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Should I write these ordinal operations in Cantor normal form?
Clash Royale CLAN TAG#URR8PPP
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I want to increment $omega^2+omega+n$ by $omega$
I.e. the operation I require is $f:omega^2cdotalpha+omegacdotbeta+ntimesomegacdotbeta_2mapsto omega^2cdotalpha+omegacdot(beta+beta_2)+n$
What's the best way to write it? I'm sure it's not:
$omega+(omega^2+omega+1)=omega^2+omegacdot2+1$
Maybe I just have to write the both in Cantor normal form and say to pairwise add terms of matching exponents?
Then I want to transform $omega+1$ to $omega^2+omega$ by increasing the exponents of every term.
but again I don't think I can just write $(omega+1)cdotomega$ so all I can come up with is to say shift the coefficients left in Cantor normal form... actually I think in this last case maybe to write $omegacdot(omega+1)$ might be okay.
ordinals
asked Jul 17 at 19:41
Robert Frost
3,8821036
add a comment |Â
up vote
-1
down vote
favorite
I want to increment $omega^2+omega+n$ by $omega$
I.e. the operation I require is $f:omega^2cdotalpha+omegacdotbeta+ntimesomegacdotbeta_2mapsto omega^2cdotalpha+omegacdot(beta+beta_2)+n$
What's the best way to write it? I'm sure it's not:
$omega+(omega^2+omega+1)=omega^2+omegacdot2+1$
Maybe I just have to write the both in Cantor normal form and say to pairwise add terms of matching exponents?
Then I want to transform $omega+1$ to $omega^2+omega$ by increasing the exponents of every term.
but again I don't think I can just write $(omega+1)cdotomega$ so all I can come up with is to say shift the coefficients left in Cantor normal form... actually I think in this last case maybe to write $omegacdot(omega+1)$ might be okay.
ordinals
asked Jul 17 at 19:41
Robert Frost
3,8821036
1
$omega+omega^2=omega^2$; recall ordinal addition is non-commutative.
– Lord Shark the Unknown
Jul 17 at 19:41
@LordSharktheUnknown as I thought
– Robert Frost
Jul 17 at 19:43
There is an ordinal calculator here sourceforge.net/projects/ord I tried it quickly seems fine to practice ordinal arithmetic.
– zwim
Jul 17 at 20:10
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I want to increment $omega^2+omega+n$ by $omega$
I.e. the operation I require is $f:omega^2cdotalpha+omegacdotbeta+ntimesomegacdotbeta_2mapsto omega^2cdotalpha+omegacdot(beta+beta_2)+n$
What's the best way to write it? I'm sure it's not:
$omega+(omega^2+omega+1)=omega^2+omegacdot2+1$
Maybe I just have to write the both in Cantor normal form and say to pairwise add terms of matching exponents?
Then I want to transform $omega+1$ to $omega^2+omega$ by increasing the exponents of every term.
but again I don't think I can just write $(omega+1)cdotomega$ so all I can come up with is to say shift the coefficients left in Cantor normal form... actually I think in this last case maybe to write $omegacdot(omega+1)$ might be okay.
ordinals
asked Jul 17 at 19:41
Robert Frost
3,8821036
I want to increment $omega^2+omega+n$ by $omega$
I.e. the operation I require is $f:omega^2cdotalpha+omegacdotbeta+ntimesomegacdotbeta_2mapsto omega^2cdotalpha+omegacdot(beta+beta_2)+n$
What's the best way to write it? I'm sure it's not:
$omega+(omega^2+omega+1)=omega^2+omegacdot2+1$
Maybe I just have to write the both in Cantor normal form and say to pairwise add terms of matching exponents?
Then I want to transform $omega+1$ to $omega^2+omega$ by increasing the exponents of every term.
but again I don't think I can just write $(omega+1)cdotomega$ so all I can come up with is to say shift the coefficients left in Cantor normal form... actually I think in this last case maybe to write $omegacdot(omega+1)$ might be okay.
ordinals
asked Jul 17 at 19:41
Robert Frost
3,8821036
edited Jul 17 at 20:22
asked Jul 17 at 19:41
Robert Frost
3,8821036
asked Jul 17 at 19:41
Robert Frost
3,8821036
asked Jul 17 at 19:41
Robert Frost
3,8821036
3,8821036
1
$omega+omega^2=omega^2$; recall ordinal addition is non-commutative.
– Lord Shark the Unknown
Jul 17 at 19:41
@LordSharktheUnknown as I thought
– Robert Frost
Jul 17 at 19:43
There is an ordinal calculator here sourceforge.net/projects/ord I tried it quickly seems fine to practice ordinal arithmetic.
– zwim
Jul 17 at 20:10
add a comment |Â
1
$omega+omega^2=omega^2$; recall ordinal addition is non-commutative.
– Lord Shark the Unknown
Jul 17 at 19:41
@LordSharktheUnknown as I thought
– Robert Frost
Jul 17 at 19:43
There is an ordinal calculator here sourceforge.net/projects/ord I tried it quickly seems fine to practice ordinal arithmetic.
– zwim
Jul 17 at 20:10
1
1
$omega+omega^2=omega^2$; recall ordinal addition is non-commutative.
– Lord Shark the Unknown
Jul 17 at 19:41
$omega+omega^2=omega^2$; recall ordinal addition is non-commutative.
– Lord Shark the Unknown
Jul 17 at 19:41
@LordSharktheUnknown as I thought
– Robert Frost
Jul 17 at 19:43
@LordSharktheUnknown as I thought
– Robert Frost
Jul 17 at 19:43
There is an ordinal calculator here sourceforge.net/projects/ord I tried it quickly seems fine to practice ordinal arithmetic.
– zwim
Jul 17 at 20:10
There is an ordinal calculator here sourceforge.net/projects/ord I tried it quickly seems fine to practice ordinal arithmetic.
– zwim
Jul 17 at 20:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I believe you are looking for the natural (or Hessenberg) sum of ordinals. In the natural sum, you simply add the coefficients in the Cantor normal form.
answered Jul 17 at 20:23
Noah Schweber
111k9140263
add a comment |Â
up vote
4
down vote
Ordinal addition and multiplication are not commutative. Incrementing $omega^2+omega+1$ by $omega$ results in $omega^2+omega+1+omega = omega^2+omega+(1+omega) =omega^2+omega+omega = omega^2+omega*2$
$(omega+1)omega$ is $omega$ copies of $omega+1$. That is not the same thing as $omega^2+omega$, which is equal to $omega(omega+1)$
answered Jul 17 at 19:53
Acccumulation
4,4732314
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I believe you are looking for the natural (or Hessenberg) sum of ordinals. In the natural sum, you simply add the coefficients in the Cantor normal form.
answered Jul 17 at 20:23
Noah Schweber
111k9140263
add a comment |Â
up vote
2
down vote
accepted
I believe you are looking for the natural (or Hessenberg) sum of ordinals. In the natural sum, you simply add the coefficients in the Cantor normal form.
answered Jul 17 at 20:23
Noah Schweber
111k9140263
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I believe you are looking for the natural (or Hessenberg) sum of ordinals. In the natural sum, you simply add the coefficients in the Cantor normal form.
answered Jul 17 at 20:23
Noah Schweber
111k9140263
I believe you are looking for the natural (or Hessenberg) sum of ordinals. In the natural sum, you simply add the coefficients in the Cantor normal form.
answered Jul 17 at 20:23
Noah Schweber
111k9140263
answered Jul 17 at 20:23
Noah Schweber
111k9140263
answered Jul 17 at 20:23
Noah Schweber
111k9140263
answered Jul 17 at 20:23
Noah Schweber
111k9140263
111k9140263
add a comment |Â
add a comment |Â
up vote
4
down vote
Ordinal addition and multiplication are not commutative. Incrementing $omega^2+omega+1$ by $omega$ results in $omega^2+omega+1+omega = omega^2+omega+(1+omega) =omega^2+omega+omega = omega^2+omega*2$
$(omega+1)omega$ is $omega$ copies of $omega+1$. That is not the same thing as $omega^2+omega$, which is equal to $omega(omega+1)$
answered Jul 17 at 19:53
Acccumulation
4,4732314
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
add a comment |Â
up vote
4
down vote
Ordinal addition and multiplication are not commutative. Incrementing $omega^2+omega+1$ by $omega$ results in $omega^2+omega+1+omega = omega^2+omega+(1+omega) =omega^2+omega+omega = omega^2+omega*2$
$(omega+1)omega$ is $omega$ copies of $omega+1$. That is not the same thing as $omega^2+omega$, which is equal to $omega(omega+1)$
answered Jul 17 at 19:53
Acccumulation
4,4732314
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Ordinal addition and multiplication are not commutative. Incrementing $omega^2+omega+1$ by $omega$ results in $omega^2+omega+1+omega = omega^2+omega+(1+omega) =omega^2+omega+omega = omega^2+omega*2$
$(omega+1)omega$ is $omega$ copies of $omega+1$. That is not the same thing as $omega^2+omega$, which is equal to $omega(omega+1)$
answered Jul 17 at 19:53
Acccumulation
4,4732314
Ordinal addition and multiplication are not commutative. Incrementing $omega^2+omega+1$ by $omega$ results in $omega^2+omega+1+omega = omega^2+omega+(1+omega) =omega^2+omega+omega = omega^2+omega*2$
$(omega+1)omega$ is $omega$ copies of $omega+1$. That is not the same thing as $omega^2+omega$, which is equal to $omega(omega+1)$
answered Jul 17 at 19:53
Acccumulation
4,4732314
answered Jul 17 at 19:53
Acccumulation
4,4732314
answered Jul 17 at 19:53
Acccumulation
4,4732314
answered Jul 17 at 19:53
Acccumulation
4,4732314
4,4732314
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
add a comment |Â
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
Thank-you. You've answered the 2nd part but for the first part I require a way to write the transformation $f(omega,omega^2+omega+1)=omega^2+2omega+1$. I'm asking if the only way is to say pairwise add corresponding coefficients in Cantor normal form or is there a better wa?.
– Robert Frost
Jul 17 at 19:56
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@RobertFrost $2omega=omega$ thus you can you add $0$.
– zwim
Jul 17 at 20:02
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
@zwim then I'm even more lost than I realised! I need to represent $omega+omega$ and it not equal $omega$ first. How do I do that? Then I can correct the question. Is it $omegacdot2$?
– Robert Frost
Jul 17 at 20:05
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
As accumulation said $(omega^2+omega+1)+(omega+1)=omega^2+omega.2+1$ is what you want.
– zwim
Jul 17 at 20:08
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
Ok I have the noncommutative part but I require a function that gives $f(omega^2cdotalpha+omegacdotbeta+n,omegacdotbeta_2)=omega^2cdotalpha+omegacdot(beta+beta_2)+n$ and the question asks whether writing out to pairwise add the coefficients in Cantor normal form is the only way.
– Robert Frost
Jul 17 at 20:15
add a comment |Â
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1
$omega+omega^2=omega^2$; recall ordinal addition is non-commutative.
– Lord Shark the Unknown
Jul 17 at 19:41
@LordSharktheUnknown as I thought
– Robert Frost
Jul 17 at 19:43
There is an ordinal calculator here sourceforge.net/projects/ord I tried it quickly seems fine to practice ordinal arithmetic.
– zwim
Jul 17 at 20:10