Is this a “good†definition of addition over the set $mathbb N_0$?
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I was thinking of defining addition and the concept of successor with the help of addition by defining addition like this:
$$+(0,1)=1$$
$$+(m,1)=+(+(m-1,1),1)=s(m)$$
and, more generally:
$$+(m,n)=+(+(m,n-1),1)$$
We could define $0$ before as the number of elements of an empty set and $1$ as the number of all empty sets.
Since in the definition of addition we encounter symbol $-$ we could define before the concept of precedessor like this:
A precedessor of $n$, $p(n)=n-1$, is a number $n-1$ for which we have $+(n-1,1)=n$ and we also have $p(1)=0$ and $p(0)$ does not exist in $mathbb N_0$.
Is everything good here? I mean, are there any circularities or something similar? How would we prove that addition is commutative and associative?
natural-numbers
asked Aug 12 at 5:30
Right
5778
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I was thinking of defining addition and the concept of successor with the help of addition by defining addition like this:
$$+(0,1)=1$$
$$+(m,1)=+(+(m-1,1),1)=s(m)$$
and, more generally:
$$+(m,n)=+(+(m,n-1),1)$$
We could define $0$ before as the number of elements of an empty set and $1$ as the number of all empty sets.
Since in the definition of addition we encounter symbol $-$ we could define before the concept of precedessor like this:
A precedessor of $n$, $p(n)=n-1$, is a number $n-1$ for which we have $+(n-1,1)=n$ and we also have $p(1)=0$ and $p(0)$ does not exist in $mathbb N_0$.
Is everything good here? I mean, are there any circularities or something similar? How would we prove that addition is commutative and associative?
natural-numbers
asked Aug 12 at 5:30
Right
5778
2
There's absolutely a circularity as you've written it, in that the definition itself doesn't say what $n-1$ is, and your definition of $p(n)$ further down is in terms of the function $+(cdot, cdot)$ itself! Instead, it's more correct to write $+(m,s(n))=s(+(m,n))$, which is of course the traditional definition.
– Steven Stadnicki
Aug 12 at 5:35
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was thinking of defining addition and the concept of successor with the help of addition by defining addition like this:
$$+(0,1)=1$$
$$+(m,1)=+(+(m-1,1),1)=s(m)$$
and, more generally:
$$+(m,n)=+(+(m,n-1),1)$$
We could define $0$ before as the number of elements of an empty set and $1$ as the number of all empty sets.
Since in the definition of addition we encounter symbol $-$ we could define before the concept of precedessor like this:
A precedessor of $n$, $p(n)=n-1$, is a number $n-1$ for which we have $+(n-1,1)=n$ and we also have $p(1)=0$ and $p(0)$ does not exist in $mathbb N_0$.
Is everything good here? I mean, are there any circularities or something similar? How would we prove that addition is commutative and associative?
natural-numbers
asked Aug 12 at 5:30
Right
5778
I was thinking of defining addition and the concept of successor with the help of addition by defining addition like this:
$$+(0,1)=1$$
$$+(m,1)=+(+(m-1,1),1)=s(m)$$
and, more generally:
$$+(m,n)=+(+(m,n-1),1)$$
We could define $0$ before as the number of elements of an empty set and $1$ as the number of all empty sets.
Since in the definition of addition we encounter symbol $-$ we could define before the concept of precedessor like this:
A precedessor of $n$, $p(n)=n-1$, is a number $n-1$ for which we have $+(n-1,1)=n$ and we also have $p(1)=0$ and $p(0)$ does not exist in $mathbb N_0$.
Is everything good here? I mean, are there any circularities or something similar? How would we prove that addition is commutative and associative?
natural-numbers
asked Aug 12 at 5:30
Right
5778
asked Aug 12 at 5:30
Right
5778
asked Aug 12 at 5:30
Right
5778
asked Aug 12 at 5:30
Right
5778
5778
2
There's absolutely a circularity as you've written it, in that the definition itself doesn't say what $n-1$ is, and your definition of $p(n)$ further down is in terms of the function $+(cdot, cdot)$ itself! Instead, it's more correct to write $+(m,s(n))=s(+(m,n))$, which is of course the traditional definition.
– Steven Stadnicki
Aug 12 at 5:35
add a comment |Â
2
There's absolutely a circularity as you've written it, in that the definition itself doesn't say what $n-1$ is, and your definition of $p(n)$ further down is in terms of the function $+(cdot, cdot)$ itself! Instead, it's more correct to write $+(m,s(n))=s(+(m,n))$, which is of course the traditional definition.
– Steven Stadnicki
Aug 12 at 5:35
2
2
There's absolutely a circularity as you've written it, in that the definition itself doesn't say what $n-1$ is, and your definition of $p(n)$ further down is in terms of the function $+(cdot, cdot)$ itself! Instead, it's more correct to write $+(m,s(n))=s(+(m,n))$, which is of course the traditional definition.
– Steven Stadnicki
Aug 12 at 5:35
There's absolutely a circularity as you've written it, in that the definition itself doesn't say what $n-1$ is, and your definition of $p(n)$ further down is in terms of the function $+(cdot, cdot)$ itself! Instead, it's more correct to write $+(m,s(n))=s(+(m,n))$, which is of course the traditional definition.
– Steven Stadnicki
Aug 12 at 5:35
add a comment |Â
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2
There's absolutely a circularity as you've written it, in that the definition itself doesn't say what $n-1$ is, and your definition of $p(n)$ further down is in terms of the function $+(cdot, cdot)$ itself! Instead, it's more correct to write $+(m,s(n))=s(+(m,n))$, which is of course the traditional definition.
– Steven Stadnicki
Aug 12 at 5:35