Determine all sets of real numbers $S$ such that: $1$ is the smallest element of $S$, for all $x,y\in S$ such that $x>y$, $\sqrt{x^2-y^2}\in S$ Adian Anibal Santos Sepcic
Problem
Source: EMC 2023 Seniors P1
Tags: algebra, easy, emc, 2023
pco
18.12.2023 19:10
rama1728 wrote: Determine all sets of real numbers $S$ such that: $1$ is the smallest element of $S$, for all $x,y\in S$ such that $x>y$, $\sqrt{x^2-y^2}\in S$ Let $T=\{x^2\quad\forall x\in S\}$ $1\in T$ and $\forall a>b$ in $ T$, $a-b\in T$ If $\exists a\in T\setminus \mathbb N$, then $a-1\in T$ and so $\{a\}\in T$, impossible since $ 1$ is the smallest element of $ T$. So $T\subseteq \mathbb Z_{>0}$ Hence the answer : $\boxed{\text{S1 : }S=\{\sqrt k\quad\forall k\in\mathbb Z_{>0}\}}$ $\boxed{\text{S2 : }S=\{\sqrt k\quad\forall k\in\{1,2,3,...,n\}\}}$ whatever is $n\in\mathbb Z_{>0}$
VicKmath7
18.12.2023 19:56
We claim the set $S$ consists of separate real numbers. Suppose otherwise - then $S$ contains a continuous interval of reals. Choose $x$ from this interval and $y=x+\epsilon$ for $\epsilon \rightarrow 0$ also from the same interval. Then $\lim_{\epsilon \rightarrow 0} \sqrt{(x+\epsilon)^2-x^2}=0$, so this element can't be in the set, as it is not greater than $1$. This means that indeed $S$ can be numerated as $1=a_1<a_2<\ldots$.
We will show by induction that $a_k=\sqrt{k} $ for all positive integers $k$. We have $a_1=1$ and $\sqrt{a_2^2-1}<a_2$, so $a_2=\sqrt{2}$ and the base case is clear. Suppose that $a_i=\sqrt{i}$ for all $i=1,2, \ldots, k$ and consider the numbers $b_i=\sqrt{a_{k+1}^2-a_i^2}=\sqrt{a_{k+1}^2-i}<a_{k+1}$ for $i=1,2, \ldots, k$. They should be among $\{a_1, a_2, \ldots, a_k\}$, and since the $b_i$ are $k$ pairwise distinct numbers with $b_i<b_j$ for $i>j$, we should have $b_i=a_{k+1-i}$ and thus $\sqrt{a_{k+1}^2-1}=b_1=a_k=\sqrt{k}$ implies $a_{k+1}=\sqrt{k+1}$ and this finishes the induction. Thus, the only working set consists of $\{\sqrt{1}, \sqrt{2}, \ldots, \sqrt{n}\}$ for some positive integer $n$, or the same set extended up to infinity.
$\textbf{Remark.}$ The first paragraph can be changed to the following: $x \in S^2 \implies x-1 \in S^2$ implies that no non-integer is member of $S^2$, otherwise there is a number smaller than $1$. Thus $S$ is subset of $\{\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots \}$ and can be numerated in the desired manner.
square_root_of_3
19.12.2023 01:23
VicKmath7 wrote:
Suppose that $S$ has elements $a_1=1 \leq a_2 \leq \ldots$.
This isn't really something you're allowed to do. Your set could in theory contain an interval of the form $[1,1+t)$ for some $t>0$, so your $a_2$ isn't really well-defined.
jasperE3
19.12.2023 02:53
same as pco Consider $T=\{s^2\mid s\in S\}$. We have that $1$ is the smallest element of $T$ and for all $x,y\in T$ with $x>y$, $x-y\in T$. Suppose that there is an element $n\notin\mathbb N$ but $n\in T$. Then since $n>1$, $n-1\in T$ as well. By induction, $n-\lfloor n\rfloor\in T$, but this contradicts that $1$ is the smallest element as $n-\lfloor n\rfloor<1$. Suppose now that there is some $n\in\mathbb N$ with $n\in T$. By a similar induction, all natural numbers less than $n$ are in $T$ as well. Therefore the answer is $T=\mathbb N$ or $T=\{t\in\mathbb N\mid t\le c\}$ for some $c\in\mathbb N$. Hence $S=\sqrt{\mathbb N}$ or $S=\{\sqrt t\in\mathbb N\mid t\le c\}$ for some $c\in\mathbb N$, which can be checked to satisfy the conditions.