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1

This is the number of partitions of a given integer into a sum. Say, $$m = 6,$$ and we would like to write it as the sum of 3 positive summands. To do this, write $$ 6 = 1 + 1 + 1 + 1 + 1 + 1. $$ A sum is formed by choosing 2 of the 5 pluses in the above expression; e.g, $$ 1 + 1 +' 1 + 1 + 1 +' 1 = 2 + 3 + 1. $$ The number of ways to choose 2 pluses is "5 ...


1

Every couple of fractions $c/a$ and $d/b$ in the Stern-Brocot tree can be represented (in the inverse notation according to Concrete Mathematics) by the matrix ${\bf M} = \left\| {\,\begin{array}{*{20}c} a & b \\ c & d \\ \end{array}\,} \right\|$, where, iff the fractions are the generators of another fraction in the tree, then the ...


2

Any $n \geq 2$. Partition according to parity so that $A = A_n$ and $B = S_n - A_n$. Take $\pi = (1, 2)$, which is an odd permutation. Since any even permutation in $A$ multiplied by the odd permutation $\pi$ will result in an odd permutation in $B$, we are done.


0

If $P$ and $Q$ are ordered sets, then the product $P\times Q$ is an ordered set, with ordering: $(p,q) \leq (u,v)$ iff $p \leq u$ and $q \leq v$. If $P$ and $Q$ are both partially ordered sets, then $P\times Q$ is also a partially ordered set (Poset).


0

Of course $R$ is reflexive. For all $(a,b)\in\mathbb{R}\times\mathbb{R}$ it holds $(a,b)R(a,b)$ because $a\leq a$ and $b\leq b$; If $(a,b)R(x,y)$ and $(x,y)R(a,b)$, then both $a\leq x,b\leq y$ and $x\leq a,y\leq b$. It follows $a=x$ and $b=y$, so $(a,b)=(x,y)$ Let $(a,b)R(x,y)$ and $(x,y)R(c,d)$. Hence, $a\leq x$ and $x\leq c$, so $a\leq c$. The same for $b\...


1

Partition of a set $X$ is a collection of pairwise disjoint subsets of $X$ such that union of all these subsets gives $X$. Any partition naturally sets an equivalence relation and vise versa. The partition in your question can be set by equivalence relation that can be described in many ways. For example like this: we say that $x$ is equivalent to $y$ if $x$ ...


2

The two things you wanted to know: THE MISTAKE: $x \not= y \land y \not= z \not \implies x \not =z$ Take any counter-example, as in Parth Kohli's comment. $(x,y)$ is an ordered pair, so essentially $(x,y) \in \mathbb{N} \times \mathbb{N}$.


2

A simple counter example will suffice. Take $x=z=1$, $y=2$. Then $x \neq y$ and $y\neq z$, but $x = z$. As to your last question, we use $x,y \in \mathbb{N}$ to mean that both $x$ and $y$ are natural numbers. This differs from $(x, y) \in \mathbb{N^2}$ where $(x,y)$ is an ordered pair.


2

There are many proofs, most of them are elementary. For example: non-intersecting lattice paths and LGV-lemma — see e.g. Bressoud. Proofs and Confirmations (ch. 3) RSK-correspondence — see e.g. Stanley. Enumerative combinatorics (vol. 2, 7.20) counting lozenge tilings using condensation


0

Let $\triangle_m \equiv \frac{m(m+1)}{2}$ be the $m$-th trangular number, and let $L(n)$ be the number of entries in the longest strict additive partition of the numbers from $1$ to $n$. Then $$ \triangle_{L(n)} \leq n $$ and $L(n)$ is the largest integer for which that relation holds. So for example, $L(8) = 3$ because $\triangle_{3} = 6 \leq 8$ but $\...


0

The best you can do is simplify your expression: $$\frac{c}{d} + \frac{e}{f} = \frac{cf + ed}{df} = \frac{a}{b}$$ The numerator and denominator will give you a system of two linear equations. With $a$ and $b$ fixed there are four unknowns $c,d,e,f$. As $4>2$ this means we have infinitely many solutions (as would be expected): $$a = cf + ed$$ $$b = df$$ ...


0

I will answer this question in the affirmative for all dimensions $d$ and numbers of points $k$. First we reduce the $d$-dimensional case to the 1-dimensional case. For every $k$ points in $\mathbb{R}^d$, there is a line onto which the projections of the points are distinct. This is because there are finitely many directions through two points and any line ...



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