Testwiki:Reference desk/Archives/Mathematics/2018 April 25

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < April 24 ! width="25%" align="center"|<< Mar | April | May >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

April 25

Non-standard "additions" on the real numbers

How many "addition" operations +' are there for which the real numbers R form a field with +' as the addition operation and the usual multiplication as the multiplication operation? Given such a non-standard "addition" operation, must there always be an automorphism of the multiplicative group R^× that extends to a field isomorphism between (R, +', *) and (R, +, *) where + is the usual addition and * the usual multiplication? Does $a{+}^{\prime }b=(a^r+b^r{)}^{\frac{1}{r}}$ work for any rational number r with both the numerator and the denominator positive odd integers? GeoffreyT2000 (talk) 03:12, 25 April 2018 (UTC)

• Some thoughts about the first/second question: this is equivalent, in functional form, to asking for a function $f(a,b)$ such that $(1)f(ac,bc)=cf(a,b);(2)f(a,b)=f(b,a);(3)f(f(a,b),c)=f(a,f(b,c))$ and for all reals a,b,c (see Field_(mathematics)#Classic_definition).

Let us denote $g(a)=f(a,0)$. From (1), $g(a)=ag(1)$. Define $K=g(1)=f(1,0)$, and assume it is nonzero in what follows (otherwise it means 0+'(anything)=0, which does not lead to an immediate contradiction that I can see, but surely that is a very different structure from the usual).

Apply (3) with c=0 and b≠0, from the previous it follows that $f(a,b)*K=g(f(a,b))=f(f(a,b),0)=f(a,f(b,0))=f(a,g(b))=bf(a/b,K)$, hence $f(a,b)=\frac{b}{K}f(a/b,K)$. Apply iteratively this relation, substituting $a,b\leftarrow a/b,K$ and it leads to $f(a,b)=\frac{b}{K}f(a/b,K)=\frac{b}{K}(\frac{K}{K}f(\frac{a/b}{K},K))=...\frac{b}{K}f(\frac{a}{b{K}^{n-1}},K)$.

Define $h(r)=f(r,K)$. From above, for all rationals r and integers n, $h(r)=h(r{K}^{-n})$. This also applies to negative n (substitute $r\leftarrow r{K}^n$). If h is continuous near 0 (which is by no means a given!) the only interesting case if $K\in \{-1,+1\}$, because otherwise for all r, a, b $h(r)=h(0)\to f(a,b)=\frac{b}{K}h(a/b)=\frac{b}{K}h(0)=f(b,a)=\frac{a}{K}h(0)$ which must thus be equal to zero for all a,b nonzero.

We also have (even if h is not continuous) $\frac{b}{K}h(a/b)=f(a,b)=f(b,a)=\frac{a}{K}h(b/a)\Rightarrow h(r)=rh(r^{-1})$

I could not go much further but that puts quite a lot of restrictions on your "addition" definition, especially if it is continuous in zero. Tigraan^{Click here to contact me} 08:56, 25 April 2018 (UTC)

It can be proven much simpler if you note that condition (1) means that $f(a,b)$ is a homogeneous function of the first degree and as such $f(a,b)=aF(a/b)$, where $F(x)$ is an arbitrary function. Condition (2) means that $F(x)=F(1/x)/x$ i.e. the function is defined but its values in $[-1,1]$ interval. So, the problem now is with condition (3). Ruslik_Zero 20:42, 25 April 2018 (UTC)

A quotient of the symmetric group

I'm defining an equivalence relation on S_n+1. The motivation is more musical than mathematical so you might find that equivalence relation is pulled out of a hat. In fact I'm really only concerned with S₁₂ but I'll phrase the question for S_n+1. So here's the equivalence relation: take a permutation A of order n+1: A(0),...,A(n). I'm saying that the permutation B is equivalent to A if for some p ≤ n B(0)=A(p),...,B(n-p)=A(n),B(n-p+1)=A(0),...,B(n)=A(p-1). In other words all those permutations equivalent to A for that equivalence relation correspond simply to one of the n+1 "shifts" of the string <A(0),...,A(n)>: <A(0),...,A(n)>,<A(1),...,A(n),A(0)>,...,<A(n),A(0),...,A(n-1)>. I have no idea how well or ill behaved that equivalence relation is with respect to the group law. I can also (set-theoretically) take the quotient of S_n+1 by that equivalence relation. Is that equivalence relation something you recognize, that has been studied? What do you get as an entity when you take that quotient? Just a set without any structure? I thought I'd ask you guys just to get an idea. Thanks. Basemetal 16:59, 25 April 2018 (UTC)

These are exactly the (left?) cosets of the cyclic (sub)group generated by the standard (n + 1)-cycle (1, 2, ..., n + 1). Equivalently, they are the orbits of the (right?) action of this group on all of S_{n + 1}. Since this subgroup is not normal, there is no group structure on its cosets. --JBL (talk) 12:46, 26 April 2018 (UTC)

Thanks. Basemetal 18:49, 26 April 2018 (UTC)

Length and width of a rectangle with only the area

There is a rectangle and its area 144ft², and its width is 4 times the length, thus the length is X and the width is 4X; but what is the formula to find the exact length and width of this rectangle? Gidev the Dood^(Talk) 17:36, 25 April 2018 (UTC)

$A=\ell w$, so (fixed per below) $144=x\cdot 4x$, which is easily solved for $x$. --Kinu ^t/_c 19:01, 25 April 2018 (UTC)

Kinu means $144=x\cdot 4x$. The square on the second $x$ is a typo. Basemetal 19:15, 25 April 2018 (UTC)

Template:Facepalm... thanks, fixed. I think my brain was one step ahead and doing the multiplication while I was typing that step. --Kinu ^t/_c 19:26, 25 April 2018 (UTC)

Alrighty, then the length is 6 and the width 24. $6\cdot 24=144f{t}^2$. Thank you! Gidev the Dood^(Talk) 19:33, 25 April 2018 (UTC)