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Proving that addition in $mathbbN$ is associative and commutative
The Next CEO of Stack OverflowProving mathematical induction with arbitrary base using (weak) inductionAre these two definitions of the natural numbers equivalent?Why this proof is incorrect?(Proof-verification) Proof that multiplication of natural numbers is commutativeIs the induction principle still viable if the induction skips over, or does not hold for some numbers?Mathematical induction from $n=1$Double inducton: Addition is commutativeProve addition is commutative using axioms, definitions, and inductionProving the commutative law for 1-based $BbbN$Proving the addition law for equality, eg. $+$ is compatible with $=$ on $mathbbN$
$begingroup$
I would like to prove the associative and commutative property of the natural numbers by using Peanos axioms only. Did I justify every step in my proofs correctly?
Definition. $forall n,m in mathbbN , n+m = underbraces(s(...s_textm (n)...)) = s^[m](n) $ where $s(n)$ is the successor function.
a) $forall n,m,k in mathbbN , n+(m+k)=(n+m)+k$
Proof: Let $SsubsetmathbbN$ be the set for which a) holds.
- Case $k=1 , forall n,m in mathbbN$
$n +(m+1) =$ (definition of $+$) $ = s^[m+1](n) = (*) = s(s^[m](n)) = s(m + n) = (m+n) + 1$
I am not sure how to justify $(*)$. I have:
$s^[m+1](n) = underbraces(s(....s_textm(underbraces_text1(n))...))=underbraces_text1(underbraces(s(...s_textm(n)...)))=s(s^[m](n))$
The composition of functions is associative but am I implicitly assuming the commutative property of the natural numbers i.e. $m+1 = 1+m$ ?
- Let $kin S$ be arbitrary $implies n+(m+k)=(n+m)+k$
$n+(m+(k+1))= $ (From 1. for $m+(k+1)$) = $n +((m+k)+1) = $ (From 1. for $n+(t+1)$ and $t=m+kinmathbbN$) $=(n+t)+1 = (n + (m+k))+1 = (kin S) = ((n+m)+k)+1=s^[k](n+m)+1=s^[k+1](n+m) = (n+m)+(k+1)$
From 1. we have $1in S$ and from 2. we have $kin S implies s(k)=k+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
b) $forall n,m in mathbbN , n+m=m+n$
Proof:
First we will show that $s(n) = s^[n](1)$
$n=1 implies s(1) = 1+1 = s^[1](1)$
Let $ninmathbbN$ be an arbitrary number for which the claim holds:
$s(n+1) = (n+1) + 1 = s(n) + 1 = s^[n](1) + 1 = s^[n+1](1)$
By the axiom of mathematical induction, the claim holds for all natural numbers.
Now let $SsubsetmathbbN$ be the set of all numbers for which the original claim holds.
- Case $m=1$
$n+1 = s(n) = s^[n](1) = 1 + n$
- Let $min S$ be arbitrary $implies n + m = m + n$
$n + (m+1) = s^[m+1](n) = s(s^[m](n)) =s(n+m) =$ ( $min S$ ) $= s(m+n) = s(s^[n](m)) = s^[n+1](m) = m + (n+1) =$ (Case 1.) $=m+(1+n)=$ (associative) $=(m+1)+n$.
From 1. we have $1in S$ and from 2. we have $min S implies s(m)=m+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
proof-verification induction natural-numbers
asked Mar 25 at 20:08
AltairACAltairAC
5591519
$endgroup$
|
show 2 more comments
$begingroup$
I would like to prove the associative and commutative property of the natural numbers by using Peanos axioms only. Did I justify every step in my proofs correctly?
Definition. $forall n,m in mathbbN , n+m = underbraces(s(...s_textm (n)...)) = s^[m](n) $ where $s(n)$ is the successor function.
a) $forall n,m,k in mathbbN , n+(m+k)=(n+m)+k$
Proof: Let $SsubsetmathbbN$ be the set for which a) holds.
- Case $k=1 , forall n,m in mathbbN$
$n +(m+1) =$ (definition of $+$) $ = s^[m+1](n) = (*) = s(s^[m](n)) = s(m + n) = (m+n) + 1$
I am not sure how to justify $(*)$. I have:
$s^[m+1](n) = underbraces(s(....s_textm(underbraces_text1(n))...))=underbraces_text1(underbraces(s(...s_textm(n)...)))=s(s^[m](n))$
The composition of functions is associative but am I implicitly assuming the commutative property of the natural numbers i.e. $m+1 = 1+m$ ?
- Let $kin S$ be arbitrary $implies n+(m+k)=(n+m)+k$
$n+(m+(k+1))= $ (From 1. for $m+(k+1)$) = $n +((m+k)+1) = $ (From 1. for $n+(t+1)$ and $t=m+kinmathbbN$) $=(n+t)+1 = (n + (m+k))+1 = (kin S) = ((n+m)+k)+1=s^[k](n+m)+1=s^[k+1](n+m) = (n+m)+(k+1)$
From 1. we have $1in S$ and from 2. we have $kin S implies s(k)=k+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
b) $forall n,m in mathbbN , n+m=m+n$
Proof:
First we will show that $s(n) = s^[n](1)$
$n=1 implies s(1) = 1+1 = s^[1](1)$
Let $ninmathbbN$ be an arbitrary number for which the claim holds:
$s(n+1) = (n+1) + 1 = s(n) + 1 = s^[n](1) + 1 = s^[n+1](1)$
By the axiom of mathematical induction, the claim holds for all natural numbers.
Now let $SsubsetmathbbN$ be the set of all numbers for which the original claim holds.
- Case $m=1$
$n+1 = s(n) = s^[n](1) = 1 + n$
- Let $min S$ be arbitrary $implies n + m = m + n$
$n + (m+1) = s^[m+1](n) = s(s^[m](n)) =s(n+m) =$ ( $min S$ ) $= s(m+n) = s(s^[n](m)) = s^[n+1](m) = m + (n+1) =$ (Case 1.) $=m+(1+n)=$ (associative) $=(m+1)+n$.
From 1. we have $1in S$ and from 2. we have $min S implies s(m)=m+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
proof-verification induction natural-numbers
asked Mar 25 at 20:08
AltairACAltairAC
5591519
$endgroup$
1
$begingroup$
You'll have to clarify lines like "Let $SsubsetmathbbN$ be the set for which a) holds." I suspect you mean that you are looking at the $k$ which make it true foe all $m,n$ but since (a) has no free variables what you say is meaningless.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
You'll find it much easier to root the inductions at $0$.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
OK, but as I said, that's not what you have written, is it?
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
I'd also recommend not using $dots$ in such proofs/definitions. I think that's where you're getting confused. If you insist in excluding $0$ from $mathbbN$ then the definition of $+$ in terms of $s$ is: $m+1=s(m)$, $m+s(n)=s(m+n)$. It's then pretty straightforward to prove associativity.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
I'm sorry, it just seems to me not to be a good enough definition as it leads to exactly this doubt. I think you are right when you say you've used $m+1=1+m$ at $(*)$; the problem is that the Peano arithmetic is escaping into the mathematical language describing it. In most standard formulations one proves associativity, then $1+m=m+1$ then full commutativity.
$endgroup$
– ancientmathematician
yesterday
|
show 2 more comments
$begingroup$
I would like to prove the associative and commutative property of the natural numbers by using Peanos axioms only. Did I justify every step in my proofs correctly?
Definition. $forall n,m in mathbbN , n+m = underbraces(s(...s_textm (n)...)) = s^[m](n) $ where $s(n)$ is the successor function.
a) $forall n,m,k in mathbbN , n+(m+k)=(n+m)+k$
Proof: Let $SsubsetmathbbN$ be the set for which a) holds.
- Case $k=1 , forall n,m in mathbbN$
$n +(m+1) =$ (definition of $+$) $ = s^[m+1](n) = (*) = s(s^[m](n)) = s(m + n) = (m+n) + 1$
I am not sure how to justify $(*)$. I have:
$s^[m+1](n) = underbraces(s(....s_textm(underbraces_text1(n))...))=underbraces_text1(underbraces(s(...s_textm(n)...)))=s(s^[m](n))$
The composition of functions is associative but am I implicitly assuming the commutative property of the natural numbers i.e. $m+1 = 1+m$ ?
- Let $kin S$ be arbitrary $implies n+(m+k)=(n+m)+k$
$n+(m+(k+1))= $ (From 1. for $m+(k+1)$) = $n +((m+k)+1) = $ (From 1. for $n+(t+1)$ and $t=m+kinmathbbN$) $=(n+t)+1 = (n + (m+k))+1 = (kin S) = ((n+m)+k)+1=s^[k](n+m)+1=s^[k+1](n+m) = (n+m)+(k+1)$
From 1. we have $1in S$ and from 2. we have $kin S implies s(k)=k+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
b) $forall n,m in mathbbN , n+m=m+n$
Proof:
First we will show that $s(n) = s^[n](1)$
$n=1 implies s(1) = 1+1 = s^[1](1)$
Let $ninmathbbN$ be an arbitrary number for which the claim holds:
$s(n+1) = (n+1) + 1 = s(n) + 1 = s^[n](1) + 1 = s^[n+1](1)$
By the axiom of mathematical induction, the claim holds for all natural numbers.
Now let $SsubsetmathbbN$ be the set of all numbers for which the original claim holds.
- Case $m=1$
$n+1 = s(n) = s^[n](1) = 1 + n$
- Let $min S$ be arbitrary $implies n + m = m + n$
$n + (m+1) = s^[m+1](n) = s(s^[m](n)) =s(n+m) =$ ( $min S$ ) $= s(m+n) = s(s^[n](m)) = s^[n+1](m) = m + (n+1) =$ (Case 1.) $=m+(1+n)=$ (associative) $=(m+1)+n$.
From 1. we have $1in S$ and from 2. we have $min S implies s(m)=m+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
proof-verification induction natural-numbers
asked Mar 25 at 20:08
AltairACAltairAC
5591519
$endgroup$
I would like to prove the associative and commutative property of the natural numbers by using Peanos axioms only. Did I justify every step in my proofs correctly?
Definition. $forall n,m in mathbbN , n+m = underbraces(s(...s_textm (n)...)) = s^[m](n) $ where $s(n)$ is the successor function.
a) $forall n,m,k in mathbbN , n+(m+k)=(n+m)+k$
Proof: Let $SsubsetmathbbN$ be the set for which a) holds.
- Case $k=1 , forall n,m in mathbbN$
$n +(m+1) =$ (definition of $+$) $ = s^[m+1](n) = (*) = s(s^[m](n)) = s(m + n) = (m+n) + 1$
I am not sure how to justify $(*)$. I have:
$s^[m+1](n) = underbraces(s(....s_textm(underbraces_text1(n))...))=underbraces_text1(underbraces(s(...s_textm(n)...)))=s(s^[m](n))$
The composition of functions is associative but am I implicitly assuming the commutative property of the natural numbers i.e. $m+1 = 1+m$ ?
- Let $kin S$ be arbitrary $implies n+(m+k)=(n+m)+k$
$n+(m+(k+1))= $ (From 1. for $m+(k+1)$) = $n +((m+k)+1) = $ (From 1. for $n+(t+1)$ and $t=m+kinmathbbN$) $=(n+t)+1 = (n + (m+k))+1 = (kin S) = ((n+m)+k)+1=s^[k](n+m)+1=s^[k+1](n+m) = (n+m)+(k+1)$
From 1. we have $1in S$ and from 2. we have $kin S implies s(k)=k+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
b) $forall n,m in mathbbN , n+m=m+n$
Proof:
First we will show that $s(n) = s^[n](1)$
$n=1 implies s(1) = 1+1 = s^[1](1)$
Let $ninmathbbN$ be an arbitrary number for which the claim holds:
$s(n+1) = (n+1) + 1 = s(n) + 1 = s^[n](1) + 1 = s^[n+1](1)$
By the axiom of mathematical induction, the claim holds for all natural numbers.
Now let $SsubsetmathbbN$ be the set of all numbers for which the original claim holds.
- Case $m=1$
$n+1 = s(n) = s^[n](1) = 1 + n$
- Let $min S$ be arbitrary $implies n + m = m + n$
$n + (m+1) = s^[m+1](n) = s(s^[m](n)) =s(n+m) =$ ( $min S$ ) $= s(m+n) = s(s^[n](m)) = s^[n+1](m) = m + (n+1) =$ (Case 1.) $=m+(1+n)=$ (associative) $=(m+1)+n$.
From 1. we have $1in S$ and from 2. we have $min S implies s(m)=m+1in S$ and by the axiom of mathematical induction it follows that $S = mathbbN$.
proof-verification induction natural-numbers
proof-verification induction natural-numbers
asked Mar 25 at 20:08
AltairACAltairAC
5591519
asked Mar 25 at 20:08
AltairACAltairAC
5591519
edited 2 days ago
AltairAC
asked Mar 25 at 20:08
AltairACAltairAC
5591519
asked Mar 25 at 20:08
AltairACAltairAC
5591519
asked Mar 25 at 20:08
AltairACAltairAC
5591519
5591519
1
$begingroup$
You'll have to clarify lines like "Let $SsubsetmathbbN$ be the set for which a) holds." I suspect you mean that you are looking at the $k$ which make it true foe all $m,n$ but since (a) has no free variables what you say is meaningless.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
You'll find it much easier to root the inductions at $0$.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
OK, but as I said, that's not what you have written, is it?
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
I'd also recommend not using $dots$ in such proofs/definitions. I think that's where you're getting confused. If you insist in excluding $0$ from $mathbbN$ then the definition of $+$ in terms of $s$ is: $m+1=s(m)$, $m+s(n)=s(m+n)$. It's then pretty straightforward to prove associativity.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
I'm sorry, it just seems to me not to be a good enough definition as it leads to exactly this doubt. I think you are right when you say you've used $m+1=1+m$ at $(*)$; the problem is that the Peano arithmetic is escaping into the mathematical language describing it. In most standard formulations one proves associativity, then $1+m=m+1$ then full commutativity.
$endgroup$
– ancientmathematician
yesterday
|
show 2 more comments
1
$begingroup$
You'll have to clarify lines like "Let $SsubsetmathbbN$ be the set for which a) holds." I suspect you mean that you are looking at the $k$ which make it true foe all $m,n$ but since (a) has no free variables what you say is meaningless.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
You'll find it much easier to root the inductions at $0$.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
OK, but as I said, that's not what you have written, is it?
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
I'd also recommend not using $dots$ in such proofs/definitions. I think that's where you're getting confused. If you insist in excluding $0$ from $mathbbN$ then the definition of $+$ in terms of $s$ is: $m+1=s(m)$, $m+s(n)=s(m+n)$. It's then pretty straightforward to prove associativity.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
I'm sorry, it just seems to me not to be a good enough definition as it leads to exactly this doubt. I think you are right when you say you've used $m+1=1+m$ at $(*)$; the problem is that the Peano arithmetic is escaping into the mathematical language describing it. In most standard formulations one proves associativity, then $1+m=m+1$ then full commutativity.
$endgroup$
– ancientmathematician
yesterday
1
1
$begingroup$
You'll have to clarify lines like "Let $SsubsetmathbbN$ be the set for which a) holds." I suspect you mean that you are looking at the $k$ which make it true foe all $m,n$ but since (a) has no free variables what you say is meaningless.
$endgroup$
– ancientmathematician
2 days ago
$begingroup$
You'll have to clarify lines like "Let $SsubsetmathbbN$ be the set for which a) holds." I suspect you mean that you are looking at the $k$ which make it true foe all $m,n$ but since (a) has no free variables what you say is meaningless.
$endgroup$
– ancientmathematician
2 days ago
1
1
$begingroup$
You'll find it much easier to root the inductions at $0$.
$endgroup$
– ancientmathematician
2 days ago
$begingroup$
You'll find it much easier to root the inductions at $0$.
$endgroup$
– ancientmathematician
2 days ago
1
1
$begingroup$
OK, but as I said, that's not what you have written, is it?
$endgroup$
– ancientmathematician
2 days ago
$begingroup$
OK, but as I said, that's not what you have written, is it?
$endgroup$
– ancientmathematician
2 days ago
1
1
$begingroup$
I'd also recommend not using $dots$ in such proofs/definitions. I think that's where you're getting confused. If you insist in excluding $0$ from $mathbbN$ then the definition of $+$ in terms of $s$ is: $m+1=s(m)$, $m+s(n)=s(m+n)$. It's then pretty straightforward to prove associativity.
$endgroup$
– ancientmathematician
2 days ago
$begingroup$
I'd also recommend not using $dots$ in such proofs/definitions. I think that's where you're getting confused. If you insist in excluding $0$ from $mathbbN$ then the definition of $+$ in terms of $s$ is: $m+1=s(m)$, $m+s(n)=s(m+n)$. It's then pretty straightforward to prove associativity.
$endgroup$
– ancientmathematician
2 days ago
1
1
$begingroup$
I'm sorry, it just seems to me not to be a good enough definition as it leads to exactly this doubt. I think you are right when you say you've used $m+1=1+m$ at $(*)$; the problem is that the Peano arithmetic is escaping into the mathematical language describing it. In most standard formulations one proves associativity, then $1+m=m+1$ then full commutativity.
$endgroup$
– ancientmathematician
yesterday
$begingroup$
I'm sorry, it just seems to me not to be a good enough definition as it leads to exactly this doubt. I think you are right when you say you've used $m+1=1+m$ at $(*)$; the problem is that the Peano arithmetic is escaping into the mathematical language describing it. In most standard formulations one proves associativity, then $1+m=m+1$ then full commutativity.
$endgroup$
– ancientmathematician
yesterday
|
show 2 more comments
0
active
oldest
votes
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1
$begingroup$
You'll have to clarify lines like "Let $SsubsetmathbbN$ be the set for which a) holds." I suspect you mean that you are looking at the $k$ which make it true foe all $m,n$ but since (a) has no free variables what you say is meaningless.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
You'll find it much easier to root the inductions at $0$.
$endgroup$
– ancientmathematician
2 days ago
1
$begingroup$
OK, but as I said, that's not what you have written, is it?
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– ancientmathematician
2 days ago
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I'd also recommend not using $dots$ in such proofs/definitions. I think that's where you're getting confused. If you insist in excluding $0$ from $mathbbN$ then the definition of $+$ in terms of $s$ is: $m+1=s(m)$, $m+s(n)=s(m+n)$. It's then pretty straightforward to prove associativity.
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– ancientmathematician
2 days ago
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I'm sorry, it just seems to me not to be a good enough definition as it leads to exactly this doubt. I think you are right when you say you've used $m+1=1+m$ at $(*)$; the problem is that the Peano arithmetic is escaping into the mathematical language describing it. In most standard formulations one proves associativity, then $1+m=m+1$ then full commutativity.
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– ancientmathematician
yesterday