left-cancellative ordinal between infinite and finite ordernals [closed]












0












$begingroup$


I am trying to prove that for any n,m∈N, ω*2+n=ω*2+m iff n=m.
proving it assuming m=n is easy, cannot prove the other direction






set-theory ordinals







share|cite|improve this question









$endgroup$



closed as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz Jan 6 at 21:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Have you tried induction on $n$?
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:35










  • $begingroup$
    how can you do induction on n when you also need to take care of m ?
    $endgroup$
    – Amit Perelman
    Jan 6 at 6:58










  • $begingroup$
    Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 9:25












  • $begingroup$
    ok. will look intp it, thx
    $endgroup$
    – Amit Perelman
    Jan 6 at 10:18
















0












$begingroup$


I am trying to prove that for any n,m∈N, ω*2+n=ω*2+m iff n=m.
proving it assuming m=n is easy, cannot prove the other direction






set-theory ordinals







share|cite|improve this question









$endgroup$



closed as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz Jan 6 at 21:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    Have you tried induction on $n$?
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:35










  • $begingroup$
    how can you do induction on n when you also need to take care of m ?
    $endgroup$
    – Amit Perelman
    Jan 6 at 6:58










  • $begingroup$
    Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 9:25












  • $begingroup$
    ok. will look intp it, thx
    $endgroup$
    – Amit Perelman
    Jan 6 at 10:18














0












0








0





$begingroup$


I am trying to prove that for any n,m∈N, ω*2+n=ω*2+m iff n=m.
proving it assuming m=n is easy, cannot prove the other direction






set-theory ordinals







share|cite|improve this question









$endgroup$




I am trying to prove that for any n,m∈N, ω*2+n=ω*2+m iff n=m.
proving it assuming m=n is easy, cannot prove the other direction






set-theory ordinals




set-theory ordinals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 23:45









Amit PerelmanAmit Perelman

1




1




closed as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz Jan 6 at 21:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz Jan 6 at 21:01


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Have you tried induction on $n$?
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:35










  • $begingroup$
    how can you do induction on n when you also need to take care of m ?
    $endgroup$
    – Amit Perelman
    Jan 6 at 6:58










  • $begingroup$
    Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 9:25












  • $begingroup$
    ok. will look intp it, thx
    $endgroup$
    – Amit Perelman
    Jan 6 at 10:18


















  • $begingroup$
    Have you tried induction on $n$?
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 0:35










  • $begingroup$
    how can you do induction on n when you also need to take care of m ?
    $endgroup$
    – Amit Perelman
    Jan 6 at 6:58










  • $begingroup$
    Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    $endgroup$
    – Asaf Karagila♦
    Jan 6 at 9:25












  • $begingroup$
    ok. will look intp it, thx
    $endgroup$
    – Amit Perelman
    Jan 6 at 10:18
















$begingroup$
Have you tried induction on $n$?
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:35




$begingroup$
Have you tried induction on $n$?
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:35












$begingroup$
how can you do induction on n when you also need to take care of m ?
$endgroup$
– Amit Perelman
Jan 6 at 6:58




$begingroup$
how can you do induction on n when you also need to take care of m ?
$endgroup$
– Amit Perelman
Jan 6 at 6:58












$begingroup$
Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
$endgroup$
– Asaf Karagila♦
Jan 6 at 9:25






$begingroup$
Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
$endgroup$
– Asaf Karagila♦
Jan 6 at 9:25














$begingroup$
ok. will look intp it, thx
$endgroup$
– Amit Perelman
Jan 6 at 10:18




$begingroup$
ok. will look intp it, thx
$endgroup$
– Amit Perelman
Jan 6 at 10:18










0






active

oldest

votes

















0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes

-

Popular posts from this blog

An IMO inspired problem

Image
14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
Read more

Management

Image
Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
Read more

Has there ever been an instance of an active nuclear power plant within or near a war zone?

Image
36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
Read more