Ordinal arithmetic with multiple brackets
So, I understand that multiplying ordinals has a distributive law on the left such that:
$alphacdot(beta+gamma)=alphacdotbeta+alphacdotgamma$
What I am struggling with is if $alpha$ is also another set of brackets, or even if there are three sets. So for example, how would you work out:
$(omegacdot2+1)(omegacdot3+3)(omegacdot4+7)$
elementary-set-theory ordinals
edited Nov 25 at 10:16
Holo
5,4962929
asked Nov 24 at 18:15
Fraiku
102
add a comment |
So, I understand that multiplying ordinals has a distributive law on the left such that:
$alphacdot(beta+gamma)=alphacdotbeta+alphacdotgamma$
What I am struggling with is if $alpha$ is also another set of brackets, or even if there are three sets. So for example, how would you work out:
$(omegacdot2+1)(omegacdot3+3)(omegacdot4+7)$
elementary-set-theory ordinals
edited Nov 25 at 10:16
Holo
5,4962929
asked Nov 24 at 18:15
Fraiku
102
2
First you check whether ordinal multiplication is associative.
– Andrés E. Caicedo
Nov 24 at 18:22
add a comment |
So, I understand that multiplying ordinals has a distributive law on the left such that:
$alphacdot(beta+gamma)=alphacdotbeta+alphacdotgamma$
What I am struggling with is if $alpha$ is also another set of brackets, or even if there are three sets. So for example, how would you work out:
$(omegacdot2+1)(omegacdot3+3)(omegacdot4+7)$
elementary-set-theory ordinals
edited Nov 25 at 10:16
Holo
5,4962929
asked Nov 24 at 18:15
Fraiku
102
So, I understand that multiplying ordinals has a distributive law on the left such that:
$alphacdot(beta+gamma)=alphacdotbeta+alphacdotgamma$
What I am struggling with is if $alpha$ is also another set of brackets, or even if there are three sets. So for example, how would you work out:
$(omegacdot2+1)(omegacdot3+3)(omegacdot4+7)$
elementary-set-theory ordinals
elementary-set-theory ordinals
edited Nov 25 at 10:16
Holo
5,4962929
asked Nov 24 at 18:15
Fraiku
102
edited Nov 25 at 10:16
Holo
5,4962929
asked Nov 24 at 18:15
Fraiku
102
edited Nov 25 at 10:16
Holo
5,4962929
edited Nov 25 at 10:16
Holo
5,4962929
edited Nov 25 at 10:16
Holo
5,4962929
5,4962929
asked Nov 24 at 18:15
Fraiku
102
asked Nov 24 at 18:15
Fraiku
102
asked Nov 24 at 18:15
Fraiku
102
102
2
First you check whether ordinal multiplication is associative.
– Andrés E. Caicedo
Nov 24 at 18:22
add a comment |
2
First you check whether ordinal multiplication is associative.
– Andrés E. Caicedo
Nov 24 at 18:22
2
2
First you check whether ordinal multiplication is associative.
– Andrés E. Caicedo
Nov 24 at 18:22
First you check whether ordinal multiplication is associative.
– Andrés E. Caicedo
Nov 24 at 18:22
add a comment |
1 Answer
1
active
oldest
votes
Simply apply it repeatedly: for example, $(wcdot 2 + 1)$ is an ordinal, so you have $(wcdot 2+1)(wcdot 3 + 3) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)$
and this is again an ordinal, so we have
$(wcdot 2 + 1)(wcdot3 + 3)(wcdot 4+7) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot (wcdot 4) + ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot 7$.
You can then simplify that using the other properties of ordinal arithmetic (it would be far more efficient to do this as you go along).
answered Nov 24 at 18:25
user3482749
2,421414
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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Simply apply it repeatedly: for example, $(wcdot 2 + 1)$ is an ordinal, so you have $(wcdot 2+1)(wcdot 3 + 3) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)$
and this is again an ordinal, so we have
$(wcdot 2 + 1)(wcdot3 + 3)(wcdot 4+7) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot (wcdot 4) + ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot 7$.
You can then simplify that using the other properties of ordinal arithmetic (it would be far more efficient to do this as you go along).
answered Nov 24 at 18:25
user3482749
2,421414
add a comment |
Simply apply it repeatedly: for example, $(wcdot 2 + 1)$ is an ordinal, so you have $(wcdot 2+1)(wcdot 3 + 3) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)$
and this is again an ordinal, so we have
$(wcdot 2 + 1)(wcdot3 + 3)(wcdot 4+7) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot (wcdot 4) + ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot 7$.
You can then simplify that using the other properties of ordinal arithmetic (it would be far more efficient to do this as you go along).
answered Nov 24 at 18:25
user3482749
2,421414
add a comment |
Simply apply it repeatedly: for example, $(wcdot 2 + 1)$ is an ordinal, so you have $(wcdot 2+1)(wcdot 3 + 3) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)$
and this is again an ordinal, so we have
$(wcdot 2 + 1)(wcdot3 + 3)(wcdot 4+7) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot (wcdot 4) + ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot 7$.
You can then simplify that using the other properties of ordinal arithmetic (it would be far more efficient to do this as you go along).
answered Nov 24 at 18:25
user3482749
2,421414
Simply apply it repeatedly: for example, $(wcdot 2 + 1)$ is an ordinal, so you have $(wcdot 2+1)(wcdot 3 + 3) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)$
and this is again an ordinal, so we have
$(wcdot 2 + 1)(wcdot3 + 3)(wcdot 4+7) = ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot (wcdot 4) + ((wcdot 2 + 1)cdot (wcdot 3) + (wcdot 2+1)cdot3)cdot 7$.
You can then simplify that using the other properties of ordinal arithmetic (it would be far more efficient to do this as you go along).
answered Nov 24 at 18:25
user3482749
2,421414
answered Nov 24 at 18:25
user3482749
2,421414
answered Nov 24 at 18:25
user3482749
2,421414
answered Nov 24 at 18:25
user3482749
2,421414
2,421414
add a comment |
add a comment |
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2
First you check whether ordinal multiplication is associative.
– Andrés E. Caicedo
Nov 24 at 18:22