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Definition

A difference is a pair of natural numbers and if $x:=(m,n)$ and $y:=(p,q)$ are differences we define $x\underset{d}\sim y$ if and only if $m+q=p+n$. In particular we say that the difference $(m,n)$ is positive if $m>n$.

Theorem

The relation $\underset{d}\sim$ between differences is a equivalence relation.

Proof. Omitted

Theorem

If $(m,n)$ is a positive difference and $(m,n)\underset{d}\sim(p,q)$ then $(p,q)$ is a positive difference too.

Proof. Omitted.

Theorem

A binary operation is defined between differences throug the condition $$ (m,n)+(p,q)=(m+p,n+q) $$ for any $m,n,p,q\in\Bbb N$

Proof. Omitted

Definition

An integer number is an equivalence class of the reletion $\underset{d}\sim$ above defined that is $x$ is an integer number if $$ x=[(m,n)]_i $$ for some $m,n\in\Bbb N$. In particular we say that the integer $x$ is positive if an its element is positive. Finally the set of all equivalence calsses is called set of the integers numeber and it is denoted by the symbol $\Bbb Z$.

Theorem

A binary operation is defined in $\Bbb Z$ by the condition $$ [x]_i+[y]_i:=[(x+y)]_i $$ for any $x,y\in\Bbb N\times\Bbb N$

Proof. Omitted.

Theorem

The sum between integers has the following properties.

  • $x+(y+z)=(x+y)+z$;
  • $x+y=y+z$;
  • $0_i+x=x$;
  • there exist and integer $y$ such that $x+y=0_i$ and it is unique so that we indicate it with the symbol $-x$;
  • if $x,y\in\Bbb Z^+$ then $(x+y)\in\Bbb Z^+$ too.

Proof. Omitted.

Definition

We define $x<y$ if and only if $(y-x)$ is positive.

So with the above formalism I ask to prove that the set $[a,b]:=\{x\in\Bbb Z:a\le x\le b\}$ is finite and has cardinality or $[(b-a)+1]$.

So could someone help me, please?

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  • $\begingroup$ Before you prove this, you have to say what it means to have cardinality $[(b-a)+1]$. Do you mean that there is a bijection between $[a,b]$ and $\{k\in\Bbb N:1\le k\le n+1\}$, where $\iota(n)=[a-b]$? $\endgroup$ – Brian M. Scott Sep 18 '20 at 1:23
  • $\begingroup$ Okay. Since $\iota$ is a bijection then the cardinality of $[(b-a)+1]$ is precisely the cardinality of $n$ but to be $\iota$ a bijection we can put $n:=[(b-a)+1]$, that's incorrect? $\endgroup$ – Antonio Maria Di Mauro Sep 18 '20 at 16:11
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Theorem

The function $\iota:\Bbb N\rightarrow\Bbb Z$ through the condition $$ \iota(n):=[(n,0)]_i $$ for all $n\in\Bbb N$ is an isomorphism of $\Bbb N$ into $\Bbb Z$.

Proof. Omitted.

Now for convenience we put $I:=[a,b]$ and and the we observe that $(c-a)\ge 0$ for any $c\in[a,b]$ so that there exist $n\in\Bbb n$ such that $\iota(n)=(c-a)$. Therefore for convenience we put $J:=(b-a)+1$

Theorem

For any set $A$, for any $a\in A$ and for any function $g:A\times\Bbb N\rightarrow A$ there exists a unique infinite sequence $f:\Bbb N\rightarrow A$ such that

  • $f(0)=a$;
  • $f(n+1)=g\big(f(n),n\big)$ for all $n\in\Bbb N$.

Proof. Omitted.

So now let be $A:=\{n\in\Bbb N:n\ge a\}$ and let be $g:A\times\Bbb N\rightarrow A$ the function defined by the condition $$ g(x,n):=a+(n+1) $$ for any $x\in A$ and for any $n\in\Bbb N$. So by the previous theorem there exist a function $f:\Bbb N\rightarrow A$ such that $$ f(n):=a+n $$ for any $n\in\Bbb N$ and we observe that if $c\in I$ then $f(n)=c$ where $n=(c-a)+1\in J$ and if $n\in J$ then $f(n)\in I$ so that $f[J]=I$ and so if we even observe that $f(n)<f(m)$ for any $n<m$ then we conclude that $|I|=|J|=[(b-a)+1]$ as I stated above.

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