Let $α$ and $β$ be ordinals such that $α>1$ and $βneq0$. Let $γ$, $ζ$, and $η$ be ordinals st...
$begingroup$
Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?
set-theory ordinals
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
$endgroup$
add a comment |
$begingroup$
Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?
set-theory ordinals
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
$endgroup$
add a comment |
$begingroup$
Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?
set-theory ordinals
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
$endgroup$
Let $alpha$ and $beta$ be ordinals such that $alpha>1$ and $betaneq0$. Let $gamma$, $zeta$, and $eta$ be ordinals such that $gamma<beta$, $zeta<alpha$, and $eta<alpha^{gamma}$. I want to prove that $alpha^{gamma}cdotzeta+eta<alpha^{beta}$. I have already observed that $alpha^{gamma}cdotzeta<alpha^{beta}$. The hard part is showing that the equality still holds if we add $eta$. Any ideas?
set-theory ordinals
set-theory ordinals
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
edited Jan 4 at 17:15
Eigenfield
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
asked Jan 4 at 16:42
EigenfieldEigenfield
627518
627518
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.
From the hypothesis, we know that:
$$
eta <alpha^gamma
$$
Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
$$
alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
$$
Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
$$
alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
$$
And thus we have shown what was needed.
Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
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add a comment |
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Note that
$$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$
Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.
From the hypothesis, we know that:
$$
eta <alpha^gamma
$$
Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
$$
alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
$$
Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
$$
alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
$$
And thus we have shown what was needed.
Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
$endgroup$
add a comment |
$begingroup$
Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.
From the hypothesis, we know that:
$$
eta <alpha^gamma
$$
Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
$$
alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
$$
Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
$$
alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
$$
And thus we have shown what was needed.
Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
$endgroup$
add a comment |
$begingroup$
Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.
From the hypothesis, we know that:
$$
eta <alpha^gamma
$$
Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
$$
alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
$$
Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
$$
alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
$$
And thus we have shown what was needed.
Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
$endgroup$
Assuming you have access to the basic rules of ordinal arithmetic, then this is a very simple exercise. If you don't, then I would work on proving those rules rather than working directly with the inequality you want to prove. For the remainder of this answer, I will assume the rules.
From the hypothesis, we know that:
$$
eta <alpha^gamma
$$
Addition is strictly increasing in the right argument and multiplication is distributive on the right argument, so we know that:
$$
alpha^gammacdotzeta+eta < alpha^gammacdotzeta+alpha^gamma = alpha^gammacdot(zeta+1)
$$
Since by our hypothesis $zeta<alpha$ we know that $zeta+1leqalpha$, and by the fact that multiplication is increasing in the right argument, that exponentiation "works as expected" on the right argument, that exponentiation is increasing in the right argument so long as the left argument is not $1$, and the hypothesis that $gamma+1leqbeta$, we know that:
$$
alpha^gammacdot(zeta+1)leqalpha^{gamma}alpha=alpha^{gamma+1}leqalpha^beta
$$
And thus we have shown what was needed.
Note, there is a lot of detail that is being taken care of by our rules. They are not hard to prove, but you must make sure that you are not taking them for granted.
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
661
add a comment |
add a comment |
$begingroup$
Note that
$$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$
Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
add a comment |
$begingroup$
Note that
$$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$
Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
add a comment |
$begingroup$
Note that
$$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$
Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
Note that
$$alpha^gammacdotgamma+eta<alpha^gammacdoteta+alpha^gamma=alpha^gammacdot(eta+1)leqalpha^gammacdotalpha=alpha^{gamma+1}leqalpha^beta.$$
Of course, you need to prove the left distributivity rules of multiplication and exponentiation. But that's a whole other topic.
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
answered Jan 4 at 17:08
Asaf Karagila♦Asaf Karagila
306k33438769
306k33438769
add a comment |
add a comment |
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