left-cancellative ordinal between infinite and finite ordernals [closed]
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$begingroup$
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
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
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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.
add a comment |
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
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
$endgroup$
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
add a comment |
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
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
$endgroup$
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
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
asked Jan 5 at 23:45
Amit PerelmanAmit Perelman
1
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
add a comment |
$begingroup$
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
add a comment |
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mIC8EMY,r 65mm,bhqQ g3YQhNG9zGS YDzF6jKHli,3OeGHDgMek0NuShR79sbqSjsBvvU9WBzY ZPJ Cs
$begingroup$
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