December 6

Does there exist any function (not as recurrence relation) for monomial order? Alireza Badali (talk) 12:16, 6 December 2017 (UTC)[reply]

I presume that you are interested in the infinite-dimensional case, and don't want to appeal to the axiom of choice? — Charles Stewart (talk) 13:36, 6 December 2017 (UTC)[reply]

I'm not interested in logic I want just make a group on ${\displaystyle \mathbb {N} }$ isomorph to the ${\displaystyle (\mathbb {Q} ,+)}$ Alireza Badali (talk) 16:33, 6 December 2017 (UTC)[reply]

IIUC, you can do what you are asking by showing that the rationals are countable and then throwing out duplicates. You don't need monomials, if that is really what you want. — Charles Stewart (talk) 16:38, 6 December 2017 (UTC)[reply]

Yes this sequence:

${\displaystyle (1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),(2,2k-3),...}$ ${\displaystyle ,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),...,(k,k),...}$ ${\displaystyle ,(2k-2,2),(2k-1,1),...}$

of course Professor Daniel Lazard gave me its formula:

${\displaystyle {\begin{cases}(a_{1},b_{1})=(1,1)\\(a_{n+1},b_{n+1})={\begin{cases}(1,a_{n}+1)&b_{n}=1\\(a_{n}+1,b_{n}-1)&{\text{else}}\end{cases}}\end{cases}}}$

but I need a formula not as recurrence relation. Alireza Badali (talk) 17:00, 6 December 2017 (UTC)[reply]

I understand what you want, but I don't see what this has to do with monomials. Also, your recurrence function does not throw out duplicates. — Charles Stewart (talk) 22:21, 6 December 2017 (UTC)[reply]

Please ignore monomial and only that sequence is important! Alireza Badali (talk) 13:11, 7 December 2017 (UTC)[reply]

I've changed the title of this question. — Charles Stewart (talk) 13:23, 7 December 2017 (UTC)[reply]

Given the recurrence you were given and thec trivial injection from the naturals into the positive rationals, you get a bijection, but you need the axiom of choice to get it directly. — Charles Stewart (talk) 13:26, 7 December 2017 (UTC)[reply]

Take a look at https://math.stackexchange.com/questions/193263/countability-of-mathbbq — Charles Stewart (talk) 13:29, 7 December 2017 (UTC)[reply]

Good title, thank you, but really what is for doing with axiom of choice here${\displaystyle \color {red}{?}\color {green}{!}\color {blue}{?}\color {fuchsia}{!}}$ Alireza Badali (talk) 13:53, 7 December 2017 (UTC)[reply]

Maybe Cantor pairing function is what you want. --JBL (talk) 00:46, 9 December 2017 (UTC)[reply]

Thank you. Alireza Badali (talk) 10:58, 9 December 2017 (UTC)[reply]

Question ${\displaystyle 1}$: Let ${\displaystyle r:\mathbb {N} \to (0,1)}$ is a function given by ${\displaystyle r(n)}$ is obtained as put a point at the beginning of ${\displaystyle n}$ like ${\displaystyle r(34880)=0.34880}$ and similarly consider ${\displaystyle \forall k\in \mathbb {N} \cup \{0\}}$ ${\displaystyle r_{k}:\mathbb {N} \to (0,1)}$ by ${\displaystyle r_{k}(n)=10^{-k}\cdot r(n)}$ and ${\displaystyle \mathbb {P} }$ is the set prime numbers now is ${\displaystyle \{t^{i}\,|\,t\in r(\mathbb {P} )\}}$ dense in the ${\displaystyle \{z\in \mathbb {C} \,|\,|z|=1\}}$ and is ${\displaystyle \{s\cdot t^{i}\,|\,s,t\in \bigcup _{k\in \mathbb {N} \cup \{0\}}r_{k}(\mathbb {P} )\}}$ dense in the ${\displaystyle \{z\in \mathbb {C} \,|\,0.1\leq |z|\leq 1\}}$.

Question ${\displaystyle 2}$: Is ${\displaystyle \{(p^{2}+q^{2})^{-0.5}\cdot (p+qi)\,|\,p,q\in \mathbb {P} \}}$ dense in the ${\displaystyle \{z\in \mathbb {C} \,|\,|z|=1,Re(z)\geq 0,Im(z)\geq 0\}}$. Alireza Badali (talk) 19:02, 6 December 2017 (UTC)[reply]

r(1)=0.1, and r(01)=0.01, but 1=01, so r is not a function. Bo Jacoby (talk) 22:41, 6 December 2017 (UTC).[reply]

That seems like a cheap shot. You could equally argue that 5 = 101₂ but r(5) = 0.5 ≠ r(101₂) = 0.101₂ = 0.625. Each natural number has a unique base-10 representation without leading zeros, and a charitable reading of this question has n represented thusly. -- ToE 02:41, 7 December 2017 (UTC)[reply]

What is ${\displaystyle 01\color {red}{?!}}$ Alireza Badali (talk) 13:39, 7 December 2017 (UTC)[reply]

Is that stated quite how you wish? Forget for a moment r(ℙ). Can you think of any t ∈ (0,1) such that tⁱ is close to -1? -- ToE 03:02, 7 December 2017 (UTC)[reply]

Or, keeping r(ℙ) in mind, is there likely to be any t ∈ r(ℙ) such that tⁱ is close to 1? -- ToE 03:18, 7 December 2017 (UTC)[reply]

And why are you complicating things by projecting this into the complex plane. Isn't what you are really getting at asking if r(ℙ) is dense in (0.1,1) and if {p/q | p, q ∈ ℙ} is dense in ℝ⁺? -- ToE 03:24, 7 December 2017 (UTC)[reply]

Complex plane wasn't my idea and a Professor (that likes be unknown) offered it to me but I don't know how should I operate with complex plane. Alireza Badali (talk) 13:39, 7 December 2017 (UTC)[reply]

How would you address the real number, "less complex", related problems I offered? How would you then draw a connection from there to the complex plane? (See Dense set#Properties.) For extra credit, determine what is wrong with your professor's statement of the first problem, correct it, then solve it. -- ToE 14:16, 7 December 2017 (UTC)[reply]

The proved theorem for reals is: Let ${\displaystyle \mathbb {P} }$ is the set prime numbers and ${\displaystyle S}$ is a set that has been made as below: put a point at the beginning of each member of ${\displaystyle \mathbb {P} }$ like ${\displaystyle 0.2}$ or ${\displaystyle 0.19}$ then ${\displaystyle S=\{0.2,0.3,0.5,0.7,...\}}$ is dense in the interval ${\displaystyle [0.1,1]}$ of real numbers. that its proved extension is: For each subinterval of ${\displaystyle [0.1,1)}$ like ${\displaystyle (a,b),\,\exists m\in \mathbb {N} }$ that ${\displaystyle \forall k\in \mathbb {N} }$ with ${\displaystyle k\geq m}$ then ${\displaystyle \exists t\in (a,b)}$ that ${\displaystyle t\cdot 10^{k}\in \mathbb {P} }$.

and this: Assume ${\displaystyle S_{1}=\{a/10^{n}\,|\,a\in S}$ for ${\displaystyle n=0,1,2,3,...\}}$ then ${\displaystyle S_{1}}$ is dense in the interval ${\displaystyle (0,1)}$ and ${\displaystyle S_{1}\times S_{1}}$ is dense in the ${\displaystyle (0,1)\times (0,1)}$.

But I want make a form for sets ${\displaystyle S}$ & ${\displaystyle S_{1}}$ & ${\displaystyle L}$ in the complex plane and below conjecture that is an equivalent to Goldbach conjectureas: Let ${\displaystyle L=\{(a,b)\,|\,a,b\in S_{1}}$ & ${\displaystyle 0.01\leq a+b<0.1}$ & ${\displaystyle \exists m\in \mathbb {N} ,\,a\cdot 10^{m},\,b\cdot 10^{m}}$ are prime numbers & ${\displaystyle a\cdot 10^{m}\neq 2\neq b\cdot 10^{m}\}}$ but conjecture: For each even natural number like ${\displaystyle t=t_{1}t_{2}t_{3}...t_{k}}$, then ${\displaystyle \exists (a,b),(b,a)\in L\cap }$ ${\displaystyle \{(x,y)\,|}$ ${\displaystyle x+y=0.0t_{1}t_{2}t_{3}...t_{k}\}}$ such that ${\displaystyle 0.0t_{1}t_{2}t_{3}...t_{k}=a+b}$ & ${\displaystyle 10^{k+1}\cdot a,10^{k+1}\cdot b}$ are prime numbers. Alireza Badali (talk) 16:44, 7 December 2017 (UTC)[reply]

My questions were rhetorical, intended as guiding hints. (We really don't need or want to see your proof.) If you have the solution in ℝ, then look and see how the problem is mapped over to ℂ and then apply Dense set#Properties. Do you understand what tⁱ does? -- ToE 17:01, 7 December 2017 (UTC)[reply]

Indeed I have no plan for complex plane but I can solve it but main difficulty for me is this question. Alireza Badali (talk) 18:24, 7 December 2017 (UTC)[reply]

• On the top of this page, it says We don't do your homework for you, though we’ll help you past the stuck point. This looks plainly like a homework question (though an advanced one). What have you tried so far? Tigraan^{Click here to contact me} 10:45, 7 December 2017 (UTC)[reply]

My notes is on real numbers but a Professor (that likes be unknown) offered it to me but I don't know how should I operate with complex plane. Alireza Badali (talk) 13:39, 7 December 2017 (UTC)[reply]

Thank you so much guys. Alireza Badali (talk) 18:24, 7 December 2017 (UTC)[reply]