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?