left-cancellative ordinal between infinite and finite ordernals [on hold]












0














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










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put on hold as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz 2 days ago


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.













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










  • how can you do induction on n when you also need to take care of m ?
    – Amit Perelman
    2 days ago










  • Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    – Asaf Karagila
    2 days ago












  • ok. will look intp it, thx
    – Amit Perelman
    2 days ago
















0














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










share|cite|improve this question













put on hold as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz 2 days ago


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.













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










  • how can you do induction on n when you also need to take care of m ?
    – Amit Perelman
    2 days ago










  • Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    – Asaf Karagila
    2 days ago












  • ok. will look intp it, thx
    – Amit Perelman
    2 days ago














0












0








0







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










share|cite|improve this question













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













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




put on hold as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz 2 days ago


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.




put on hold as off-topic by Andrés E. Caicedo, Cesareo, José Carlos Santos, amWhy, A. Pongrácz 2 days ago


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.












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










  • how can you do induction on n when you also need to take care of m ?
    – Amit Perelman
    2 days ago










  • Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    – Asaf Karagila
    2 days ago












  • ok. will look intp it, thx
    – Amit Perelman
    2 days ago


















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










  • how can you do induction on n when you also need to take care of m ?
    – Amit Perelman
    2 days ago










  • Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
    – Asaf Karagila
    2 days ago












  • ok. will look intp it, thx
    – Amit Perelman
    2 days ago
















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




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












how can you do induction on n when you also need to take care of m ?
– Amit Perelman
2 days ago




how can you do induction on n when you also need to take care of m ?
– Amit Perelman
2 days ago












Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
– Asaf Karagila
2 days ago






Prove by induction on $n$ that for all $m$, if $omegacdot 2+n=omegacdot 2+m$, then $n=m$.
– Asaf Karagila
2 days ago














ok. will look intp it, thx
– Amit Perelman
2 days ago




ok. will look intp it, thx
– Amit Perelman
2 days ago










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