Let $gamma,alphaneq 0$ be ordinals. Then there exists a unique ordinal $beta$ and a unique $rho<alpha$...
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Let $gamma,alphaneq 0$ be ordinals. Then there exists a unique ordinal $beta$ and a unique $rho<alpha$ such that $gamma=alphacdotbeta+rho$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $beta=max{delta in {rm Ord} midalphacdotdeltalegamma}$. Since $alphacdotbetalegamma$, there is a unique $rho$ such that $gamma=alphacdotbeta+rho$. We have $rho<alpha$. If not, $alphalerhoimpliesalphacdot(beta+1)=alphacdotbeta+alphalealphacdotbeta+rho=gammaimplies$ $alphacdot(beta+1)legamma$. This contradicts the maximality of $beta$.
To prove uniqueness, let $gamma=alphacdotbeta_1+rho_1 =alphacdotbeta_2+rho_2$ with $rho_1,rho_2<alpha$. Assume that $beta_1<beta_2impliesbeta_1+1lebeta_2impliesalphacdot(beta_1+1)lealphacdotbeta_2$. Then $alphacdotbeta_1+rho_1<alphacdotbeta_1+alpha=$ $alphacdot(beta_1+1)lealphacdotbeta_2lealphacdotbeta_2+rho_2impliesalphacdotbeta_1+rho_1<alphacdotbeta_2lealphacdotbeta_2impliesgamma<gamma$. This is a contradiction and thus $beta_1=beta_2$. It follows immediately that $rho_1=rho_2$.
proof-verification elementary-set-theory ordinals
asked Nov 18 at 10:49
Le Anh Dung
9701421
add a comment |
up vote
0
down vote
favorite
Let $gamma,alphaneq 0$ be ordinals. Then there exists a unique ordinal $beta$ and a unique $rho<alpha$ such that $gamma=alphacdotbeta+rho$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $beta=max{delta in {rm Ord} midalphacdotdeltalegamma}$. Since $alphacdotbetalegamma$, there is a unique $rho$ such that $gamma=alphacdotbeta+rho$. We have $rho<alpha$. If not, $alphalerhoimpliesalphacdot(beta+1)=alphacdotbeta+alphalealphacdotbeta+rho=gammaimplies$ $alphacdot(beta+1)legamma$. This contradicts the maximality of $beta$.
To prove uniqueness, let $gamma=alphacdotbeta_1+rho_1 =alphacdotbeta_2+rho_2$ with $rho_1,rho_2<alpha$. Assume that $beta_1<beta_2impliesbeta_1+1lebeta_2impliesalphacdot(beta_1+1)lealphacdotbeta_2$. Then $alphacdotbeta_1+rho_1<alphacdotbeta_1+alpha=$ $alphacdot(beta_1+1)lealphacdotbeta_2lealphacdotbeta_2+rho_2impliesalphacdotbeta_1+rho_1<alphacdotbeta_2lealphacdotbeta_2impliesgamma<gamma$. This is a contradiction and thus $beta_1=beta_2$. It follows immediately that $rho_1=rho_2$.
proof-verification elementary-set-theory ordinals
asked Nov 18 at 10:49
Le Anh Dung
9701421
Your first step requires proving a set is not empty. There is only one $beta$ that is the maximum of that set. With ordinals a + b = a + c implies b = c is not simply a matter of subtraction.
– William Elliot
Nov 18 at 11:58
Hi @WilliamElliot! I actually proved those results before and i take that for granted. Besides these points, have you found any mistake/gap in my proof?
– Le Anh Dung
Nov 18 at 12:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $gamma,alphaneq 0$ be ordinals. Then there exists a unique ordinal $beta$ and a unique $rho<alpha$ such that $gamma=alphacdotbeta+rho$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $beta=max{delta in {rm Ord} midalphacdotdeltalegamma}$. Since $alphacdotbetalegamma$, there is a unique $rho$ such that $gamma=alphacdotbeta+rho$. We have $rho<alpha$. If not, $alphalerhoimpliesalphacdot(beta+1)=alphacdotbeta+alphalealphacdotbeta+rho=gammaimplies$ $alphacdot(beta+1)legamma$. This contradicts the maximality of $beta$.
To prove uniqueness, let $gamma=alphacdotbeta_1+rho_1 =alphacdotbeta_2+rho_2$ with $rho_1,rho_2<alpha$. Assume that $beta_1<beta_2impliesbeta_1+1lebeta_2impliesalphacdot(beta_1+1)lealphacdotbeta_2$. Then $alphacdotbeta_1+rho_1<alphacdotbeta_1+alpha=$ $alphacdot(beta_1+1)lealphacdotbeta_2lealphacdotbeta_2+rho_2impliesalphacdotbeta_1+rho_1<alphacdotbeta_2lealphacdotbeta_2impliesgamma<gamma$. This is a contradiction and thus $beta_1=beta_2$. It follows immediately that $rho_1=rho_2$.
proof-verification elementary-set-theory ordinals
asked Nov 18 at 10:49
Le Anh Dung
9701421
Let $gamma,alphaneq 0$ be ordinals. Then there exists a unique ordinal $beta$ and a unique $rho<alpha$ such that $gamma=alphacdotbeta+rho$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $beta=max{delta in {rm Ord} midalphacdotdeltalegamma}$. Since $alphacdotbetalegamma$, there is a unique $rho$ such that $gamma=alphacdotbeta+rho$. We have $rho<alpha$. If not, $alphalerhoimpliesalphacdot(beta+1)=alphacdotbeta+alphalealphacdotbeta+rho=gammaimplies$ $alphacdot(beta+1)legamma$. This contradicts the maximality of $beta$.
To prove uniqueness, let $gamma=alphacdotbeta_1+rho_1 =alphacdotbeta_2+rho_2$ with $rho_1,rho_2<alpha$. Assume that $beta_1<beta_2impliesbeta_1+1lebeta_2impliesalphacdot(beta_1+1)lealphacdotbeta_2$. Then $alphacdotbeta_1+rho_1<alphacdotbeta_1+alpha=$ $alphacdot(beta_1+1)lealphacdotbeta_2lealphacdotbeta_2+rho_2impliesalphacdotbeta_1+rho_1<alphacdotbeta_2lealphacdotbeta_2impliesgamma<gamma$. This is a contradiction and thus $beta_1=beta_2$. It follows immediately that $rho_1=rho_2$.
proof-verification elementary-set-theory ordinals
proof-verification elementary-set-theory ordinals
asked Nov 18 at 10:49
Le Anh Dung
9701421
asked Nov 18 at 10:49
Le Anh Dung
9701421
asked Nov 18 at 10:49
Le Anh Dung
9701421
asked Nov 18 at 10:49
Le Anh Dung
9701421
asked Nov 18 at 10:49
Le Anh Dung
9701421
9701421
Your first step requires proving a set is not empty. There is only one $beta$ that is the maximum of that set. With ordinals a + b = a + c implies b = c is not simply a matter of subtraction.
– William Elliot
Nov 18 at 11:58
Hi @WilliamElliot! I actually proved those results before and i take that for granted. Besides these points, have you found any mistake/gap in my proof?
– Le Anh Dung
Nov 18 at 12:27
add a comment |
Your first step requires proving a set is not empty. There is only one $beta$ that is the maximum of that set. With ordinals a + b = a + c implies b = c is not simply a matter of subtraction.
– William Elliot
Nov 18 at 11:58
Hi @WilliamElliot! I actually proved those results before and i take that for granted. Besides these points, have you found any mistake/gap in my proof?
– Le Anh Dung
Nov 18 at 12:27
Your first step requires proving a set is not empty. There is only one $beta$ that is the maximum of that set. With ordinals a + b = a + c implies b = c is not simply a matter of subtraction.
– William Elliot
Nov 18 at 11:58
Your first step requires proving a set is not empty. There is only one $beta$ that is the maximum of that set. With ordinals a + b = a + c implies b = c is not simply a matter of subtraction.
– William Elliot
Nov 18 at 11:58
Hi @WilliamElliot! I actually proved those results before and i take that for granted. Besides these points, have you found any mistake/gap in my proof?
– Le Anh Dung
Nov 18 at 12:27
Hi @WilliamElliot! I actually proved those results before and i take that for granted. Besides these points, have you found any mistake/gap in my proof?
– Le Anh Dung
Nov 18 at 12:27
add a comment |
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Your first step requires proving a set is not empty. There is only one $beta$ that is the maximum of that set. With ordinals a + b = a + c implies b = c is not simply a matter of subtraction.
– William Elliot
Nov 18 at 11:58
Hi @WilliamElliot! I actually proved those results before and i take that for granted. Besides these points, have you found any mistake/gap in my proof?
– Le Anh Dung
Nov 18 at 12:27