What actually "forces" us to define the operations on $\mathbb{Q}$ the way they are?

Question

Given the integers, we have the usual operations which behave the way they are expected to behave. Now suppose we want to create $\mathbb{Q}$ which are pairs $\cfrac{a}{b\neq 0}$. We want that $\mathbb{Q}$ have a copy of $\mathbb{Z}$ that behaves exactly as $\mathbb{Z}$ itself, that is: We want:

$$a_\mathbb{Z}+b_\mathbb{Z}:=a_\mathbb{Q}+b_\mathbb{Q}\quad \quad \quad \quad \quad \quad a_\mathbb{Z}\cdot b_\mathbb{Z}:=a_\mathbb{Q}\cdot b_\mathbb{Q}$$

Which could be defined as:

$$a_\mathbb{Z}+b_\mathbb{Z}:=\cfrac{a}{1}+\cfrac{b}{1}=\cfrac{a+b}{1} \quad\quad \quad \quad a_\mathbb{Z}+b_\mathbb{Z}:=\cfrac{a}{1}\cdot\cfrac{b}{1}=\cfrac{a\cdot b}{1}\tag{$\star$}$$

That is: $\cfrac{a}{1}$ is the representation of $a$ in $\mathbb{Q}$. Given only addition and multiplication, and taking into account the definitions made in $(\star)$, is it possible to define addition and multiplication in $\mathbb{Q}$ in a way that for all $a,b\in \mathbb{Q}$, $(\star)$ is true but the elements $\frac{c}{d\neq 1}$ behave in a way different than we expect for $\mathbb{Q}$?

That is: Could we define it in a way that $1+1=2,\; 2+1=3\; \dots$ but $\cfrac{1}{2}+\cfrac{1}{3}\neq \cfrac{5}{6},\; \cfrac{1}{4}+\cfrac{1}{5}\neq\cfrac{9}{20},\; \dots$?

My intentions: I want to figure out if having only the definition $(\star)$, addition and multiplication forces us to define $\mathbb{Q}$'s operations the way it is defined, that is:

$$\cfrac{a}{b}+\cfrac{c}{d}:=\cfrac{ad+cb }{bd}\quad\quad \quad \cfrac{a}{b}\cdot \cfrac{c}{d}:=\cfrac{a\cdot c}{b\cdot d} $$

Obviously, we could do it in a completely retarded way like saying that $\cfrac{a}{b}+\cfrac{c}{d}:= \cfrac{a+c}{1}$ if $b=d=1$ and $\cfrac{a}{b}+\cfrac{c}{d}:= \cfrac{a^2b^3c^4d^5}{1}$ if $b\neq 1$ or $d\neq 1$ and my original intention is to avoid these but I'm not sure if I described it good enough to really avoid it.

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It reminds me of a talk I had with a professor and asked if there are continuous everywhere differentiable nowhere functions other than Weierstrass function. He replied:

• Yes, there are infinitely many. Let $W$ be the Weierstrass function, take $f(c)=W+c$. For each $c\neq 0$, you have a different continuous everywhere differentiable nowhere function!

Answer 1

NOTE: In order to make this shorter, I will say $\frac a b$ is $ab^{-1}$.

Let's say we want to add division to $\Bbb{Z}$. This means we have to figure out how to add and multiply $ab^{-1}$ and $cd^{-1}$ while still satisfying:

• Associative Property of Addition and Multiplication
• Commutative Property of Addition and Multiplication
• Distributive Property of Multiplication over Addition
• Identity of Multiplication: $a\cdot 1=a$
• Division is Same as Multiplying By Reciprocal: $ab^{-1}=a\cdot b^{-1}$
• Dividing Number by the Same Number Cancels: $aa^{-1}=1$
• Distributive Property of Exponents over Multiplication: More specifically, we want $a^{-1}b^{-1}=(ab)^{-1}$ (Technically, this property is redundant because it can be proven from the above properties, but it's still an important property to know.)

Multiplication is actually pretty easy. By Commutative Property, just have: $$ab^{-1}\cdot cd^{-1}=acb^{-1}d^{-1}=(ac)(bd)^{-1}$$

Now, addition is kind of odd, but the key here is to find common denominator by multiplying $ab^{-1}$ by $dd^{-1}$ and $cd^{-1}$ by $bb^{-1}$: $$ab^{-1}+cd^{-1}=ab^{-1}\cdot dd^{-1}+cd^{-1}\cdot bb^{-1}=ad(bd)^{-1}+bc(bd)^{-1}=(ad+bc)(bd)^{-1}$$

Thus, by just trying to satisfy the properties listed at the top. Even though it might seem like a lot of assumptions, they are pretty normal, regular properties that come up a lot in abstract algebra, we get these definitions of addition and multiplication for fractions. There are also other important properties that you can check out by looking at the field axioms.

Answer 2

A "high-level" explanation was hinted at in the comments already:

If you care about fields and integral domains then you can think of the "process" (functor) of forgetting everything about a field, except that it is an integral domain.

The essentially unique "inverse process" (left-adjoint) takes integral domains to their field of fractions, in particular $\mathbb{Z}$ to $\mathbb{Q}$. That's not just some definition, that's actually a theorem basically telling you that this is "the best you can do" if you want to turn an integral domain into a field.

Answer 3

Suppose you want to have the following relatively simple conditions:

• $\mathbb{Z}$ embeds into $\mathbb{Q}$, namely that $\frac a1 + \frac b1 = \frac{a+b}1$ and $\frac a1 \times \frac b1 = \frac{a \times b}1$, for any integers $a,b$.

• $\mathbb{Q}$ is closed under $+,\times$ each of which is associative and commutative.

• $\sum_{k=1}^n \frac{m}n \overset{Δ}= \underbrace{\frac{m}n+\frac{m}n+\cdots+\frac{m}n}_\text{$n$ terms} = \frac{m}1$, for any integers $m,n$ such that $n > 0$.

• $\sum_{k=1}^n ( \frac{a}b \times \frac{c}d ) = \frac{a}b \times \sum_{k=1}^n \frac{c}d$, for any integers $a,b,c,d,n$ such that $b,d,n > 0$.

• $\sum_{k=1}^n \frac{a}b = \sum_{k=1}^n \frac{c}d$ implies $\frac{a}b = \frac{c}d$, for any integers $a,b,c,d,n$ such that $b,d,n > 0$.

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Then we can obtain the following.

Take any integers $a,b,c,d$ such that $b,d > 0$.

If $a = 0$, then $ad = 0 = a$, and hence $\frac{ad}{bd} = \frac01 = \frac{a}{b}$.

If $a \ne 0$, then $ad \ne 0$ and $bd > 0$ and $\sum_{k=1}^{bd} \frac{ad}{bd} = ad = d \sum_{k=1}^b \frac{a}{b} = \sum_{k=1}^{bd} \frac{a}{b}$, and hence $\frac{ad}{bd} = \frac{a}{b}$.

By symmetry $\frac{cb}{db} = \frac{c}{d}$.

Thus $\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd}$.

Thus $\sum_{k=1}^{bd} ( \frac{a}{b} + \frac{c}{d} ) = d \sum_{k=1}^b \frac{a}{b} + b \sum_{k=1}^d \frac{c}{d} = da + bc = \sum_{k=1}^{bd} \frac{ad+bc}{bd}$, and hence $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$.

Also $\sum_{k=1}^{bd} ( \frac{a}{b} \times \frac{c}{d} ) = \sum_{k=1}^b \sum_{k=1}^d ( \frac{a}{b} \times \frac{c}{d} ) = \sum_{k=1}^b ( \frac{a}{b} \times \sum_{k=1}^d \frac{c}{d} ) = \sum_{k=1}^b ( \frac{a}{b} \times \frac{c}1 )$

$\quad = \sum_{k=1}^b ( \frac{c}1 \times \frac{a}{b} ) = \frac{c}1 \times \sum_{k=1}^b \frac{a}{b} = \frac{c}1 \times \frac{a}1 = \frac{ca}1 = \sum_{k=1}^{bd} \frac{ac}{bd}$, and hence $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$.

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So indeed we are forced to have the usual rules for addition and multiplication of fractions if we want the five conditions listed above. Now I purposely did not state the usual ring/field axioms as the conditions because I wanted to show that 'much less' is already sufficient. Note also that the above does not prove that the fractions form a field. One still needs to show that there is some structure that satisfies the conditions.

But at least the above tells you exactly how to go about constructing such a structure. Namely, you know that a fraction should be equivalent if you divide both numerator and denominator by some common factor. So you might as well enforce that they do not have any common factor, or you could take the alternative route of defining the equivalence relations. Both ways are very well explained in any good textbooks, and the latter is in particular the one to use when constructing fraction fields of general integral domains.