Equality of positive rational numbers, Part-2 I am reading this answer.  I have some doubts which I want to clarify.
Question 1.
The author defines a rational number $\dfrac ab$ as, $$b\times\left(\dfrac{a}{b}\right) = a$$
He presumes that $\dfrac{a}{b}$ is a solution to $a=xb$. Then he states the definition of equivalence of two rational numbers as:

We also say that $\dfrac{a}{b}$ is "the same fraction" as $\dfrac{c}{d}$ if and only if $ad=bc$.  

After this he gives other definitions of addition, multiplication and subtraction of rational numbers. I claim that he could omit the definition of equivalence of rational numbers. User Pgatti pointed out that in my previous version of this question the justification of my claim was incorrect.
User Pgatti showed me an elegant way to justify my claim. Here I present it:
We need three definitions,  


*

*$\dfrac{a}{1}=a$  

*$\dfrac ab \times \dfrac cd = \dfrac{ac}{bd} $  

*$b \times \dfrac ab =a $ 
From these three definitions we can prove that $\dfrac ab = \dfrac cd $ iff $ad=bc$. The proof is:
By definition 3,    $$b \left( \dfrac ab \right)=a $$
by definition 1,   $$\dfrac b1 \dfrac ab = a$$
by definition 2,    $$\dfrac {ba}b = a $$
For $a=1$ we have,  $$\dfrac bb =1=\dfrac 11 \tag{property 1}$$
Let four positive integers $a,b,c$ and $d$ such that $ad=bc$. Now, 
$$ \dfrac ab = \dfrac ab \dfrac dd = \dfrac{ad}{bd} = \dfrac{bc}{bd} = \dfrac bb \dfrac cd = \dfrac cd$$     
So we have proved that "$\dfrac ab = \dfrac cd$  if  $ad=bc$".
Edit: 
User HTFB explained in his/her answer I still have to show when two rational numbers are not equal. The proof in the previous version was that 
$\dfrac ab = \dfrac cd $ if $ad=bc$ --- half of the job. But I have to prove that $\dfrac ab = \dfrac cd $ if and only if $ad=bc$. The remaining task is to prove that  $\dfrac ab \neq \dfrac cd $ if $ad \neq bc$. I've tried to use proof by contradiction, which I think is satisfactory.
Let $ad \neq bc$ and $\dfrac ab = \dfrac cd$, both are true.


*

*Given that $a,b,c,d \in \mathbb{N}$ and have such values that $ad \neq bc$.  

*Assume that $\dfrac ab =\dfrac cd$.  

*We have, $\dfrac{ad}{bc} \times bc=ad$ (Definition-3)     
$ \implies \dfrac ab \dfrac dc bc = ad$  (Definition-2)  
$ \implies \dfrac cd \dfrac dc bc = ad$   
$ \implies \dfrac{cd}{cd} bc = ad$  
$ \implies bc=ad$. (By property 1)
We have reached at the contradiction of statement 1. Hence statement 2 has to be false, given that statement one is true. So now, we have proved that if $ad \neq bc$ then $\dfrac ab \neq \dfrac cd $. For any given natural numbers $a,b,c$ and $d$, there can only be two possible cases, either $ad=bc$ or $ad \neq bc$. So the only way $\dfrac ab$ can be equal to $\dfrac cd$ is that $ad=bc$ otherwise $\dfrac ab \neq \dfrac cd$. So we conclude that:
$$\dfrac ab = \dfrac cd \text{ if and only if }ad=bc.$$

  
*
  
*So, could the author of that answer omit the definition of equality of rational numbers?
  

Question 2
The author of that answer doesn't define $\dfrac a1 = a$, rather he states it as a corollary of the definition, $b \dfrac ab = a$.  The reasoning given is that since $1 \times a=a$ and $1\times \dfrac a1 = a$  so $\dfrac a1$ should be equivalent to $a$. 

  
*
  
*Is he doing it correctly? Does the definition $b \left(\dfrac ab \right) = a$ implies $\dfrac a1=a$? Don't we have to define $\dfrac a1 = 1$
  

Question 3.
I have an idea. The idea is that $\dfrac rr = 1$, means $\dfrac rr = 1$ should act as a multiplicative identity. E.g. in $1*t=t$, I should be able to replace 1 with $\dfrac rr$. I can do this as:
$$ \dfrac rr \times t = \dfrac rr \dfrac t1 = \dfrac{rt}{r1} = \dfrac{tr}{r1} =  \dfrac tr \dfrac r1 = \dfrac tr r=t$$.
 So any number $\dfrac rr$ acts as an multiplicative identity.

  
*
  
*Is this argument enough to justify that $\dfrac rr=1$?
  


P.S: Another way to show that $\dfrac cc =1$ is to prove that $\dfrac cc - 1$ is 0. This can be done with the definition of subtraction  for rational numbers. The reason this can be done is that division is equivalent to subtraction, e.g. $\left( \dfrac {1}{c} \right) c =1$ is equivalent to $c-c=0$
P.P.S: This question was a part of my another similar question. I wanted to isolate this question from that so as to get a satisfactory answer, that's why I've re-asked it as a new question.
 A: Question 1. Yes, you do need to define your relations. In particular, your three axioms are consistent with a further axiom "$\mathrm{frac}(a,b) = \mathrm{frac}(c,d)$ for all $a,b,c,d$: i.e, that all fractions are equal to each other. 
None of your axioms say anything about when fractions are not equal. So you need something else. The suggested axiom
$$ \frac{a}{b} = \frac{c}{d} \text{ if and only if }ab=cd$$
does this in the "only if" direction. 
There are other problems with your axioms. In particular, numbers (1) and (3) mention integers as members of your new class of fractions. But they aren't! Really what you mean is that there is a mapping from the integers into the fractions. You also want this mapping to be 1-1, but an axiom saying so is equivalent to the "only if" direction of the one you are trying to avoid.
Indeed if you are axiomatising the class of fractions you also need to say that every fraction is of form $\mathrm{frac}(a,b)$ for some pair of integers $a,b$. Otherwise the ring of real numbers, for example, satisfies the axioms.
All this would become clearer if you kept an eye on the distinction between a definition and an axiom. Your list (1), (2), (3) is of axioms, things you want to be true in any class of fractions however they're defined. But they don't on their own define what fractions are. With only your axioms we can't look at a thing and say "this is a fraction" without finding a whole ring of all other things-that-might-be-fractions and checking the axioms (and the other ones you need but haven't listed). But with the definition "a fraction is a member of an equivalence class of ordered pairs of integers $(a,b)$ under the equivalence $(a,b) \equiv (c,d) \text{ iff } ad = bc$", just looking at the one object will tell us if it's a fraction. Is it such an equivalence class? No, then it's not a fraction. Yes, then it is.
Question 2. I think you have got in a muddle writing this out. You keep saying "$\mathrm{frac}(a,1) = 1$" but this is obviously a typo. Unfortunately it's a bit too hard to tidy this up for you to see what you mean to say. But you are right that the obvious thing to do is to define the embedding $\pi$ of the integers into the fractions by $\pi(a) = \mathrm{frac}(a,1)$, which is what you are really doing with your axiom (1), and not to try to infer this from elsewhere.
