# The interval $[a,b]:=\{x\in\Bbb Z:a\le x\le b\}$ is finite and has cardinality $[(b-a)+1]$

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 $$x if and only if $$(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?

• 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 $\iota(n)=[a-b]$? – Brian M. Scott Sep 18 '20 at 1:23
• 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? – Antonio Maria Di Mauro Sep 18 '20 at 16:11

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 $$f(n) for any $$n then we conclude that $$|I|=|J|=[(b-a)+1]$$ as I stated above.