Proof Critique: For any $bin mathbf{N}$, $b++$ is unique.
$begingroup$
As I was trying to answer this question (using only the peano axioms), I came up with the following "proof". On a second look, I noticed a flaw. But I'm struggling to articulate why or what the flaw is. Could somebody take a look, and let me know if I'm on the right track? Thanks!
$textbf{Incorrect Proof}$: Take any arbitrary $b in mathbf{N}$. Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
$textbf{Self-critique}$: I am assuming that $b++$ doesn't have two different successors. What if we're working with some number system where it does? The right way to show this is induction on $b$ that uses the axiom "if $nneq m$, then $n++ neq m++$."
proof-explanation
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
asked Dec 17 '18 at 6:35
skmskm
12
$endgroup$
|
show 5 more comments
$begingroup$
As I was trying to answer this question (using only the peano axioms), I came up with the following "proof". On a second look, I noticed a flaw. But I'm struggling to articulate why or what the flaw is. Could somebody take a look, and let me know if I'm on the right track? Thanks!
$textbf{Incorrect Proof}$: Take any arbitrary $b in mathbf{N}$. Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
$textbf{Self-critique}$: I am assuming that $b++$ doesn't have two different successors. What if we're working with some number system where it does? The right way to show this is induction on $b$ that uses the axiom "if $nneq m$, then $n++ neq m++$."
proof-explanation
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
asked Dec 17 '18 at 6:35
skmskm
12
$endgroup$
1
$begingroup$
You ought to learn logic first...
$endgroup$
– user21820
Dec 17 '18 at 6:38
2
$begingroup$
How is $btext{++}$ defined in your context? If this is the successor function in Peano Axioms, then it is assumed to have a unique value for each $bin mathbf{N}$.
$endgroup$
– Sangchul Lee
Dec 17 '18 at 6:51
$begingroup$
We have the following definitions: 1) $n in mathbb{N} to n++ in mathbb{N}$, 2) for all $n$, $n++ neq 0$, and 3) $forall n.(nneq m to n++neq m++)$.
$endgroup$
– skm
Dec 17 '18 at 7:02
$begingroup$
@skm Sure. But usually, right before you get to that part, the successor function is defined as a function. And part of the definition of a function is exactly that for any input the output is unique. Is this the case for you? And if not, then what is your $++$?
$endgroup$
– Arthur
Dec 17 '18 at 7:13
$begingroup$
@Arthur I don't see a successor function in the definition for incrementation. What I do have is a proposition that follows from the axiom stating induction about recursive definitions: Suppose for each $n in N$, we have some function $f_n :Nto N$. Take some $c in N$. Then we can assign a unique natural number $a_n$ to each natural number $n$, such that $a_0 := c$ and $a_n++ := f_n(a_n)$. Is this what you're referring to?
$endgroup$
– skm
Dec 17 '18 at 7:28
|
show 5 more comments
$begingroup$
As I was trying to answer this question (using only the peano axioms), I came up with the following "proof". On a second look, I noticed a flaw. But I'm struggling to articulate why or what the flaw is. Could somebody take a look, and let me know if I'm on the right track? Thanks!
$textbf{Incorrect Proof}$: Take any arbitrary $b in mathbf{N}$. Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
$textbf{Self-critique}$: I am assuming that $b++$ doesn't have two different successors. What if we're working with some number system where it does? The right way to show this is induction on $b$ that uses the axiom "if $nneq m$, then $n++ neq m++$."
proof-explanation
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
asked Dec 17 '18 at 6:35
skmskm
12
$endgroup$
As I was trying to answer this question (using only the peano axioms), I came up with the following "proof". On a second look, I noticed a flaw. But I'm struggling to articulate why or what the flaw is. Could somebody take a look, and let me know if I'm on the right track? Thanks!
$textbf{Incorrect Proof}$: Take any arbitrary $b in mathbf{N}$. Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
$textbf{Self-critique}$: I am assuming that $b++$ doesn't have two different successors. What if we're working with some number system where it does? The right way to show this is induction on $b$ that uses the axiom "if $nneq m$, then $n++ neq m++$."
proof-explanation
proof-explanation
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
asked Dec 17 '18 at 6:35
skmskm
12
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
asked Dec 17 '18 at 6:35
skmskm
12
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
edited Dec 17 '18 at 6:51
Hans Hüttel
3,3572921
3,3572921
asked Dec 17 '18 at 6:35
skmskm
12
asked Dec 17 '18 at 6:35
skmskm
12
asked Dec 17 '18 at 6:35
skmskm
12
12
1
$begingroup$
You ought to learn logic first...
$endgroup$
– user21820
Dec 17 '18 at 6:38
2
$begingroup$
How is $btext{++}$ defined in your context? If this is the successor function in Peano Axioms, then it is assumed to have a unique value for each $bin mathbf{N}$.
$endgroup$
– Sangchul Lee
Dec 17 '18 at 6:51
$begingroup$
We have the following definitions: 1) $n in mathbb{N} to n++ in mathbb{N}$, 2) for all $n$, $n++ neq 0$, and 3) $forall n.(nneq m to n++neq m++)$.
$endgroup$
– skm
Dec 17 '18 at 7:02
$begingroup$
@skm Sure. But usually, right before you get to that part, the successor function is defined as a function. And part of the definition of a function is exactly that for any input the output is unique. Is this the case for you? And if not, then what is your $++$?
$endgroup$
– Arthur
Dec 17 '18 at 7:13
$begingroup$
@Arthur I don't see a successor function in the definition for incrementation. What I do have is a proposition that follows from the axiom stating induction about recursive definitions: Suppose for each $n in N$, we have some function $f_n :Nto N$. Take some $c in N$. Then we can assign a unique natural number $a_n$ to each natural number $n$, such that $a_0 := c$ and $a_n++ := f_n(a_n)$. Is this what you're referring to?
$endgroup$
– skm
Dec 17 '18 at 7:28
|
show 5 more comments
1
$begingroup$
You ought to learn logic first...
$endgroup$
– user21820
Dec 17 '18 at 6:38
2
$begingroup$
How is $btext{++}$ defined in your context? If this is the successor function in Peano Axioms, then it is assumed to have a unique value for each $bin mathbf{N}$.
$endgroup$
– Sangchul Lee
Dec 17 '18 at 6:51
$begingroup$
We have the following definitions: 1) $n in mathbb{N} to n++ in mathbb{N}$, 2) for all $n$, $n++ neq 0$, and 3) $forall n.(nneq m to n++neq m++)$.
$endgroup$
– skm
Dec 17 '18 at 7:02
$begingroup$
@skm Sure. But usually, right before you get to that part, the successor function is defined as a function. And part of the definition of a function is exactly that for any input the output is unique. Is this the case for you? And if not, then what is your $++$?
$endgroup$
– Arthur
Dec 17 '18 at 7:13
$begingroup$
@Arthur I don't see a successor function in the definition for incrementation. What I do have is a proposition that follows from the axiom stating induction about recursive definitions: Suppose for each $n in N$, we have some function $f_n :Nto N$. Take some $c in N$. Then we can assign a unique natural number $a_n$ to each natural number $n$, such that $a_0 := c$ and $a_n++ := f_n(a_n)$. Is this what you're referring to?
$endgroup$
– skm
Dec 17 '18 at 7:28
1
1
$begingroup$
You ought to learn logic first...
$endgroup$
– user21820
Dec 17 '18 at 6:38
$begingroup$
You ought to learn logic first...
$endgroup$
– user21820
Dec 17 '18 at 6:38
2
2
$begingroup$
How is $btext{++}$ defined in your context? If this is the successor function in Peano Axioms, then it is assumed to have a unique value for each $bin mathbf{N}$.
$endgroup$
– Sangchul Lee
Dec 17 '18 at 6:51
$begingroup$
How is $btext{++}$ defined in your context? If this is the successor function in Peano Axioms, then it is assumed to have a unique value for each $bin mathbf{N}$.
$endgroup$
– Sangchul Lee
Dec 17 '18 at 6:51
$begingroup$
We have the following definitions: 1) $n in mathbb{N} to n++ in mathbb{N}$, 2) for all $n$, $n++ neq 0$, and 3) $forall n.(nneq m to n++neq m++)$.
$endgroup$
– skm
Dec 17 '18 at 7:02
$begingroup$
We have the following definitions: 1) $n in mathbb{N} to n++ in mathbb{N}$, 2) for all $n$, $n++ neq 0$, and 3) $forall n.(nneq m to n++neq m++)$.
$endgroup$
– skm
Dec 17 '18 at 7:02
$begingroup$
@skm Sure. But usually, right before you get to that part, the successor function is defined as a function. And part of the definition of a function is exactly that for any input the output is unique. Is this the case for you? And if not, then what is your $++$?
$endgroup$
– Arthur
Dec 17 '18 at 7:13
$begingroup$
@skm Sure. But usually, right before you get to that part, the successor function is defined as a function. And part of the definition of a function is exactly that for any input the output is unique. Is this the case for you? And if not, then what is your $++$?
$endgroup$
– Arthur
Dec 17 '18 at 7:13
$begingroup$
@Arthur I don't see a successor function in the definition for incrementation. What I do have is a proposition that follows from the axiom stating induction about recursive definitions: Suppose for each $n in N$, we have some function $f_n :Nto N$. Take some $c in N$. Then we can assign a unique natural number $a_n$ to each natural number $n$, such that $a_0 := c$ and $a_n++ := f_n(a_n)$. Is this what you're referring to?
$endgroup$
– skm
Dec 17 '18 at 7:28
$begingroup$
@Arthur I don't see a successor function in the definition for incrementation. What I do have is a proposition that follows from the axiom stating induction about recursive definitions: Suppose for each $n in N$, we have some function $f_n :Nto N$. Take some $c in N$. Then we can assign a unique natural number $a_n$ to each natural number $n$, such that $a_0 := c$ and $a_n++ := f_n(a_n)$. Is this what you're referring to?
$endgroup$
– skm
Dec 17 '18 at 7:28
|
show 5 more comments
1 Answer
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$begingroup$
You write
Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
We cannot conclude this, as we are assuming that $b!+!!+$ is not unique.
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
$endgroup$
add a comment |
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$begingroup$
You write
Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
We cannot conclude this, as we are assuming that $b!+!!+$ is not unique.
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
$endgroup$
add a comment |
$begingroup$
You write
Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
We cannot conclude this, as we are assuming that $b!+!!+$ is not unique.
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
$endgroup$
add a comment |
$begingroup$
You write
Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
We cannot conclude this, as we are assuming that $b!+!!+$ is not unique.
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
$endgroup$
You write
Suppose for the sake of contradiction that there is some $a,a' in mathbf{N}$ such that $b++=a$ and $b++ = a'$. Then we can write $a = b++ = a'$ yielding $a=a'$ as required.
We cannot conclude this, as we are assuming that $b!+!!+$ is not unique.
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
answered Dec 17 '18 at 6:55
Hans HüttelHans Hüttel
3,3572921
3,3572921
add a comment |
add a comment |
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1
$begingroup$
You ought to learn logic first...
$endgroup$
– user21820
Dec 17 '18 at 6:38
2
$begingroup$
How is $btext{++}$ defined in your context? If this is the successor function in Peano Axioms, then it is assumed to have a unique value for each $bin mathbf{N}$.
$endgroup$
– Sangchul Lee
Dec 17 '18 at 6:51
$begingroup$
We have the following definitions: 1) $n in mathbb{N} to n++ in mathbb{N}$, 2) for all $n$, $n++ neq 0$, and 3) $forall n.(nneq m to n++neq m++)$.
$endgroup$
– skm
Dec 17 '18 at 7:02
$begingroup$
@skm Sure. But usually, right before you get to that part, the successor function is defined as a function. And part of the definition of a function is exactly that for any input the output is unique. Is this the case for you? And if not, then what is your $++$?
$endgroup$
– Arthur
Dec 17 '18 at 7:13
$begingroup$
@Arthur I don't see a successor function in the definition for incrementation. What I do have is a proposition that follows from the axiom stating induction about recursive definitions: Suppose for each $n in N$, we have some function $f_n :Nto N$. Take some $c in N$. Then we can assign a unique natural number $a_n$ to each natural number $n$, such that $a_0 := c$ and $a_n++ := f_n(a_n)$. Is this what you're referring to?
$endgroup$
– skm
Dec 17 '18 at 7:28