$Q(n)$: “$P(k)$ holds for all $k<n$”. Then why is $Q(0)$ clearly true? [duplicate]
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This question already has an answer here:
Strong induction and vacuous truth
2 answers
Strong Induction Requires No Base Case?
2 answers
Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?
6 answers
It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.
In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.
$Q(0)$ means that "$P(k)$ holds for all $k<0$".
I understood there are no $k<0$.
And then I couldn't proceed.
induction proof-explanation
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
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marked as duplicate by Asaf Karagila♦ set-theory
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Jan 4 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Strong induction and vacuous truth
2 answers
Strong Induction Requires No Base Case?
2 answers
Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?
6 answers
It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.
In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.
$Q(0)$ means that "$P(k)$ holds for all $k<0$".
I understood there are no $k<0$.
And then I couldn't proceed.
induction proof-explanation
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
$endgroup$
marked as duplicate by Asaf Karagila♦ set-theory
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Jan 4 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05
add a comment |
$begingroup$
This question already has an answer here:
Strong induction and vacuous truth
2 answers
Strong Induction Requires No Base Case?
2 answers
Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?
6 answers
It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.
In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.
$Q(0)$ means that "$P(k)$ holds for all $k<0$".
I understood there are no $k<0$.
And then I couldn't proceed.
induction proof-explanation
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
$endgroup$
This question already has an answer here:
Strong induction and vacuous truth
2 answers
Strong Induction Requires No Base Case?
2 answers
Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?
6 answers
It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.
In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.
$Q(0)$ means that "$P(k)$ holds for all $k<0$".
I understood there are no $k<0$.
And then I couldn't proceed.
This question already has an answer here:
Strong induction and vacuous truth
2 answers
Strong Induction Requires No Base Case?
2 answers
Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?
6 answers
induction proof-explanation
induction proof-explanation
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
edited Jan 4 at 17:54
Andrés E. Caicedo
65.7k8160250
65.7k8160250
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
asked Jan 4 at 12:02
Doyun NamDoyun Nam
66119
66119
marked as duplicate by Asaf Karagila♦ set-theory
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Jan 4 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Asaf Karagila♦ set-theory
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Jan 4 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05
add a comment |
$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05
$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05
$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05
add a comment |
2 Answers
2
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oldest
votes
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$Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
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add a comment |
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A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.
If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
$endgroup$
add a comment |
$begingroup$
$Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
$endgroup$
add a comment |
$begingroup$
$Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
$endgroup$
$Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
answered Jan 4 at 12:23
WuestenfuxWuestenfux
5,0871513
5,0871513
add a comment |
add a comment |
$begingroup$
A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.
If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
$endgroup$
add a comment |
$begingroup$
A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.
If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
$endgroup$
add a comment |
$begingroup$
A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.
If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
$endgroup$
A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.
If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
answered Jan 4 at 12:09
ajotatxeajotatxe
53.9k24090
53.9k24090
add a comment |
add a comment |
$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05