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How to prove that doesnt exist a natural number such that is equal to it successor from Peano axioms?
$begingroup$
Im getting a hard time trying to prove the general for any natural number $n$ such that
$$nexists ninBbb N: S(n)= n$$
From the second Peano axiom we know that
$$nexists ninBbb N: S(1)= n$$
and from the third axiom that
$$S(a)=S(b)to a=b$$
I really dont have a clear clue about how to prove this. Someone tell me that I must use the "axiom" number 5, the induction. But the induction need the existence of a second natural number more than $1$ (or more than $0$ using the axioms that use it) that hold the implicit difference on the axiom 2. Example:
$$S(1)ne 1 to S(1)=2$$
But now, how to prove that $S(2)ne 2$? Thank you in advance.
analysis proof-explanation peano-axioms
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
Im getting a hard time trying to prove the general for any natural number $n$ such that
$$nexists ninBbb N: S(n)= n$$
From the second Peano axiom we know that
$$nexists ninBbb N: S(1)= n$$
and from the third axiom that
$$S(a)=S(b)to a=b$$
I really dont have a clear clue about how to prove this. Someone tell me that I must use the "axiom" number 5, the induction. But the induction need the existence of a second natural number more than $1$ (or more than $0$ using the axioms that use it) that hold the implicit difference on the axiom 2. Example:
$$S(1)ne 1 to S(1)=2$$
But now, how to prove that $S(2)ne 2$? Thank you in advance.
analysis proof-explanation peano-axioms
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
Im getting a hard time trying to prove the general for any natural number $n$ such that
$$nexists ninBbb N: S(n)= n$$
From the second Peano axiom we know that
$$nexists ninBbb N: S(1)= n$$
and from the third axiom that
$$S(a)=S(b)to a=b$$
I really dont have a clear clue about how to prove this. Someone tell me that I must use the "axiom" number 5, the induction. But the induction need the existence of a second natural number more than $1$ (or more than $0$ using the axioms that use it) that hold the implicit difference on the axiom 2. Example:
$$S(1)ne 1 to S(1)=2$$
But now, how to prove that $S(2)ne 2$? Thank you in advance.
analysis proof-explanation peano-axioms
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
$endgroup$
Im getting a hard time trying to prove the general for any natural number $n$ such that
$$nexists ninBbb N: S(n)= n$$
From the second Peano axiom we know that
$$nexists ninBbb N: S(1)= n$$
and from the third axiom that
$$S(a)=S(b)to a=b$$
I really dont have a clear clue about how to prove this. Someone tell me that I must use the "axiom" number 5, the induction. But the induction need the existence of a second natural number more than $1$ (or more than $0$ using the axioms that use it) that hold the implicit difference on the axiom 2. Example:
$$S(1)ne 1 to S(1)=2$$
But now, how to prove that $S(2)ne 2$? Thank you in advance.
analysis proof-explanation peano-axioms
analysis proof-explanation peano-axioms
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
edited Feb 25 '16 at 3:01
Masacroso
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
asked Feb 25 '16 at 2:53
MasacrosoMasacroso
13.1k41747
13.1k41747
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The successor map $sigma : mathbb{N} rightarrow mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $sigma(1) neq 1$; this is also an axiom since $1$ is not the successor of another number.
Now, we employ induction. Suppose for some $n$, $sigma(n) neq {n}$. This implies $$sigma(sigma(n)) neq sigma(n)$$ since otherwise would contradict the function being injective.
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
$endgroup$
add a comment |
$begingroup$
Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)ne 2$.
After all it was so simple but I didnt see in first place. What a dumb.
To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).
So, necesarily, if $S(2)ne 2$ and $S(1)ne 1$ then $S(n)ne n$ by induction.
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
Proof
Indeed, let $X := {n in mathbb{N}: s(n) neq n}$
$Rightarrow 1 in X$ (by Peano's Axiom 2)
And let $n in X Rightarrow s(n) neq n$
$Rightarrow s(s(n)) neq s(n)$ (Since s is injective, Peano's Axiom 1)
$Rightarrow s(n) in X, forall n in X$
$Rightarrow X = mathbb{N}$ (By Peano's Axiom 3)
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
$endgroup$
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
The successor map $sigma : mathbb{N} rightarrow mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $sigma(1) neq 1$; this is also an axiom since $1$ is not the successor of another number.
Now, we employ induction. Suppose for some $n$, $sigma(n) neq {n}$. This implies $$sigma(sigma(n)) neq sigma(n)$$ since otherwise would contradict the function being injective.
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
$endgroup$
add a comment |
$begingroup$
The successor map $sigma : mathbb{N} rightarrow mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $sigma(1) neq 1$; this is also an axiom since $1$ is not the successor of another number.
Now, we employ induction. Suppose for some $n$, $sigma(n) neq {n}$. This implies $$sigma(sigma(n)) neq sigma(n)$$ since otherwise would contradict the function being injective.
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
$endgroup$
add a comment |
$begingroup$
The successor map $sigma : mathbb{N} rightarrow mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $sigma(1) neq 1$; this is also an axiom since $1$ is not the successor of another number.
Now, we employ induction. Suppose for some $n$, $sigma(n) neq {n}$. This implies $$sigma(sigma(n)) neq sigma(n)$$ since otherwise would contradict the function being injective.
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
$endgroup$
The successor map $sigma : mathbb{N} rightarrow mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $sigma(1) neq 1$; this is also an axiom since $1$ is not the successor of another number.
Now, we employ induction. Suppose for some $n$, $sigma(n) neq {n}$. This implies $$sigma(sigma(n)) neq sigma(n)$$ since otherwise would contradict the function being injective.
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
edited Feb 25 '16 at 4:39
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
answered Feb 25 '16 at 3:55
MathematicsStudent1122MathematicsStudent1122
8,65122467
8,65122467
add a comment |
add a comment |
$begingroup$
Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)ne 2$.
After all it was so simple but I didnt see in first place. What a dumb.
To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).
So, necesarily, if $S(2)ne 2$ and $S(1)ne 1$ then $S(n)ne n$ by induction.
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)ne 2$.
After all it was so simple but I didnt see in first place. What a dumb.
To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).
So, necesarily, if $S(2)ne 2$ and $S(1)ne 1$ then $S(n)ne n$ by induction.
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
$endgroup$
add a comment |
$begingroup$
Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)ne 2$.
After all it was so simple but I didnt see in first place. What a dumb.
To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).
So, necesarily, if $S(2)ne 2$ and $S(1)ne 1$ then $S(n)ne n$ by induction.
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
$endgroup$
Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)ne 2$.
After all it was so simple but I didnt see in first place. What a dumb.
To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).
So, necesarily, if $S(2)ne 2$ and $S(1)ne 1$ then $S(n)ne n$ by induction.
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
edited Feb 25 '16 at 3:55
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
answered Feb 25 '16 at 3:37
MasacrosoMasacroso
13.1k41747
13.1k41747
add a comment |
add a comment |
$begingroup$
Proof
Indeed, let $X := {n in mathbb{N}: s(n) neq n}$
$Rightarrow 1 in X$ (by Peano's Axiom 2)
And let $n in X Rightarrow s(n) neq n$
$Rightarrow s(s(n)) neq s(n)$ (Since s is injective, Peano's Axiom 1)
$Rightarrow s(n) in X, forall n in X$
$Rightarrow X = mathbb{N}$ (By Peano's Axiom 3)
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
$endgroup$
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
add a comment |
$begingroup$
Proof
Indeed, let $X := {n in mathbb{N}: s(n) neq n}$
$Rightarrow 1 in X$ (by Peano's Axiom 2)
And let $n in X Rightarrow s(n) neq n$
$Rightarrow s(s(n)) neq s(n)$ (Since s is injective, Peano's Axiom 1)
$Rightarrow s(n) in X, forall n in X$
$Rightarrow X = mathbb{N}$ (By Peano's Axiom 3)
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
$endgroup$
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
add a comment |
$begingroup$
Proof
Indeed, let $X := {n in mathbb{N}: s(n) neq n}$
$Rightarrow 1 in X$ (by Peano's Axiom 2)
And let $n in X Rightarrow s(n) neq n$
$Rightarrow s(s(n)) neq s(n)$ (Since s is injective, Peano's Axiom 1)
$Rightarrow s(n) in X, forall n in X$
$Rightarrow X = mathbb{N}$ (By Peano's Axiom 3)
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
$endgroup$
Proof
Indeed, let $X := {n in mathbb{N}: s(n) neq n}$
$Rightarrow 1 in X$ (by Peano's Axiom 2)
And let $n in X Rightarrow s(n) neq n$
$Rightarrow s(s(n)) neq s(n)$ (Since s is injective, Peano's Axiom 1)
$Rightarrow s(n) in X, forall n in X$
$Rightarrow X = mathbb{N}$ (By Peano's Axiom 3)
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
answered Jan 15 at 13:37
John Medina DiazJohn Medina Diaz
1
1
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
add a comment |
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P1. $s: mathbb{N}rightarrow mathbb{N}$ is injective
$endgroup$
– John Medina Diaz
Jan 15 at 13:38
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P2. $mathbb{N}setminus s(mathbb{N})$ is made of one element. There exists only one number that is not succesor of anything. This is called one: 1.
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
$begingroup$
P3. (Principle of Induction): If $X subseteq mathbb{N}:1 in X$ and $s(n) in X, forall n in X Rightarrow X = mathbb{N}$
$endgroup$
– John Medina Diaz
Jan 15 at 13:39
add a comment |
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