# Why does the law of distributivity use the distributive property to prove it exists?

I was looking at the Law of Distributivity and disagreed with how it was proven. It proved that $$\frac ab\bigg(\frac cd + \frac ef\bigg) = \frac ab\cdot \frac cd + \frac ab\cdot \frac ef.$$ It multiplied $(c\div d)$ by $(f\div f)$ and it multiplied $(e\div f)$ by $(d\div d)$. Of course, the left hand side of the equation remains unchanged, i.e. $$\frac ab\bigg(\frac cd+\frac ef\bigg)=\frac ab\bigg(\frac cd\cdot \frac ff+\frac ef\cdot \frac dd\bigg)=\frac ab\bigg(\frac{cf}{df}+\frac{ed}{fd}\bigg).$$ This is because $(f\div f)$ and $(d\div d)$ each equal $1$. Now notice that $df=fd$, so since the fraction pair in the brackets have the same denominator, it can be combined, i.e. $$\frac ab\bigg(\frac{cf}{df}+\frac{ed}{fd}\bigg)=\frac ab\cdot\frac{cf+ed}{df}=\frac{a(cf+ed)}{bdf}.$$ The proof now follows by stating that $a(cf+ed)=acf+aed$, but this is the kind of property it is trying to prove in the first place. It is trying to prove this sense of distribution, and in it, it uses it to prove itself.

How is this allowed?

Edit:

This post is very much related, but it does not answer my question.

• It proves distributivity for fractions (i.e. rationals) assuming the distributivity for integers. – Mauro ALLEGRANZA Jun 15 '18 at 6:45
• @MauroALLEGRANZA ok, thank you.... but aren't integers rational themselves? – Mr Pie Jun 15 '18 at 6:47
• See Number system. – Mauro ALLEGRANZA Jun 15 '18 at 6:52
• @MauroALLEGRANZA I am fully aware of the different number systems, but say $a$ was an integer, then there exist values $A$ and $a_0$ such that $A\div a_0=a$. This makes $a$ rational, e.g. $5=10\div 2$. Do we have to use Peano Axioms to prove distributive property for strictly naturals (or integers, but we can just swap the sign)? I mean, this distribute law is proving the property for rationals... assuming that integers (being rational themselves) also fit the law. That's where I don't understand. – Mr Pie Jun 15 '18 at 6:56
• @MauroALLEGRANZA Thank you very much! You can submit that as an answer if you wish :) – Mr Pie Jun 15 '18 at 7:21

Then we have integers, defined as pair $(n,m)$ of naturals: in this way, $(n,0)$ is $n$ and $(0,m)$ is $−m$.
Then we have rationals that are (equivalence classes) of integers : $(r,s)$ with $s≠0$.