Is the set $\Bbb Q$ a quotient set of $\Bbb Q^*$?

Question

Let $\Bbb Q^*=\{\frac a b: a\in \Bbb Z, b\in \Bbb N\}$.

From this definition we can see $c=\frac 2 3$ and $d=\frac 4 6$ are elements of $\Bbb Q^*$.

Claim:

$$\frac 2 3\neq \frac 4 6$$

Proof:

Suppose $a=b$. By the definition of equality, $P(a)\iff P(b)$ for all properties.

Let $f:\Bbb Q^*\to \Bbb Z/f \left (\frac \xi \psi \right )=\xi$.

We then have $f(c)=2\neq4=f(d)$ a contradiction. Thus $c\neq d._\square$

Now let's define the relation $$\simeq\subseteq \Bbb Q^*\times\Bbb Q^*: \left(\frac a b,\frac c d\right)\in \space \simeq\iff \text {ad=bc.}$$

Is the set we normally call the rational numbers, $\Bbb Q=\Bbb Q^*/\simeq$?

Answer 1

You should exclude the case $b=0$. It's enough to let $a$ belong to $\mathbb {Z} $ and $b$ belong to $\mathbb{N}$. You can define the equivalence relation you need by $(a, b) \sim (c, d)$ if and only if $ad = bc $. Then yes, $\mathbb {Q}$ is the quotient of the big set by the equivalence relation.

Answer 2

In the definition of $\simeq$ the elements of $\mathbb Q^*$ are said to have a decimal expansion.

This reveals that these elements are supposed to be numbers (not some notation for pairs, or something else).

Consequently $f$ is not a well defined function.