Wikipedia:Reference desk/Archives/Mathematics/2009 April 27

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April 27

The Greatest Integer Function

Hello, I am trying to prove that

$[{\sqrt {n}}+{\sqrt {n+1}}]=[{\sqrt {n}}+{\sqrt {n+2}}]$

is true for all positive integers where the square brackets represent the greatest integer function. I reasoned that if I can show that

$m<{\sqrt {n}}+{\sqrt {n+1}}<m+1\Leftrightarrow m<{\sqrt {n}}+{\sqrt {n+2}}<m+1$

for an integer m, then I am done because that is a definition of the floor function. So in order to prove this, I have shown that the difference between these functions is always between zero and one. Furthermore, two of the inequalities are easy to show but the other two are hard. Any ideas?--68.121.32.160 (talk) 03:08, 27 April 2009 (UTC)

If the claim is not true, there is some integer m and some value a between 1 and 2 such that

${\sqrt {n}}+{\sqrt {n+a}}=m.$

The solution to this equation is

$n={\frac {(m^{2}-a)^{2}}{4m^{2}}}.$

For a in (1,2) the fractional part of the right side lies in

$\left({\frac {1}{m^{2}}},{\frac {1}{2}}+{\frac {1}{4m^{2}}}\right){\text{ and }}\left({\frac {1}{4}}+{\frac {1}{m^{2}}},{\frac {3}{4}}+{\frac {1}{4m^{2}}}\right),$

for even and odd m, respectively, so it can't be an integer. McKay (talk) 04:12, 27 April 2009 (UTC)

I understand everything perfectly except for the fractional part. How did you arrive at those bounds for the fractional part? Why does it matter if the integer is odd or even and how did you get those intervals? Thanks!68.126.127.36 (talk) 07:51, 28 April 2009 (UTC)

If m is even, say m=2k, then

${\frac {(m^{2}-a)^{2}}{4m^{2}}}=k^{2}-{\frac {a}{2}}+{\frac {a^{2}}{4m^{2}}}.$

Now you can check that for a in (1,2), that value is always strictly between k²-1 and k².

If m is odd, say m=2k+1, then

${\frac {(m^{2}-a)^{2}}{4m^{2}}}=k^{2}+k-{\frac {a}{2}}+{\frac {1}{4}}+{\frac {a^{2}}{4m^{2}}},$

which is strictly between k²+k-1 and k²+k. McKay (talk) 08:20, 28 April 2009 (UTC)

Why is "dense-in-itself" a useful notion?

After committing the embarassing rookie's mistake of linking the phrase "dense in itself" in the sentence "a nowhere dense set is always dense in itself" to dense-in-itself I got myself thinking: why is the notion of a set being dense-in-itself at all useful? Are there any interesting non-trivial properties of topological spaces without isolated points? The topology books I know may define the notion but only to never mention it again. — Tobias Bergemann (talk) 07:06, 27 April 2009 (UTC)

Perfect sets are, by definition, closed dense-in-itself sets, and they appear in various contexts, see e.g. Cantor–Bendixson theorem. — Emil J. 10:27, 27 April 2009 (UTC)

A complete metric space with no isolated points essentially has a subspace that is the continuous injective image of the cantor space. A similar result that (possibly) applies to a larger class of spaces asserts that any locally compact Hausdorff space without isolated points, has cardinality at least that of the continuum. The proofs of these facts are relatively simple if you were to attempt the proofs yourself. The idea embedded within the previous assertions is that spaces with no isolated points, having certain properties, must essentially be "large". --PST 10:39, 27 April 2009 (UTC)

Thank you both for your answers. I really should have thought of perfect sets and the Cantor-Bendixson theorem myself. I just couldn't think of anything interesting to say about a topological space about which it only is known that it has no isolated points and nothing else (a perfect space). — Tobias Bergemann (talk) 12:29, 27 April 2009 (UTC)

Proving or disproving a homeomorphism with [0,1]

Hi there guys - I was wondering about how to go about showing that $[0,1]$ is or is not homeomorphic to $\{0,1,...,9\}^{\mathbb {N} }$? I don't want you to tell me how to do it, but what would you suggest to get started? I know the 2 sets have the same cardinality so that won't rule the possibility of bijection out, but I'm less certain about continuity - could anyone suggest anything to get me going? I imagine if they aren't homeomorphic I'll simply want to find a topological property they don't share, but I'm not sure where to start looking or whether that's even the case...

Also, I'm trying to find a homeomorphism between $\mathbb {R} \setminus \{0\}$ and $\mathbb {R} \setminus \{x:|x|\leq 1\}$ - is the function $\left\{{\begin{array}{lr}{\frac {1}{x-1}}&:x>1\\{\frac {1}{x+1}}&:x<-1\end{array}}\right.$ continuous in the topological sense between these 2 sets?

I hope I'm not asking too much - Thanks a lot! Spamalert101 (talk) 08:39, 27 April 2009 (UTC)

(On a formatting note, how do I get my 2 sets to display in the same sized font?) Spamalert101 (talk) 08:40, 27 April 2009 (UTC)

To begin my response, let me stress that much of topology is intuitive. I do not think that it is worth it to worry too much about proving whether two spaces are homeomorphic or not if you absolutely see it intuitively - an exception being when the equivalence of two spaces may be of crucial importance in a theorem. If you have first learnt the concept however, it is nice to construct a few homeomorphisms.

To expand on my previous point is equivalent to solving the first problem. Initially, the idea is recall the connection between the factors of a product space (assuming the product topology - not that it matters in this case, even if you were to choose the box topology) and the product itself. Often one can say a lot about the product given information about its factors; this assertion lies within the continuous projection maps onto the factors. Therefore, it is necessary to find a property that is preserved under continuous maps, and that is shared by [0,1] but not by a finite discrete space.

Secondly, to check the continuity of the map given, is equivalent to checking continuity of its restriction to each "piece". This is because, essentially, the two pieces are "far from each other" (or more precisely, their closures are disjoint), and since continuity is "points close together get mapped to points close together", we only need to consider the map defined on each piece separately. Doing so is simply basic calculus.

Hope this helps. Let me add that it is nice to have a question on topology, once in a while! Rarely are questions on fields outside calculus, asked. --PST 10:24, 27 April 2009 (UTC)

To the first question: [0,1] is connected, whereas $\{0,1,...,9\}^{\mathbb {N} }$ is disconnected (indeed, totally disconnected), hence they are not homeomorphic.

To the second question: yes, your function is continuous, and in fact a homeomorphism. However, you are making it unnecessarily complicated: ${\begin{cases}x-1&x>1\\x+1&x<-1\end{cases}}$ works just as well. — Emil J. 10:19, 27 April 2009 (UTC)

Let me note, Emil J., that the OP requested specifically that only a hint be given (to get him/her started) rather than the answer. --PST 10:25, 27 April 2009 (UTC)

Don't worry, having the answers will be useful to check my own suggestions against, having (oddly) read upwards from the bottom of the post I managed to avoid the given solutions themselves whilst reading, but will certainly come back to them after attempting the rest of the problem - Thankyou both for the help, and I'll be sure to bring a couple more topology questions your way in the future! ;) Spamalert101 (talk) 11:26, 27 April 2009 (UTC)

Can every unit algebraic number be expressed as a root of unity?

What I mean by "unit algebraic number" is an algebraic number which has an absolute value of 1. Root of unity, of course, means a solution to Zⁿ-1=0 for some positive integer n.

I believe this is equivalent (in light of the fundamental theorem of algebra and closure of algebraic numbers under multiplication) to saying that every polynomial with rational coefficients can be expressed, by multiplying it by some other polynomial and then factoring, as a product of polynomials of the form (aZ)ⁿ-1 for some quadratic a and positive integer n, in addition to some polynomial of the form Zⁿ for positive integer n, for the zero roots. Of course, it doesn't matter whether a is a coefficient of Zⁿ or directly with Z, but in the latter case, it has the more intuitive meaning of being the reciprocal of the magnitude of Z.

Is that statement true for polynomials with any complex coefficients, letting a be any real number?

All responses appreciated. --COVIZAPIBETEFOKY (talk) 12:47, 27 April 2009 (UTC)

I think the answer to your first question is "no". Consider the ring of algebraic integers in Q(sqrt(2)). Then 3+sqrt(2) is a unit in this ring, because its minimal polynomial is x² - 6x + 1 (its associate is 3-sqrt(2)). But 3+sqrt(2) is clearly not a root of unity - all its integer powers are greater than 1. Gandalf61 (talk) 13:17, 27 April 2009 (UTC)

Note that the OP uses nonstandard terminology. The absolute value of 3+sqrt(2) is not 1, so it is not a "unit number" the way he defined it. — Emil J. 13:21, 27 April 2009 (UTC)

Nevertheless, the answer is still "no". The algebraic number (3 + 4i)/5 has absolute value 1, but it is not a root of unity, as its minimal polynomial is 5x² − 6x + 5. — Emil J. 13:34, 27 April 2009 (UTC)

This result gives a large number of counterexamples, namely any a+bi where (a/b)² is rational and not in the set {0, 1/3, 1, 3}. -- BenRG (talk) 14:00, 27 April 2009 (UTC)

A simple example is u:=2+i. It is easy to see (induction) that for all even natural number n, 5 divides uⁿ+u (i.e. it divides both the real and the imaginary part). Therefore no positive integer power of u is a real number. Hence u/|u| is an algebraic number of modulus 1, not a root of unity. --pma (talk) 14:32, 27 April 2009 (UTC)

I'm not sure I understand how you can use the minimal polynomial to predict whether a number will be a root of unity; after all, the minimal polynomial of -1/2+i√(3)/2 is X² + X + 1, but it is a 3^rd root of unity.

However, I do have an understanding of why (3+4i)/5 wouldn't be a root of unity, unrelated to its minimal polynomial: because its angle (roughly 0.927295 radians or 53.130102 degrees) is an irrational multiple of 2π, adding the angle to itself many times will never give a multiple of 2π. Thus, multiplying the number by itself will never yield 1. This sheds some light on BenRG's set of counterexamples. I also might understand pma's explanation if I mull it over a bit.

Also, apologies for using "nonstandard terminology". I haven't taken an actual class on the material, and I thought I could get away with it if I actually explained what I meant thoroughly in the text. But who reads the actual text? Silly me...

Thanks for the help. I suppose I should have been able to figure this out by myself, but I was thinking about it in terms of polynomials (the latter half of my question) rather than the numbers, which seems to have muddied things up a bit. Any pointers to getting a better understanding of the behavior of the polynomial side of the question? --COVIZAPIBETEFOKY (talk) 15:00, 27 April 2009 (UTC)

Well, yes, the angle atan(4/3) is an irrational multiple of 2π, but how do you prove that? It's no easier than showing that (3 + 4i)/5 is not a root of unity in the first place.

As for minimal polynomials: sorry I wasn't more clear on this point. Roots of unity are algebraic integers, hence their primitive minimal polynomials are monic (or equivalently, their monic minimal polynomials have integer coefficients). 5x² − 6x + 5 is a primitive irreducible polynomial and it is not monic, hence its roots are not roots of unity. — Emil J. 15:13, 27 April 2009 (UTC)

There's a bit more to say here. The OP's idea is salvageable. Properly interpreted, "unit algebraic numbers" - algebraic numbers with absolute value 1 - are necessarily roots of unity. It's easier not to work with the field of all algebraic numbers, but with finite extensions of the rationals, e.g. the number field generated by some given algebraic number u : K = Q(u). The statement as given above is clearly true of rational numbers, or of any field with an embedding into the real numbers, but not, as it is stated, of arbitrary extensions - examples above. In general you need to consider other absolute values (or units in the more usual sense if you want to avoid these) and all real and complex embeddings / absolute values of the number field K. It is easy to see that an element u of a number field, all of whose non-archimedean absolute values are 1 is a unit in the ring of integers of the number field. Dirichlet's unit theorem says that the group of units of this ring of integers is mapped onto a lattice in a real vector space by taking logs of all the real and complex absolute values, with the kernel being precisely the roots of unity in the ring of integers. log (abs val) = 0 means abs val = 1, so it says that if all the archimedean abs vals of u are 1 also, i.e. all of the AVs of u = 1, then it must be a root of unity.John Z (talk) 11:25, 2 May 2009 (UTC)

Disjoint balls into a ball in R^n

Hi there, since you enjoyed the last topology question so much I figured I might send another one or two your way! I'm revising it right now so you might end up getting a good few if you don't mind lending me a little more help!

I've showed that there DNE 2 closed balls of radius 1 inside a closed ball of radius 2 in the Euclidean space $\mathbb {R} ^{n}$ but I'm now trying to find how many closed unit balls there are in the space inside balls of radii 3.001 and 2.001 - the first is apparently $(1+k)^{n}$ for some k>0, but how do I go about beginning to prove it? Thanks very much again for the help and if I'm asking too much just say!

Spamalert101 (talk) 14:54, 27 April 2009 (UTC)

So you want to pack $k$ unit $n$-dimensional Euclidean balls into a ball $B(r)$ of radius $r$. I took the liberty of using open unit balls instead of closed, since the problem remains essentially the same, and things are easier to describe. Now, although these problems are generally difficult, at least in the case $\scriptstyle k\leq n+1$ the situation is quite simple: take the balls pairwise tangent, that is, put their centers at a distance 2 from each other. With a small computation this gives a radius $\scriptstyle 1+{\sqrt {2\left(1-{\frac {1}{k}}\right)}}$ for the minimal ball containing them. Also, it is not immediate but not even hard to prove that this is actually the least number $r$ such that there are $k$ disjoint unit open balls inside $B(r)$: in other words, the minimizing configuration for the $k$ balls is the one above, where they are pairwise tangent. In particular, in dimension $n>1$, there are three disjoint open unit balls inside $B(r)$ iff $\scriptstyle r\geq 1+{\frac {2}{\sqrt {3}}}$. Another consequence is that if $\scriptstyle r<1+{\sqrt {2}}$ the number of unit balls inside $B(r)$ is bounded independently from the dimension. On the other hand, as soon as $\scriptstyle r\geq 1+{\sqrt {2}}$, in dimension $n$ there are at least $n+1$ unit balls in $B(r)$, so the number of balls is unbounded as dimension increases, and very difficult to count exacltly. If $r=3$ I guess that the maximum number of unit balls in $B(3)$ is obtained with one ball concentric with $B(3)$ and all the other tangent to this one (as far as I see, it could be trivially true or trivially false, or an open problem). If this is true, the max number of unit balls in $B(3)$ is 1 plus the kissing number, which is still a topic of current research. PS: Your question has the following nice Hilbert space version: there are infinitely many disjoint unit open balls inside the ball $\scriptstyle B(0,1+{\sqrt {2}})$ of an infinite dimensional Hilbert space (just take them centered in $\scriptstyle \{{\sqrt {2}}e_{j}\}_{j}$, where $\scriptstyle \{e_{j}\}_{j}$ is an orthonormal basis). But if you take the radius any smaller, $\scriptstyle r<1+{\sqrt {2}}$, then only finitely many disjoint unit open balls can be located in a ball of radius $r$: precisely, at most $\scriptstyle {\big \lfloor }{\frac {2}{2-(r-1)^{2}}}{\big \rfloor }$. Life in Hilbert space is curious... (PPS: of course, feel free to ask for details if needed) pma (talk) 21:47, 27 April 2009 (UTC)