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

= 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 $$a +' b = (a^{r} + b^{r})^\frac{1}{r}$$ 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 $$(1) f(ac,bc) = c f(a,b);\ (2) f(a,b)=f(b,a);\ (3) f(f(a,b),c) = f(a,f(b,c))$$ and for all reals a,b,c (see Field_(mathematics)).
 * Let us denote $$g(a)=f(a,0)$$. From (1), $$g(a) = a g(1)$$. 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 $$f(a,b)*K = g(f(a,b)) = f(f(a,b),0)=f(a,f(b,0)) = f(a,g(b)) = b f(a/b,K)$$, 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 $$f(a,b) = \frac{b}{K} f(a/b,K) = \frac{b}{K} \left( \frac{K}{K} f(\frac{a/b}{K},K) \right)= ... \frac{b}{K} f\left(\frac{a}{b\ K^{n-1}},K\right)$$.
 * Define $$h(r) = f(r,K)$$. From above, for all rationals r and integers n, $$h(r) = h(rK^{-n})$$. This also applies to negative n (substitute $$r \leftarrow rK^n$$). 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 $$h(r) = h(0) \rightarrow f(a,b) = \frac{b}{K} h(a/b) = \frac{b}{K} h(0) = f(b,a) = \frac{a}{K} h(0)$$ which must thus be equal to zero for all a,b nonzero.
 * We also have (even if h is not continuous) $$\frac{b}{K} h(a/b) = f(a,b) = f(b,a) = \frac{a}{K} h(b/a) \Rightarrow h(r) = r h(r^{-1}) $$
 * 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 Sn+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 Sn+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 : ,,...,. 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 Sn+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)
 * ... 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. $$6 \cdot 24=144ft^2$$. Thank you! Gidev the Dood(Talk) 19:33, 25 April 2018 (UTC)