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Proof for Strong Induction Principle
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I am currently studying analysis and I came across the following exercise.
Proposotion 2.2.14
Let $m_0$ be a natural number and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $mgeq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0leq m'< m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true, since in this case the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $mgeq m_0$.
Prove Proposition 2.2.14. (Hint: define $Q(n)$ to be the property that $P(m)$ is true for all $m_0leq m < n$; note that $Q(n)$ is vacuously true when $n<m_0$.)
I have difficulty understanding how I should use the hint and in general what the framework of this proof would look like (probably an inductive proof; but on what variable do we induct, what will be the induction hypothesis and how would I go about proving the inductive step etc.?). Could anyone please provide me with some hints to help me get started?
analysis elementary-set-theory induction
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
asked Jul 30 '13 at 13:27
dreamer
1,61132857
add a comment |Â
up vote
3
down vote
favorite
I am currently studying analysis and I came across the following exercise.
Proposotion 2.2.14
Let $m_0$ be a natural number and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $mgeq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0leq m'< m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true, since in this case the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $mgeq m_0$.
Prove Proposition 2.2.14. (Hint: define $Q(n)$ to be the property that $P(m)$ is true for all $m_0leq m < n$; note that $Q(n)$ is vacuously true when $n<m_0$.)
I have difficulty understanding how I should use the hint and in general what the framework of this proof would look like (probably an inductive proof; but on what variable do we induct, what will be the induction hypothesis and how would I go about proving the inductive step etc.?). Could anyone please provide me with some hints to help me get started?
analysis elementary-set-theory induction
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
asked Jul 30 '13 at 13:27
dreamer
1,61132857
1
Should the last sentence have $mge m_o$?
– Stefan Hamcke
Jul 30 '13 at 13:37
@StefanH. Yes, you're right. Silly typo.
– dreamer
Jul 30 '13 at 13:40
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle?
– skyking
Sep 7 '15 at 7:53
An edit was proposed to change the last line of Proposition 2.2.14 to read "...is vacuously true when $n le m_0$," with a citation to the errata. Given that this is an error in the text, it seems like some attention should be drawn to the change.
– Xander Henderson
Dec 28 '17 at 0:59
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am currently studying analysis and I came across the following exercise.
Proposotion 2.2.14
Let $m_0$ be a natural number and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $mgeq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0leq m'< m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true, since in this case the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $mgeq m_0$.
Prove Proposition 2.2.14. (Hint: define $Q(n)$ to be the property that $P(m)$ is true for all $m_0leq m < n$; note that $Q(n)$ is vacuously true when $n<m_0$.)
I have difficulty understanding how I should use the hint and in general what the framework of this proof would look like (probably an inductive proof; but on what variable do we induct, what will be the induction hypothesis and how would I go about proving the inductive step etc.?). Could anyone please provide me with some hints to help me get started?
analysis elementary-set-theory induction
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
asked Jul 30 '13 at 13:27
dreamer
1,61132857
I am currently studying analysis and I came across the following exercise.
Proposotion 2.2.14
Let $m_0$ be a natural number and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $mgeq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0leq m'< m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true, since in this case the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $mgeq m_0$.
Prove Proposition 2.2.14. (Hint: define $Q(n)$ to be the property that $P(m)$ is true for all $m_0leq m < n$; note that $Q(n)$ is vacuously true when $n<m_0$.)
I have difficulty understanding how I should use the hint and in general what the framework of this proof would look like (probably an inductive proof; but on what variable do we induct, what will be the induction hypothesis and how would I go about proving the inductive step etc.?). Could anyone please provide me with some hints to help me get started?
analysis elementary-set-theory induction
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
asked Jul 30 '13 at 13:27
dreamer
1,61132857
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
edited Jul 30 '13 at 13:49
Stefan Hamcke
20.9k42475
20.9k42475
asked Jul 30 '13 at 13:27
dreamer
1,61132857
asked Jul 30 '13 at 13:27
dreamer
1,61132857
asked Jul 30 '13 at 13:27
dreamer
1,61132857
1,61132857
1
Should the last sentence have $mge m_o$?
– Stefan Hamcke
Jul 30 '13 at 13:37
@StefanH. Yes, you're right. Silly typo.
– dreamer
Jul 30 '13 at 13:40
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle?
– skyking
Sep 7 '15 at 7:53
An edit was proposed to change the last line of Proposition 2.2.14 to read "...is vacuously true when $n le m_0$," with a citation to the errata. Given that this is an error in the text, it seems like some attention should be drawn to the change.
– Xander Henderson
Dec 28 '17 at 0:59
add a comment |Â
1
Should the last sentence have $mge m_o$?
– Stefan Hamcke
Jul 30 '13 at 13:37
@StefanH. Yes, you're right. Silly typo.
– dreamer
Jul 30 '13 at 13:40
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle?
– skyking
Sep 7 '15 at 7:53
An edit was proposed to change the last line of Proposition 2.2.14 to read "...is vacuously true when $n le m_0$," with a citation to the errata. Given that this is an error in the text, it seems like some attention should be drawn to the change.
– Xander Henderson
Dec 28 '17 at 0:59
1
1
Should the last sentence have $mge m_o$?
– Stefan Hamcke
Jul 30 '13 at 13:37
Should the last sentence have $mge m_o$?
– Stefan Hamcke
Jul 30 '13 at 13:37
@StefanH. Yes, you're right. Silly typo.
– dreamer
Jul 30 '13 at 13:40
@StefanH. Yes, you're right. Silly typo.
– dreamer
Jul 30 '13 at 13:40
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle?
– skyking
Sep 7 '15 at 7:53
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle?
– skyking
Sep 7 '15 at 7:53
An edit was proposed to change the last line of Proposition 2.2.14 to read "...is vacuously true when $n le m_0$," with a citation to the errata. Given that this is an error in the text, it seems like some attention should be drawn to the change.
– Xander Henderson
Dec 28 '17 at 0:59
An edit was proposed to change the last line of Proposition 2.2.14 to read "...is vacuously true when $n le m_0$," with a citation to the errata. Given that this is an error in the text, it seems like some attention should be drawn to the change.
– Xander Henderson
Dec 28 '17 at 0:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Let $B$ be the subset of $N=m_0,m_0+1,...$ such that $P(m)iff min B$. This $B$ is not empty since for all $m_0le m'<m_0$ the property $P$ is satisfied, thus also $P(m_0)$. We want to show that $B=N$. So assume that $A:=Nsetminus Bneemptyset$. Then there is an $min A$ such that each $m_0le n<m$ is in $B$, in other words $m$ is the minimum of $A$. But if $n<m$ implies $P(n)$, then by hypothesis $P(m)$ and so $min B$, a contradiction. Hence $B=N$.
Remark: This works for all sets $N$ where each non-empty subset has a minimal element with respect to a relation $R$. These sets are called well-founded.
If you want to use the hint, show that $Q(n)$ implies $Q(n+1)$ and that $Q(m_0)$: Since $Q(m')$ is true for all $m_0le m'< m_0$, it is also true for $m_0$. Assume $n$ is a natural number $ge m_0$ such that $Q(n)$. This means that $P(m) forall m_ole m<n$. By hypothesis this implies $P(n)$, thus $P(m) forall m_0le m<n+1$, so again by definition of $Q$ we have $Q(n+1)$. Now apply the induction principle.
So we can proof the strong induction principle via the induction principle. However, the normal induction principle itself requires a proof, it that is the proof I wrote in the first paragraph. As mentioned it works for all well-founded sets ($mathbb N$ is such a set.)
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
1
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
1
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
1
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Let $B$ be the subset of $N=m_0,m_0+1,...$ such that $P(m)iff min B$. This $B$ is not empty since for all $m_0le m'<m_0$ the property $P$ is satisfied, thus also $P(m_0)$. We want to show that $B=N$. So assume that $A:=Nsetminus Bneemptyset$. Then there is an $min A$ such that each $m_0le n<m$ is in $B$, in other words $m$ is the minimum of $A$. But if $n<m$ implies $P(n)$, then by hypothesis $P(m)$ and so $min B$, a contradiction. Hence $B=N$.
Remark: This works for all sets $N$ where each non-empty subset has a minimal element with respect to a relation $R$. These sets are called well-founded.
If you want to use the hint, show that $Q(n)$ implies $Q(n+1)$ and that $Q(m_0)$: Since $Q(m')$ is true for all $m_0le m'< m_0$, it is also true for $m_0$. Assume $n$ is a natural number $ge m_0$ such that $Q(n)$. This means that $P(m) forall m_ole m<n$. By hypothesis this implies $P(n)$, thus $P(m) forall m_0le m<n+1$, so again by definition of $Q$ we have $Q(n+1)$. Now apply the induction principle.
So we can proof the strong induction principle via the induction principle. However, the normal induction principle itself requires a proof, it that is the proof I wrote in the first paragraph. As mentioned it works for all well-founded sets ($mathbb N$ is such a set.)
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
1
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
1
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
1
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
add a comment |Â
up vote
5
down vote
accepted
Let $B$ be the subset of $N=m_0,m_0+1,...$ such that $P(m)iff min B$. This $B$ is not empty since for all $m_0le m'<m_0$ the property $P$ is satisfied, thus also $P(m_0)$. We want to show that $B=N$. So assume that $A:=Nsetminus Bneemptyset$. Then there is an $min A$ such that each $m_0le n<m$ is in $B$, in other words $m$ is the minimum of $A$. But if $n<m$ implies $P(n)$, then by hypothesis $P(m)$ and so $min B$, a contradiction. Hence $B=N$.
Remark: This works for all sets $N$ where each non-empty subset has a minimal element with respect to a relation $R$. These sets are called well-founded.
If you want to use the hint, show that $Q(n)$ implies $Q(n+1)$ and that $Q(m_0)$: Since $Q(m')$ is true for all $m_0le m'< m_0$, it is also true for $m_0$. Assume $n$ is a natural number $ge m_0$ such that $Q(n)$. This means that $P(m) forall m_ole m<n$. By hypothesis this implies $P(n)$, thus $P(m) forall m_0le m<n+1$, so again by definition of $Q$ we have $Q(n+1)$. Now apply the induction principle.
So we can proof the strong induction principle via the induction principle. However, the normal induction principle itself requires a proof, it that is the proof I wrote in the first paragraph. As mentioned it works for all well-founded sets ($mathbb N$ is such a set.)
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
1
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
1
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
1
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let $B$ be the subset of $N=m_0,m_0+1,...$ such that $P(m)iff min B$. This $B$ is not empty since for all $m_0le m'<m_0$ the property $P$ is satisfied, thus also $P(m_0)$. We want to show that $B=N$. So assume that $A:=Nsetminus Bneemptyset$. Then there is an $min A$ such that each $m_0le n<m$ is in $B$, in other words $m$ is the minimum of $A$. But if $n<m$ implies $P(n)$, then by hypothesis $P(m)$ and so $min B$, a contradiction. Hence $B=N$.
Remark: This works for all sets $N$ where each non-empty subset has a minimal element with respect to a relation $R$. These sets are called well-founded.
If you want to use the hint, show that $Q(n)$ implies $Q(n+1)$ and that $Q(m_0)$: Since $Q(m')$ is true for all $m_0le m'< m_0$, it is also true for $m_0$. Assume $n$ is a natural number $ge m_0$ such that $Q(n)$. This means that $P(m) forall m_ole m<n$. By hypothesis this implies $P(n)$, thus $P(m) forall m_0le m<n+1$, so again by definition of $Q$ we have $Q(n+1)$. Now apply the induction principle.
So we can proof the strong induction principle via the induction principle. However, the normal induction principle itself requires a proof, it that is the proof I wrote in the first paragraph. As mentioned it works for all well-founded sets ($mathbb N$ is such a set.)
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
Let $B$ be the subset of $N=m_0,m_0+1,...$ such that $P(m)iff min B$. This $B$ is not empty since for all $m_0le m'<m_0$ the property $P$ is satisfied, thus also $P(m_0)$. We want to show that $B=N$. So assume that $A:=Nsetminus Bneemptyset$. Then there is an $min A$ such that each $m_0le n<m$ is in $B$, in other words $m$ is the minimum of $A$. But if $n<m$ implies $P(n)$, then by hypothesis $P(m)$ and so $min B$, a contradiction. Hence $B=N$.
Remark: This works for all sets $N$ where each non-empty subset has a minimal element with respect to a relation $R$. These sets are called well-founded.
If you want to use the hint, show that $Q(n)$ implies $Q(n+1)$ and that $Q(m_0)$: Since $Q(m')$ is true for all $m_0le m'< m_0$, it is also true for $m_0$. Assume $n$ is a natural number $ge m_0$ such that $Q(n)$. This means that $P(m) forall m_ole m<n$. By hypothesis this implies $P(n)$, thus $P(m) forall m_0le m<n+1$, so again by definition of $Q$ we have $Q(n+1)$. Now apply the induction principle.
So we can proof the strong induction principle via the induction principle. However, the normal induction principle itself requires a proof, it that is the proof I wrote in the first paragraph. As mentioned it works for all well-founded sets ($mathbb N$ is such a set.)
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
edited Sep 7 '15 at 7:30
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
answered Jul 30 '13 at 13:48
Stefan Hamcke
20.9k42475
20.9k42475
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
1
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
1
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
1
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
add a comment |Â
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
1
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
1
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
1
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all?
– dreamer
Jul 30 '13 at 13:53
1
1
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer...
– Stefan Hamcke
Jul 30 '13 at 14:02
1
1
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
@rbm: No I used the original Induction Principle ($(Q(n)to Q(n+1))implies Q(m) forall minBbb N$) and the hypothesis $P(n)forall m_0le n<m to P(m)$.
– Stefan Hamcke
Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
Your edit clarified it :). Thanks again :)!
– dreamer
Jul 30 '13 at 14:18
1
1
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there?
– Aseem Dua
May 23 '14 at 7:47
add a comment |Â
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1
Should the last sentence have $mge m_o$?
– Stefan Hamcke
Jul 30 '13 at 13:37
@StefanH. Yes, you're right. Silly typo.
– dreamer
Jul 30 '13 at 13:40
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle?
– skyking
Sep 7 '15 at 7:53
An edit was proposed to change the last line of Proposition 2.2.14 to read "...is vacuously true when $n le m_0$," with a citation to the errata. Given that this is an error in the text, it seems like some attention should be drawn to the change.
– Xander Henderson
Dec 28 '17 at 0:59