Ordinal multiplication
$begingroup$
I understand why $(omega+1)cdot2 = omegacdot2+1$ and why $(omega+1)cdotomega = omega^2$
What I am struggling with is something along the lines of:
$(omegacdot3+4)cdot3$ which I think is = $omegacdot9+4$
And:
$(omegacdot3+4)(omegacdot3)$ which I think is = $omega^2cdot9$
Am I correct?
elementary-set-theory ordinals
edited Nov 25 '18 at 10:15
Holo
5,60321030
asked Nov 24 '18 at 21:41
FraikuFraiku
102
$endgroup$
add a comment |
$begingroup$
I understand why $(omega+1)cdot2 = omegacdot2+1$ and why $(omega+1)cdotomega = omega^2$
What I am struggling with is something along the lines of:
$(omegacdot3+4)cdot3$ which I think is = $omegacdot9+4$
And:
$(omegacdot3+4)(omegacdot3)$ which I think is = $omega^2cdot9$
Am I correct?
elementary-set-theory ordinals
edited Nov 25 '18 at 10:15
Holo
5,60321030
asked Nov 24 '18 at 21:41
FraikuFraiku
102
$endgroup$
$begingroup$
Could you explain how to evaluate them?
$endgroup$
– Hanul Jeon
Nov 25 '18 at 5:08
$begingroup$
(w.3 + 4).3 = (w.3 + 4) + (w.3 + 4) + (w.3 + 4) = w.3 + (4 + w.3) + (4 + w.3) + 4 = w.3 + w.3 + w.3 + 4 = w.9 + 4 ??
$endgroup$
– Fraiku
Nov 25 '18 at 9:22
$begingroup$
Your first evaluation is correct; however the second one is not valid.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 9:27
$begingroup$
Thank you. I was following this explanation for (w + 1).w^2 and trying to extrapolate, but I'm not sure where I've gone wrong. Can you clarify?
$endgroup$
– Fraiku
Nov 25 '18 at 9:41
add a comment |
$begingroup$
I understand why $(omega+1)cdot2 = omegacdot2+1$ and why $(omega+1)cdotomega = omega^2$
What I am struggling with is something along the lines of:
$(omegacdot3+4)cdot3$ which I think is = $omegacdot9+4$
And:
$(omegacdot3+4)(omegacdot3)$ which I think is = $omega^2cdot9$
Am I correct?
elementary-set-theory ordinals
edited Nov 25 '18 at 10:15
Holo
5,60321030
asked Nov 24 '18 at 21:41
FraikuFraiku
102
$endgroup$
I understand why $(omega+1)cdot2 = omegacdot2+1$ and why $(omega+1)cdotomega = omega^2$
What I am struggling with is something along the lines of:
$(omegacdot3+4)cdot3$ which I think is = $omegacdot9+4$
And:
$(omegacdot3+4)(omegacdot3)$ which I think is = $omega^2cdot9$
Am I correct?
elementary-set-theory ordinals
elementary-set-theory ordinals
edited Nov 25 '18 at 10:15
Holo
5,60321030
asked Nov 24 '18 at 21:41
FraikuFraiku
102
edited Nov 25 '18 at 10:15
Holo
5,60321030
asked Nov 24 '18 at 21:41
FraikuFraiku
102
edited Nov 25 '18 at 10:15
Holo
5,60321030
edited Nov 25 '18 at 10:15
Holo
5,60321030
edited Nov 25 '18 at 10:15
Holo
5,60321030
5,60321030
asked Nov 24 '18 at 21:41
FraikuFraiku
102
asked Nov 24 '18 at 21:41
FraikuFraiku
102
asked Nov 24 '18 at 21:41
FraikuFraiku
102
102
$begingroup$
Could you explain how to evaluate them?
$endgroup$
– Hanul Jeon
Nov 25 '18 at 5:08
$begingroup$
(w.3 + 4).3 = (w.3 + 4) + (w.3 + 4) + (w.3 + 4) = w.3 + (4 + w.3) + (4 + w.3) + 4 = w.3 + w.3 + w.3 + 4 = w.9 + 4 ??
$endgroup$
– Fraiku
Nov 25 '18 at 9:22
$begingroup$
Your first evaluation is correct; however the second one is not valid.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 9:27
$begingroup$
Thank you. I was following this explanation for (w + 1).w^2 and trying to extrapolate, but I'm not sure where I've gone wrong. Can you clarify?
$endgroup$
– Fraiku
Nov 25 '18 at 9:41
add a comment |
$begingroup$
Could you explain how to evaluate them?
$endgroup$
– Hanul Jeon
Nov 25 '18 at 5:08
$begingroup$
(w.3 + 4).3 = (w.3 + 4) + (w.3 + 4) + (w.3 + 4) = w.3 + (4 + w.3) + (4 + w.3) + 4 = w.3 + w.3 + w.3 + 4 = w.9 + 4 ??
$endgroup$
– Fraiku
Nov 25 '18 at 9:22
$begingroup$
Your first evaluation is correct; however the second one is not valid.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 9:27
$begingroup$
Thank you. I was following this explanation for (w + 1).w^2 and trying to extrapolate, but I'm not sure where I've gone wrong. Can you clarify?
$endgroup$
– Fraiku
Nov 25 '18 at 9:41
$begingroup$
Could you explain how to evaluate them?
$endgroup$
– Hanul Jeon
Nov 25 '18 at 5:08
$begingroup$
Could you explain how to evaluate them?
$endgroup$
– Hanul Jeon
Nov 25 '18 at 5:08
$begingroup$
(w.3 + 4).3 = (w.3 + 4) + (w.3 + 4) + (w.3 + 4) = w.3 + (4 + w.3) + (4 + w.3) + 4 = w.3 + w.3 + w.3 + 4 = w.9 + 4 ??
$endgroup$
– Fraiku
Nov 25 '18 at 9:22
$begingroup$
(w.3 + 4).3 = (w.3 + 4) + (w.3 + 4) + (w.3 + 4) = w.3 + (4 + w.3) + (4 + w.3) + 4 = w.3 + w.3 + w.3 + 4 = w.9 + 4 ??
$endgroup$
– Fraiku
Nov 25 '18 at 9:22
$begingroup$
Your first evaluation is correct; however the second one is not valid.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 9:27
$begingroup$
Your first evaluation is correct; however the second one is not valid.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 9:27
$begingroup$
Thank you. I was following this explanation for (w + 1).w^2 and trying to extrapolate, but I'm not sure where I've gone wrong. Can you clarify?
$endgroup$
– Fraiku
Nov 25 '18 at 9:41
$begingroup$
Thank you. I was following this explanation for (w + 1).w^2 and trying to extrapolate, but I'm not sure where I've gone wrong. Can you clarify?
$endgroup$
– Fraiku
Nov 25 '18 at 9:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have evaluated $(omegacdot 3+4)cdot 3$ correctly. However, $(omegacdot 3+4)cdot(omegacdot 3)$ is not correctly evaluated. I do not know which step of your evaluation is incorrect, but I guess you think adding $(omegacdot 3)$ $(omegacdot 3)$ times is $omega^2cdot 9$.
This is not true, however, as
$$(omegacdot 3)cdot (omegacdot 3) = omegacdot (3cdotomega)cdot 3 = omega^2cdot 3.$$
The core part of the evaluation is $3cdotomega=omega$ that can be proven in the way of proving $3+omega =omega$ (and the linked question in your comment.)
We would think evaluating $(omegacdot 3+4)cdot(omegacdot 3)$ is not too different from that of $(omegacdot 3)cdot (omegacdot 3)$. The problem is, we have extra $+4$ and we cannot apply the distribution law in the case.
However, the addition in the middle of the expression would disappear when adding $(omegacdot 3+4)$ many times, so we can guess the answer is $omega^2cdot 3$.
You can make our discussion concretely by following the steps of the proof you have linked; that is, use the transfinite induction for $betale omegacdot 3$ to prove
$$(omegacdot 3+4)cdotbeta = begin{cases}
omegacdot (3cdotbeta) &text{if $beta$ is a limit ordinal,} \ omegacdot (3cdotbeta) + 4 & text{if $beta$ is a successor ordinal.}end{cases}$$
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
$endgroup$
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
|
show 3 more comments
Your Answer
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1 Answer
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$begingroup$
You have evaluated $(omegacdot 3+4)cdot 3$ correctly. However, $(omegacdot 3+4)cdot(omegacdot 3)$ is not correctly evaluated. I do not know which step of your evaluation is incorrect, but I guess you think adding $(omegacdot 3)$ $(omegacdot 3)$ times is $omega^2cdot 9$.
This is not true, however, as
$$(omegacdot 3)cdot (omegacdot 3) = omegacdot (3cdotomega)cdot 3 = omega^2cdot 3.$$
The core part of the evaluation is $3cdotomega=omega$ that can be proven in the way of proving $3+omega =omega$ (and the linked question in your comment.)
We would think evaluating $(omegacdot 3+4)cdot(omegacdot 3)$ is not too different from that of $(omegacdot 3)cdot (omegacdot 3)$. The problem is, we have extra $+4$ and we cannot apply the distribution law in the case.
However, the addition in the middle of the expression would disappear when adding $(omegacdot 3+4)$ many times, so we can guess the answer is $omega^2cdot 3$.
You can make our discussion concretely by following the steps of the proof you have linked; that is, use the transfinite induction for $betale omegacdot 3$ to prove
$$(omegacdot 3+4)cdotbeta = begin{cases}
omegacdot (3cdotbeta) &text{if $beta$ is a limit ordinal,} \ omegacdot (3cdotbeta) + 4 & text{if $beta$ is a successor ordinal.}end{cases}$$
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
$endgroup$
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
|
show 3 more comments
$begingroup$
You have evaluated $(omegacdot 3+4)cdot 3$ correctly. However, $(omegacdot 3+4)cdot(omegacdot 3)$ is not correctly evaluated. I do not know which step of your evaluation is incorrect, but I guess you think adding $(omegacdot 3)$ $(omegacdot 3)$ times is $omega^2cdot 9$.
This is not true, however, as
$$(omegacdot 3)cdot (omegacdot 3) = omegacdot (3cdotomega)cdot 3 = omega^2cdot 3.$$
The core part of the evaluation is $3cdotomega=omega$ that can be proven in the way of proving $3+omega =omega$ (and the linked question in your comment.)
We would think evaluating $(omegacdot 3+4)cdot(omegacdot 3)$ is not too different from that of $(omegacdot 3)cdot (omegacdot 3)$. The problem is, we have extra $+4$ and we cannot apply the distribution law in the case.
However, the addition in the middle of the expression would disappear when adding $(omegacdot 3+4)$ many times, so we can guess the answer is $omega^2cdot 3$.
You can make our discussion concretely by following the steps of the proof you have linked; that is, use the transfinite induction for $betale omegacdot 3$ to prove
$$(omegacdot 3+4)cdotbeta = begin{cases}
omegacdot (3cdotbeta) &text{if $beta$ is a limit ordinal,} \ omegacdot (3cdotbeta) + 4 & text{if $beta$ is a successor ordinal.}end{cases}$$
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
$endgroup$
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
|
show 3 more comments
$begingroup$
You have evaluated $(omegacdot 3+4)cdot 3$ correctly. However, $(omegacdot 3+4)cdot(omegacdot 3)$ is not correctly evaluated. I do not know which step of your evaluation is incorrect, but I guess you think adding $(omegacdot 3)$ $(omegacdot 3)$ times is $omega^2cdot 9$.
This is not true, however, as
$$(omegacdot 3)cdot (omegacdot 3) = omegacdot (3cdotomega)cdot 3 = omega^2cdot 3.$$
The core part of the evaluation is $3cdotomega=omega$ that can be proven in the way of proving $3+omega =omega$ (and the linked question in your comment.)
We would think evaluating $(omegacdot 3+4)cdot(omegacdot 3)$ is not too different from that of $(omegacdot 3)cdot (omegacdot 3)$. The problem is, we have extra $+4$ and we cannot apply the distribution law in the case.
However, the addition in the middle of the expression would disappear when adding $(omegacdot 3+4)$ many times, so we can guess the answer is $omega^2cdot 3$.
You can make our discussion concretely by following the steps of the proof you have linked; that is, use the transfinite induction for $betale omegacdot 3$ to prove
$$(omegacdot 3+4)cdotbeta = begin{cases}
omegacdot (3cdotbeta) &text{if $beta$ is a limit ordinal,} \ omegacdot (3cdotbeta) + 4 & text{if $beta$ is a successor ordinal.}end{cases}$$
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
$endgroup$
You have evaluated $(omegacdot 3+4)cdot 3$ correctly. However, $(omegacdot 3+4)cdot(omegacdot 3)$ is not correctly evaluated. I do not know which step of your evaluation is incorrect, but I guess you think adding $(omegacdot 3)$ $(omegacdot 3)$ times is $omega^2cdot 9$.
This is not true, however, as
$$(omegacdot 3)cdot (omegacdot 3) = omegacdot (3cdotomega)cdot 3 = omega^2cdot 3.$$
The core part of the evaluation is $3cdotomega=omega$ that can be proven in the way of proving $3+omega =omega$ (and the linked question in your comment.)
We would think evaluating $(omegacdot 3+4)cdot(omegacdot 3)$ is not too different from that of $(omegacdot 3)cdot (omegacdot 3)$. The problem is, we have extra $+4$ and we cannot apply the distribution law in the case.
However, the addition in the middle of the expression would disappear when adding $(omegacdot 3+4)$ many times, so we can guess the answer is $omega^2cdot 3$.
You can make our discussion concretely by following the steps of the proof you have linked; that is, use the transfinite induction for $betale omegacdot 3$ to prove
$$(omegacdot 3+4)cdotbeta = begin{cases}
omegacdot (3cdotbeta) &text{if $beta$ is a limit ordinal,} \ omegacdot (3cdotbeta) + 4 & text{if $beta$ is a successor ordinal.}end{cases}$$
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
answered Nov 25 '18 at 9:51
Hanul JeonHanul Jeon
17.5k42780
17.5k42780
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
|
show 3 more comments
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
Ahh I see, thank you very much. So would I right in saying then that something like (w⋅2+2)⋅(w⋅3) would = (w^2).3 ?
$endgroup$
– Fraiku
Nov 25 '18 at 10:25
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
@Fraiku Yes. $!$
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:26
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
Sorry, you are being so helpful I hope you don't mind if I ask you another question. Given the lack of distributive law on the right, I am struggling to know where to start to tackle more complex multiplications such as: ((w^2).2)+(w.3+1))5 or ((w^2).2)+(w.3+1))(w.5)
$endgroup$
– Fraiku
Nov 25 '18 at 10:32
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
@Fraiku The easiest strategy is evaluate $alphacdotbeta$ for small $beta$ and guess the pattern.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 10:34
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
$begingroup$
So ((w^2).2)+(w.3+1))2 = ((w^2).2)+(w.3+1)+((w^2).2)+(w.3+1) = ((w^2).2)+((w^2).2)+(w.3+1) = ((w^2).4)+w.3+1?
$endgroup$
– Fraiku
Nov 25 '18 at 10:40
|
show 3 more comments
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$begingroup$
Could you explain how to evaluate them?
$endgroup$
– Hanul Jeon
Nov 25 '18 at 5:08
$begingroup$
(w.3 + 4).3 = (w.3 + 4) + (w.3 + 4) + (w.3 + 4) = w.3 + (4 + w.3) + (4 + w.3) + 4 = w.3 + w.3 + w.3 + 4 = w.9 + 4 ??
$endgroup$
– Fraiku
Nov 25 '18 at 9:22
$begingroup$
Your first evaluation is correct; however the second one is not valid.
$endgroup$
– Hanul Jeon
Nov 25 '18 at 9:27
$begingroup$
Thank you. I was following this explanation for (w + 1).w^2 and trying to extrapolate, but I'm not sure where I've gone wrong. Can you clarify?
$endgroup$
– Fraiku
Nov 25 '18 at 9:41