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Somewhat stupid question, but why do the lines y1=x and y2=x+1 never meet in a point? Is it just because of algebra? Whatever y1 is equal to, y2 will be y1+1? Thanks

Definition: A space $X$ is $\Delta$-normal if for every $A \subset X^2 \setminus \Delta_X$ closed in $X^2$ there exist disjoint open $U$ and $V$ in $X^2$ such that $A \subset U$ and $\Delta_X \subset V$, where $\Delta_X$ is the diagonal of $X$, i.e., $\{(x,x): x\in X\}$. Question: Is Sorgenfrey plane $\mathbb R_l^2$ $\Delta$-normal? Note: What I asked is $\mathbb R_l^2$, not $\mathbb R_l$. So if you want to go on with diagonal, its a closed subspace of $\mathbb R_l^4$, not $\mathbb R_l^2$ itself. Here is some partial results: $\mathbb R_l^2$ is not $\Delta$-paracompact. See Corollary 16 in https://topology.nipissingu.ca/tp/reprints/v39/tp39010.pdf. However, $\Delta$-normality does not directly imply $\Delta$-paracompactness, despite Tychonoff ($T_{3\frac12}$) spaces. $\mathbb R_l$ is $\Delta$-paracompact and $\Delta$-normal, for being a generalized ordered space through Theorem 2.5 in https://www.sciencedirect.com/science/article/pii/S0166864109004271. $\Delta_{\mathbb R_l^2} \subset \mathbb R_l^4$ and the closed (discrete) set $(\Delta')^2 \setminus \Delta_{\mathbb R_l^2} \subset \mathbb R_l^4$ is separable by open (even clopen) sets, where $\Delta' = \left\{ (x, -x) : x \in \mathbb R_l \right\}$ is the anti-diagonal. Explicitly, let $A = \left\{ (x_1, y_1), (x_2, y_2) \mid x_1 + y_1 < 0 \vee x_2 + y_2 < 0 \vee \min\{x_1, x_2\} + \min\{y_1, y_2\} \geq 0\right\}$, it is easily seen that $A$ is clopen in $\mathbb R_l^4$, and also we have (i) $\Delta_{\mathbb R_l^2} \subset A$; (ii) $A \cap \left( \Delta' \right)^2 = \varnothing$. So I'm stuck with construct direct witness. Thanks for your help all.

Prove that $$x^{2m+1} - a^{2m+1} = (x-a)\prod_{r=1}^{m} \left(x^{2} - 2ax \cos \left( \frac{2 \pi r}{2m+1}\right) + a^{2}\right)$$ where $m$ is positive integer $\ge 1$. I decided to use induction to prove the above identity. The base case is easy ($m=1$). Now here is my progress on the inductive hypothesis step. $x^{2m+3} - a^{2m+3} = (x-a)(x^{2m+2} + x^{2m+1}a +x^{2m}a^{2}+ \dots + xa^{2m+1} + a^{2m+2}) \tag{1}$ Now from the assumption of our hypothesis, we find that $$\prod_{r=1}^{m} \left(x^{2} - 2ax \cos \left( \frac{2 \pi r}{2m+1}\right) + a^{2}\right) = (x^{2m}+x^{2m-1}a + \dots + xa^{2m-1} + a^{2m})$$ So by factoring $ax$ in $(1)$ we get $$(x-a)(x^{2m+2} +ax\prod_{r=1}^{m} \left(x^{2} - 2ax \cos \left( \frac{2 \pi r}{2m+1}\right) + a^{2}\right) +a^{2m+2}) \tag{2}$$ How can I transform the product in $(2)$ into a desirable form?

Emir, Cillian, and Lasha are avid collectors of vintage postage stamps. During a local exhibition, they decided to donate parts of their collections to a museum. The ratio of the number of stamps Emir and Cillian gave away was 4:7, while the ratio of the number of stamps Cillian and Lasha gave away was 3:5. The number of stamps Emir gave away was 20 fewer than the number of stamps he kept for himself. Cillian kept 3/2 as many stamps as Emir kept, and Lasha kept 4/3 as many stamps as Cillian kept. After their donations, they had a combined total of 738 stamps left in their collections. How many stamps did Lasha give away? This is a ratio and proportion problem from Algebra.com. I am practicing for ratio manipulation, systems of linear equations, and variable modeling I started by setting up the ratios for the stamps given away: E:C = 4:7 C:L = 3:5 To find a common ratio, I multiplied them to get E:C:L = 12:21:35. I’m having trouble linking the "stamps kept" total (738) back to the "stamps given" ratios. I know Emir gave away 20 fewer than he kept (Eg = Ek - 20), but substituting the fractional relationships for Cillian and Lasha's kept stamps into the total sum is getting messy. How do I bridge the "kept" total to find the value of one "part" in the "given away" ratio?

Let $(F_n)$ be the Fibonacci sequence defined by $F_1=F_2=1$. Fix an even index $m=2r$, and let $a=F_{2r}$ and $M=F_{2r+1}$. For $1 \le b < M$, define the modular residues:$$R_b \equiv ab \pmod M, \quad 0 \le R_b < M$$Let $\rho_b$ be the rank of $R_b$ among $\{R_1, \dots, R_b\}$ when arranged in increasing order:$$\rho_b = 1 + \#\{k < b : R_k < R_b\}$$Using the symmetry $R_{b-k} \equiv R_b - R_k \pmod M$, one obtains the exact identity:$$S_b := \sum_{k=1}^{b} R_k = \frac{(b+1)R_b + (b-\rho_b)M}{2}$$The problem of computing $S_b$ thus reduces to determining the rank $\rho_b$. Numerical experiments strongly suggest that $\rho_b$ is governed by the Zeckendorf decomposition of the final residue $R_b$, rather than behaving randomly. Case 1: Single Fibonacci numberIf $R = F_n$, then $F_n$ is a residue-record (a new minimum or maximum), depending on the parity of $n$. Its rank is always extremal. Case 2: Sum of two Fibonacci numbersFor $R = F_i+F_j$ ($i-j \ge 2$), I consistently obtain:$$\rho(F_i+F_j) = \begin{cases} F_{i-j}+1+\varepsilon_i, & j \text{ even} \\ F_i+F_j-F_{i-j}+\varepsilon_i, & j \text{ odd} \end{cases}$$where $\varepsilon_n = +1$ if $n$ is even, and $-1$ if $n$ is odd. The parity of the smallest index $j$ decides whether the rank is "small" or "large", while the parity of $i$ contributes only a small correction. Case 3: Sum of three Fibonacci numbersLet $A=F_{\ell-i}$, $B=F_{i-j}$, and $C=F_{\ell-j}$. For $R = F_\ell+F_i+F_j$ ($\ell-i \ge 2, i-j \ge 2$):$$\rho(F_\ell+F_i+F_j) = \begin{cases} \varepsilon_i A + B + C + 1 + \varepsilon_\ell + \varepsilon_i, & j \text{ even} \\ R + \varepsilon_i A - B - C + \varepsilon_\ell + \varepsilon_i, & j \text{ odd} \end{cases}$$The parity of the smallest index $j$ determines the regime, while the other parities only modify signs and the constant term. Case 4: Four termsFor $R = F_n+F_\ell+F_i+F_j$ ($n > \ell > i > j$, with standard gap conditions), the pattern persists: the smallest index $j$ dictates the regime, and the rank is a linear combination of differences $F_{n_u-n_v}$ plus a small constant. The "All-Even" Case Suppose the Zeckendorf decomposition is $R = \sum_{t=1}^{s} F_{n_t}$ ($n_1 > \dots > n_s$ and $n_t - n_{t+1} \ge 2$). If all indices $n_t$ are even, I obtain the exact formula:$$\boxed{\rho(R) = \sum_{1\le u < v \le s} F_{n_u-n_v} + s}$$This also suggests the recurrence:$$\rho(F_m+R) - \rho(R) = 1 + \sum_{t=1}^s F_{m-n_t}$$when adding a new even Fibonacci term $F_m$ to the decomposition. Examples: Two terms: $\rho(F_i+F_j) = F_{i-j} + 2$ Three terms: $\rho(F_\ell+F_i+F_j) = F_{\ell-i} + F_{\ell-j} + F_{i-j} + 3$ Four terms: $\rho(R) = \sum_{u<v} F_{n_u-n_v} + 4$ Conjectural Unified FormFor a general Zeckendorf decomposition $R = \sum_{t=1}^{s} F_{n_t}$, the rank appears to always satisfy:$$\boxed{\rho(R) = L(R) \quad \text{or} \quad \rho(R) = R + L(R)}$$. where $L(R)$ is defined as:$$\boxed{L(R) = \sum_{1\le u < v \le s} \varepsilon_{n_v} F_{n_u-n_v} + \mathbf{1}_{n_s \text{ even}} + \sum_{u=1}^{s-1} \varepsilon_{n_u}}$$ Experimentally, the smallest index parity determines the regime: $j$ even $\implies$ small rank, $j$ odd $\implies$ large rank. Questions Is this rank law known in the literature?