Raja Oktovin wrote: Let $ \mathbb{Z}$ be the set of all integers. Define the set $ \mathbb{H}$ as follows: (1). $ \dfrac{1}{2} \in \mathbb{H}$, (2). if $ x \in \mathbb{H}$, then $ \dfrac{1}{1 + x} \in \mathbb{H}$ and also $ \dfrac{x}{1 + x} \in \mathbb{H}$. Prove that there exists a bijective function $ f: \mathbb{Z} \rightarrow \mathbb{H}$. There likely miss some elements in your problem since $ \mathbb H = \mathbb R^ +$ clearly respects the two conditions $ (1)$ and $ (2)$ although there is no bijection from $ \mathbb Z\to\mathbb R^ +$ If you mean that $ \mathbb H$ is the minimum set generated by successive applications of rule $ (2)$ on $ \{\frac 12\}$, then the result is obvious since all elements of $ \mathbb H$ are then positive rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result.

pco wrote: Raja Oktovin wrote: Let $ \mathbb{Z}$ be the set of all integers. Define the set $ \mathbb{H}$ as follows: (1). $ \dfrac{1}{2} \in \mathbb{H}$, (2). if $ x \in \mathbb{H}$, then $ \dfrac{1}{1 + x} \in \mathbb{H}$ and also $ \dfrac{x}{1 + x} \in \mathbb{H}$. Prove that there exists a bijective function $ f: \mathbb{Z} \rightarrow \mathbb{H}$. There likely miss some elements in your problem since $ \mathbb H = \mathbb R^ +$ clearly respects the two conditions $ (1)$ and $ (2)$ although there is no bijection from $ \mathbb Z\to\mathbb R^ +$ If you mean that $ \mathbb H$ is the minimum set generated by successive applications of rule $ (2)$ on $ \{\frac 12\}$, then the result is obvious since all elements of $ \mathbb H$ are then positive rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result. $ \mathbb{H}\in(0,1)$ and $ \mathbb{H}\in\mathbb{Q}$ is it? thus by setting $ x=\frac{m}{n}$ with $ m<n$ and $ m,n\in\mathbb{N}$ we can change the form in (2) become if $ \frac{m}{n} \in \mathbb{H}$, then $ \dfrac{m}{m+n} \in \mathbb{H}$ and also $ \dfrac{n}{m+n} \in \mathbb{H}$. If $ m=1$ then since $ \frac12\in\mathbb{H}$ then all numbers $ \frac{1}{x}$ are in $ \mathbb{H}$ If $ m=2$ then since we can get $ \frac23\in\mathbb{H}$ from $ \frac{n}{m+n}=\frac{2}{1+2}=\frac23$ ($ (m,n)=(1,2)$) then all numbers $ \frac{2}{x}$ that $ x\equiv_1 2$ are in $ \mathbb{H}$ If $ m=3$ we divide into 2 cases -we can get $ \frac34\in\mathbb{H}$ from $ \frac{n}{m+n}=\frac{3}{1+3}=\frac34$ ($ (m,n)=(1,3)$) then all numbers $ \frac{3}{x}$ that $ x\equiv_1 3$ are in $ \mathbb{H}$ -we can get $ \frac35\in\mathbb{H}$ from $ \frac{n}{m+n}=\frac{3}{2+3}=\frac35$ ($ (m,n)=(2,3)$) then all numbers $ \frac{3}{x}$ that $ x\equiv_2 3$ are in $ \mathbb{H}$ And so on so we conclude that $ \mathbb{H}\in\mathbb{Q}$ and $ \mathbb{H}\in(0,1)$

That was a perfect piece of spam. matrix41 ! based on reading too quickly the previous post(s), then adding some completely inconsequential remarks. pco pointed out some set(s) that fulfill both requirements, then guessed that the problem refers to the minimal such set. What you did was some wishy-washy attempt by incomplete induction to prove $ \mathbb{H} \subseteq \mathbb{Q} \cap (0,1)$. But this is obvious, by the fact the starting seed is $ 1/2$, while functions $ f(x) = 1/(1 + x)$ and $ g(x) = x/(1 + x)$ have both the property that they take rational values at rational points, while they move the $ (0,1)$ interval to $ (1/2, 1)$, respectively $ (0, 1/2)$. Also, don't use $ \in$ for inclusion of sets, and fight the urge to quote entirely the post you refer to, especially if it is the one just above

pco wrote: rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result. I believe that this statement should be well-known, but can you write the details here? Thank you.

pco wrote: Raja Oktovin wrote: Let $ \mathbb{Z}$ be the set of all integers. Define the set $ \mathbb{H}$ as follows: (1). $ \dfrac{1}{2} \in \mathbb{H}$, (2). if $ x \in \mathbb{H}$, then $ \dfrac{1}{1 + x} \in \mathbb{H}$ and also $ \dfrac{x}{1 + x} \in \mathbb{H}$. Prove that there exists a bijective function $ f: \mathbb{Z} \rightarrow \mathbb{H}$. There likely miss some elements in your problem since $ \mathbb H = \mathbb R^ +$ clearly respects the two conditions $ (1)$ and $ (2)$ although there is no bijection from $ \mathbb Z\to\mathbb R^ +$ If you mean that $ \mathbb H$ is the minimum set generated by successive applications of rule $ (2)$ on $ \{\frac 12\}$, then the result is obvious since all elements of $ \mathbb H$ are then positive rational numbers, and so $ \mathbb H\subseteq\mathbb Q$ is countable, hence the result. I can't understand your idea I think that two condition of problem only to prove $ H = (0,1)\cap Q$ and we need prove that:there exist a bijective function $ f: \mathbb{Z}\rightarrow\mathbb{H}$

@math10 : My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result. @Raja Oktovin : I think this is a basic definition of countable : an infinite set A is countable $ \iff$ $ \exists$ a bijection between A and $ \mathbb N$ And so $ \exists$ bijections between any two infinite countable sets, so between $ \mathbb Z$ and $ \mathbb H$.

pco wrote: @math10 : My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result. We only have $ H$ is infinite and $ H\in Q$. How do you to get the results? (sorry for my stupid ) please write clear

math10 wrote: pco wrote: @math10 : My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result. We only have $ H$ is infinite and $ H\in Q$. How do you to get the results? (sorry for my stupid ) please write clear I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ? Is this what you want I explain more ?

pco wrote: math10 wrote: pco wrote: @math10 : My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result. We only have $ H$ is infinite and $ H\in Q$. How do you to get the results? (sorry for my stupid ) please write clear I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ? Is this what you want I explain more ? I can't understand step: *)If $ H$ is infinite and $ H\in Q$ then there exist a bijection from $ H$ to $ Z$

math10 wrote: pco wrote: math10 wrote: pco wrote: @math10 : My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result. We only have $ H$ is infinite and $ H\in Q$. How do you to get the results? (sorry for my stupid ) please write clear I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ? Is this what you want I explain more ? math10 wrote: I can't understand step: *)If $ H$ is infinite and $ H\in Q$ then there exist a bijection from $ H$ to $ Z$ Step 1 : every infinite subset of a countable set is infinite countable. Is step 1 OK ? Step 2 : $ \mathbb H$ is an infinite subset of the countable set $ \mathbb Q$, and so $ \mathbb H$ is infinite countable. Is step 2 OK ? Step 3 : If set A is infinite countable, there exists a bijection from A$ \to\mathbb N$ Is step 3 OK ? Step 4 : if there exists a bijection from $ A\to C$ and a bijection from $ B\to C$, there exists a bijection from $ A\to B$ Is step 4 OK ? Step 5 : $ \mathbb Z$ is infinite countable, so $ \exists$ a bijection from $ \mathbb Z\to\mathbb N$ (step 3) $ \mathbb H$ is infinite countable, so $ \exists$ a bijection from $ \mathbb H\to\mathbb N$ (step 3) So $ \exists$ a bijection from $ \mathbb H\to\mathbb Z$ OK ?

mavropnevma wrote: That was a perfect piece of spam. matrix41 ! based on reading too quickly the previous post(s), then adding some completely inconsequential remarks. pco pointed out some set(s) that fulfill both requirements, then guessed that the problem refers to the minimal such set. What you did was some wishy-washy attempt by incomplete induction to prove $ \mathbb{H} \subseteq \mathbb{Q} \cap (0,1)$. But this is obvious, by the fact the starting seed is $ 1/2$, while functions $ f(x) = 1/(1 + x)$ and $ g(x) = x/(1 + x)$ have both the property that they take rational values at rational points, while they move the $ (0,1)$ interval to $ (1/2, 1)$, respectively $ (0, 1/2)$. Also, don't use $ \in$ for inclusion of sets, and fight the urge to quote entirely the post you refer to, especially if it is the one just above Hahaha... I'm sorry my friend mavropnevma.... I just want to show that $ \mathbb{H}$ is infinite countable set. But maybe I did it in wrong way , can you please tell me what makes it wrong please, I'm just a newbie in mathematics Olympiad . Thank you @pco Yeal , all OK

pco wrote: math10 wrote: pco wrote: math10 wrote: pco wrote: @math10 : My idea is just to show that $ \mathbb H$ is infinite (obviously) and countable (since $ \subseteq Q$). Hence the result. We only have $ H$ is infinite and $ H\in Q$. How do you to get the results? (sorry for my stupid ) please write clear I'm sorry : I dont understand what you dont understand : isn't clear that "every infinite subset of a countable subset is countable" ? Is this what you want I explain more ? math10 wrote: I can't understand step: *)If $ H$ is infinite and $ H\in Q$ then there exist a bijection from $ H$ to $ Z$ pco wrote: Step 1 : every infinite subset of a countable set is infinite countable. Is step 1 OK ? Step 2 : $ \mathbb H$ is an infinite subset of the countable set $ \mathbb Q$, and so $ \mathbb H$ is infinite countable. Is step 2 OK ? Step 3 : If set A is infinite countable, there exists a bijection from A$ \to\mathbb N$ Is step 3 OK ? Step 4 : if there exists a bijection from $ A\to C$ and a bijection from $ B\to C$, there exists a bijection from $ A\to B$ Is step 4 OK ? Step 5 : $ \mathbb Z$ is infinite countable, so $ \exists$ a bijection from $ \mathbb Z\to\mathbb N$ (step 3) $ \mathbb H$ is infinite countable, so $ \exists$ a bijection from $ \mathbb H\to\mathbb N$ (step 3) So $ \exists$ a bijection from $ \mathbb H\to\mathbb Z$ OK ? For me,step 3 not clear .

math10 wrote: pco wrote: Step 3 : If set A is infinite countable, there exists a bijection from A$ \to\mathbb N$ Is step 3 OK ? For me,step 3 not clear . I think it's the definition of infinite countable. See http://en.wikipedia.org/wiki/Countable_set for example

since $H$ is a subset of $Q$,it's countable. the solution is trivial......

littletush wrote: since $H$ is a subset of $Q$,it's countable. the solution is trivial...... Ohhhh Yessssssss! Quite new very interesting idea which solve this problem in a very nice way ! How was it possible that we miss this quite nice proof ! Thanks a lot, littletush, for your so interesting posts always adding brand new proofs