Proving equal cardinality of sets using Cantor-Bernstein theorem

Hello,

I have to prove, that the sets $\displaystyle Z \times N^{+}$, which is a set containing Cartesian product of elements from the set of integers and from the set of positive natural numbers, has the same cardinality as $\displaystyle Q $, a set of rational numbers using Cantor-Bernstein theorem, which states that two sets have the same cardinality, if we can create two injective functions, one from the first set to the other, and the second from the second set to the first.

So in my opinion these are the two functions:

$\displaystyle f1: Z \times N^{+} \rightarrow Q$

For $\displaystyle <a,b> \in Z \times N^{+} $ the function gives a $\displaystyle q \in Q $ such that:

1)if $\displaystyle a>0$ then $\displaystyle q=\frac{(b+1)^{a}}{b}}$

2)if $\displaystyle a<0$ then $\displaystyle q=\frac{-(b+1)^{-a}}{b}}$

Because if $\displaystyle b=1$ we will get integers, but no other $\displaystyle b$ in natural numbers divides $\displaystyle b+1$, so we will get fractions, and because every time $\displaystyle a$ is different, so the prime factors of the whole thing will be different, so we will get a different rational number every time.

3)if $\displaystyle a=0$ then $\displaystyle q=\frac{(1)}{b}}$

Because $\displaystyle b \in N^{+} $ so $\displaystyle b+1>=2$ every time in my formula, so when $\displaystyle a=0$ we can put $\displaystyle 1$ in the numerator.

$\displaystyle f2: Q \rightarrow Z \times N^{+}$

For any $\displaystyle q \in Q $ we choose $\displaystyle m,n \in N$ such that $\displaystyle \frac{m}{n}=|q|$ and :

1) if $\displaystyle q>=0$ we return $\displaystyle <m,n>$

2) if $\displaystyle q<0$ we return $\displaystyle <-m,n>$.

Are those two function correct?

Re: Proving equal cardinality of sets using Cantor-Bernstein theorem

Your mapping from $\displaystyle \mathbb{Z} \times \mathbb{N^{+}} \to \mathbb{Q}$ is good.

The mapping for $\displaystyle \mathbb{Q} \to \mathbb{Z} \times \mathbb{N^{+}} $ also looks good.

Re: Proving equal cardinality of sets using Cantor-Bernstein theorem

For the first injection is following map also correct?

$\displaystyle f1: Z \times N^{+} \rightarrow Q$

For $\displaystyle <a,b> \in Z \times N^{+} $ the function gives a $\displaystyle q \in Q $ such that:

$\displaystyle q=\frac{2^{a}}{3^{b}}}$

Re: Proving equal cardinality of sets using Cantor-Bernstein theorem

Quote:

Originally Posted by

**MachinePL1993** For the first injection is following map also correct?

$\displaystyle f1: Z \times N^{+} \rightarrow Q$

For $\displaystyle <a,b> \in Z \times N^{+} $ the function gives a $\displaystyle q \in Q $ such that:

$\displaystyle q=\frac{2^{a}}{3^{b}}}$

That is correct. Probably the most natural thing to have done here would be to define $\displaystyle \mathbb{Z}\times\mathbb{N}\to\mathbb{Q}$ by exactly $\displaystyle (a,b)\mapsto 2^a 3^b$ which is an injection by the fundamental theorem of arithmetic. Then, the backwards injection could have been $\displaystyle \frac{a}{b}\mapsto (a,b)$ where $\displaystyle a,b$ are coprime and $\displaystyle b\in\mathbb{N}$.