Let $α$ and $β$ be ordinals such that $α>1$ and $βneq0$. Let $γ$, $ζ$, and $η$ be ordinals st...
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
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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
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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
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
edited Jan 4 at 17:15
Eigenfield
asked Jan 4 at 16:42
EigenfieldEigenfield
624518
624518
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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.
New contributor
add a comment |
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.
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
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.
New contributor
add a comment |
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.
New contributor
add a comment |
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.
New contributor
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.
New contributor
New contributor
answered Jan 4 at 17:17
ItsJustSomeOrdinalsBroItsJustSomeOrdinalsBro
661
661
New contributor
New contributor
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add a comment |
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.
add a comment |
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.
add a comment |
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.
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
302k32427757
302k32427757
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add a comment |
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