $(k_1,k_2,k_3)=(1980, 2016, 2556)$ with $k_1<k_2<k_3$ we have $I(k_1)>I(k_2)>I(k_3)$. So here on (harmonic) average the abundancy of the three $k_i$ is $3$. So we could have that all three are abundant. (Or maybe the first two, $k_1$ and $k_2$, are very abundant, and the last one, $k_3$, is perfect/deficient, in a way so that the average abundancy I describe is still $3$.) Again this example triple is not "total". For an example of an amicable (not total) quadruple ("average" abundancy is $4$), check $(k_1,k_2,k_3,k_4)=(3270960,3361680,3461040,3834000)$.
1d
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
What you denote by $I(k)$ I call the abundancy (index) of $k$. If $(k_1,k_2)$ is an amicable pair, and we agree that it is sorted ascendingly by $k$, so $k_1<k_2$, then clearly the abundancy is descending, $I(k_1)>I(k_2)$. Since $k_1+k_2=n$, we have $\frac{k_1}{n}+\frac{k_2}{n}=1$ or $\frac12 ( \frac{k_1}{n}+\frac{k_2}{n} ) =\frac12$, so on (harmonic) average the abundancy $\frac{n}{k}$ is $2$. So $k_1$ is abundant and $k_2$ is deficient. Example: $(k_1,k_2)=(220,284)$. However this pair is not "total" by my definition from the question.
1d
comment Caden has 4/3 kg of sand which fills 2/3 ​​ of his bucket. How many buckets will 1kg sand fill?
This is a division question, specifically $\frac{\frac23}{\frac43} = \frac12$.
2d
comment Proving a strictly decreasing sequence which tends to zero is positive
Exactly. And I think your proof is incomplete because it does not include the zero case.
2d
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
@boldbrandywine I am not sure that was my exact reason. I guess one could have chosen modulo $10$ or something else also. I just wanted to illustrate that $\Sigma_n$ is more often large when $n$ has many (small) divisors (it seems). I chose $6$ because it is not too huge for a table, and because it is the product of the first two prime numbers. First I considered quoting odd and even $n$ only, but then I decided for a bit more detail.
2d
comment Proving a strictly decreasing sequence which tends to zero is positive
You suppose an $a_N$ is negative. But what if one is zero? Do you consider zero a positive number? In the usual terminology (in English at least) zero is not positive. That is what I meant. The opposite of "positive" is "negative or zero".
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comment Why 1 is not considered to be a prime number?
This answer almost leads one to ask: Why is 0 not a prime number? (thread from 2010 Aug 31).
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awarded  Good Question
2d
comment Volume of a parallelepiped, given 8 vertices
It is natural to focus on the vertex at $(0,0,0)$. What are the three other vertices whose squared distances from $(0,0,0)$ are smallest? Like @SimonS says, your fourth point has $3^2+5^2+1^2$ which is not on the "top-3" of smallest squared distances. Addition: Wait, for some very long and slim parallelepipeds it may happen that the surface diagonal on the tiny side on the end is shorter than the long edge.
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comment Proving a strictly decreasing sequence which tends to zero is positive
Are you not asked to prove $(a_n)$ is strictly positive?
Apr
12
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
@AleksVlasev In how many of those cases is $n-1$ a prime number? Because in that case the value could arise from just one numer, $A_n=\{ n-1 \}$. Of course, for a prime $p$, we have $\sigma(p)=p+1$.
Apr
12
comment a measurable set intersect a compact set
@MAM Of course this assumes you are allowed to use that fact at your current "level". If you have not yet proved this about the measurable, but are in the process of doing it, the argument could be circular.
Apr
12
revised a measurable set intersect a compact set
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Apr
12
answered a measurable set intersect a compact set
Apr
10
comment Is there a relationship between trigonometric functions and their “co” functions?
To elaborate on @SteveJessop's comment (see Wikipedia Right-angled triangle definitions): We can first define sine as $\frac{\text{opposite}}{\text{hypotenuse}}$, define tangent as $\frac{\text{opposite}}{\text{adjacent}}$, and secant as $\frac{\text{hypotenuse}}{\text{adjacent}}$. After that we can define the three "co-" functions by exchanging the words "opposite" and "adjacent". As Steve says, going "co-" corresponds to shifting to the other acute angle in the right triangle.
Apr
9
comment A number $n$ which is the sum of all numbers $k$ with $\sigma(k)=n$?
@celtschk This is the sum of all divisors of $k$, called $\sigma_1$ in this Wikipedia article. As an example, $\sigma(54)=1+2+3+6+9+18+27+54=120$.