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July 16 edit

Solitary numbers edit

Why is 18 a solitary number even though (18, σ(18)) ≠ 1 ? Same for 45, 48, etc. Thanks, --Think Fast (talk) 01:51, 16 July 2008 (UTC)[reply]

According to Friendly number, a number that is coprime to the sum of its divisors is solitary, but the converse is not true, 18 being a counterexample. --Tango (talk) 01:57, 16 July 2008 (UTC)[reply]

I read the article. Can you explain why 18 is a counterexample? --Think Fast (talk) 02:31, 16 July 2008 (UTC)[reply]

The formulation in Friendly number seemed easy to misunderstand so I tweaked it.[1] PrimeHunter (talk) 02:39, 16 July 2008 (UTC)[reply]

The converse (which is false) would be: If a number is solitary then it's coprime to the sum of its divisors. It is this for which 18 is a counterexample: 18 is solitary (not proved here), but 18 is not coprime to its sum of divisors 39. PrimeHunter (talk) 02:46, 16 July 2008 (UTC)[reply]

This is all good, but why is 18 a solitary number? It is not prime, it is not a prime power, and (18, σ(18)) ≠ 1 (i.e., it is not coprime to the sum of its divisors). But nonetheless, it is a solitary number. Can anyone explain what makes it a solitary number even though it does not satisfy any of these criteria? --Think Fast (talk) 03:04, 16 July 2008 (UTC)[reply]

Presumably it is a solitary number because it is the only positive integer n such that σ(n)/n = 13/6. In other words, your question is "Why are there no positive numbers n (other than 18) such that the sum of their divisors is 13n/6?" or so. JackSchmidt (talk) 03:42, 16 July 2008 (UTC)[reply]

I don't know if it helps, but this was discussed in 2004 at sci.math:

We know that 2*3=6 divides x, so we know that x=2^a * 3^b * n where n is 
relatively prime to 6.  Now it is easy to see that if sigma(x)/x = 13/6 
and x != 18 then a=b=1. (Can you see why?)  Now if x=6*n with n 
relatively prime to 6 then the numerator of sigma(x)/x reduced to lowest 
terms will be even, and the denominator will be relatively prime to 6, 
in particular it will not equal 13/6.

The only other reference appears to be some sort of closed mailing list. JackSchmidt (talk) 04:01, 16 July 2008 (UTC)[reply]

Thank you for finding this. It answers my question. --Think Fast (talk) 05:30, 16 July 2008 (UTC)[reply]

Visitors from the future: If you come across this page and the proof is not clear enough, some additional details are provided here. -- Meni Rosenfeld (talk) 17:39, 20 July 2015 (UTC)[reply]

That's exactly what I wanted (although my math fu is too weak for it to make sense) Ataru (talk) 01:42, 7 January 2019 (UTC)[reply]

Divisor function edit

This isn't really a reference desk question, but I hoped one of you maths guys could check the table of values I just added to divisor function. Is it ok? (There - I managed to ask a question...) -- SGBailey (talk) 08:24, 16 July 2008 (UTC)[reply]

It doesn't contain any errors, if that's what you're asking. Algebraist 10:09, 16 July 2008 (UTC)[reply]

I added Table of divisors to Divisor function#See also. Your table lists the first 15 of 1000 entries. PrimeHunter (talk) 12:32, 16 July 2008 (UTC)[reply]

For future reference, if you want help with writing/editing/proofreading mathematics articles, the best place to go is Wikipedia talk:WikiProject Mathematics. --Trovatore (talk) 00:58, 17 July 2008 (UTC)[reply]

Numbers edit

Hello, I do not speak english well and I do not understand the next number: How much is thirty hundred thousand in number? Thank you --Humberto (talk) 14:22, 16 July 2008 (UTC)[reply]

"Thirty hundred thousand" is not how one would phrase that number correctly, so I can't blame you for not understanding it. I think that it is supposed to be 3,000,000; which is three million. Paragon12321 (talk) 14:27, 16 July 2008 (UTC)[reply]

I do not speak English well, either, however... Simple 'googling' returns lots of results:

http://www.google.com/search?q=%22thirty+hundred+thousand%22

which may indicate it is not an ordinary number, but rather a synonym for "lots of", "a large number". --CiaPan (talk) 14:42, 16 July 2008 (UTC)[reply]

It actually sounds like english as spoken by a non-perfect foreign language speaker of english, or possibly a translation from another language written in psuedo-archaic terms.. If that makes sense.

Also is 'thirty-hundred' a normal term in sanskrit or something, as many of the pages turned up by google seem to have a vedic connection?87.102.86.73 (talk) 17:01, 16 July 2008 (UTC)[reply]

Numbers in the teens are often used before "hundred" to indicated thousands, so "thirteen hundred" is the same as "one thousand three hundred". I've never heard numbers greater than 19 used in this sense, though, nor have I heard this technique used on longer numbers. « Aaron Rotenberg « Talk « 16:46, 17 July 2008 (UTC)[reply]

This may be a bad translation of "30 lakh". A lakh equals 100,000 and is a widely used term in India. --169.230.94.28 (talk) 02:56, 22 July 2008 (UTC)[reply]

Dedekind cuts and the proff that 0.999.... = 1 edit

I have a question about these two statements in the proof:

"Every element of 0.999… is less than 1, so it is an element of the real number 1."

I feel that while this states that every cut belonging to 0.999... includes values < 1 -(1/10)^n, it does not necessary exclude the value 1-(1/10)^n on each set. If that is the case, how does the proof jump to the conclusion that "Every element of 0.999… is less than 1, so it is an element of the real number 1" so immediately. Are there any missing steps?

There are some missing steps, but they are arithmetic in nature. If we want to fill in the missing steps we would say something like: What are the elements of 0.999...? Each is some rational smaller than some number of the form 0.999..99. Since the latter number is smaller than one, also the rational is smaller than one. Oded (talk) 19:01, 16 July 2008 (UTC)[reply]

The next question I have regarding this quote: "Conversely, an element of 1 is a rational number \begin{align}\tfrac{a}{b}<1\end{align}, which implies \begin{align}\tfrac{a}{b}<1-(\tfrac{1}{10})^b\end{align}"

How does the first imply the 2nd? If I test out that statement with values for a and b, it could work. But won't I have to test it infinitely to prove to myself that it does work, therefore defeating the purpose of this proof. And then the would "implies" would have a really loose meaning.

You are right: the proof should be improved at this point, ~~and I will improve it~~. (It was not incorrect, but assumed that the reader could fill in some missing steps, which is quite reasonable. But it is better if these steps are more elementary.) Oded (talk) 19:01, 16 July 2008 (UTC)[reply]

Sorry. I tried to improve, but if I try to put in all the details, I would have to prove that

1/b>10^{-b}

. While this can certainly be done, I think it is better to accept on faith that the reader knows these things. Otherwise, this would be just too involved. Oded (talk) 19:19, 16 July 2008 (UTC)[reply]

I am only a student so maybe the proof is rigorous and I just can't understand the jumps. However, these are serious questions and the proof seems to take leaps instead of steps, leaving gaps in the proof. If it is doing so, I hope the author can continue his proof. If it isn't skipping steps, I hope my question would be answered in a way I can understand.

98.210.254.18 (talk) 16:33, 16 July 2008 (UTC)Quang Pham[reply]

Where is this proof you are referring to? I looked at the article Dedekind cut and found no such proof. Oded (talk) 18:08, 16 July 2008 (UTC)[reply]

It's in the 0.999... article - [2]. --LarryMac | Talk 18:11, 16 July 2008 (UTC)[reply]

The infinite set that corresponds to real number

1-(1/10)^{n}

is

U_{n}=\{x\in \mathbb {Q} |x<1-(1/10)^{n}\}

Real number 0.999... is the union of all sets

U_{n}

U_{\infty }=\bigcup _{n=1}^{\infty }U_{n}

and real number 1 is the set

V=\{x\in \mathbb {Q} |x<1\}

What needs to be proven is that the two sets are equal

U_{\infty }=V

.

Clearly every

U_{n}

is a subset of V, thus so is their union

U_{\infty }\subset V

.

For the other inclusion

{\tfrac {a}{b}}<1

implies

{\tfrac {a}{b}}\leq {\tfrac {b-1}{b}}=1-{\tfrac {1}{b}}<1-({\tfrac {1}{10}})^{b}

. Thus a/b is an element of the set

U_{b}

.

{\tfrac {a}{b}}\in U_{b}\subset U_{\infty }

This is true for every element a/b in V, thus V is a subset of

U_{\infty }

. Tlepp (talk) 19:04, 16 July 2008 (UTC)[reply]

Intersecting Spheres edit

Imagine an (n+1)-simplex in n-dimensional space, with each edge of unit length. I'm pretty sure that the intersection of the unit (n-1)-spheres around all but three of its vertices would form a 2-sphere containing the remaining three. Is this true? And if so, how could I estimate the radius of this sphere? Black Carrot (talk) 20:06, 16 July 2008 (UTC)[reply]

Yes, it is true. The intersection of a sphere with a sphere is a sphere (or a point or empty) you known that each sphere in your intersection reduces the dimension by 1. By symmetry, the center of the 2-sphere will be the average of the vertices which are the centers of the spheres in the intersection. That should facilitate the calculation of the radius. Oded (talk) 20:19, 16 July 2008 (UTC) PS: in calculations involving the m-simplex, it is usually easiest to think of it as the convex hull of the m unit vectors in m-space, rather than embed it in (m-1)-space. Oded (talk) 20:23, 16 July 2008 (UTC)[reply]

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