I am studying quantitative reasoning and I am wondering how to prove formally that division is not commutative. Everyone knows that 7/3 is not the same as 3/7, but is there a formal way to conceptualize it? Thank you
How to prove formally that division is not commutative?
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Let us define $$\frac{a}{b}:=a b^{-1}$$ for $b\neq0$.. Here, $b^{-1}$ just denotes the element that produces unity upon multiplication by $b$ (stated succinctly, $b\cdot b^{-1}=1$). Since $ab^{-1}\neq a^{-1}b$, we have that division is not commutative. Indeed, take $a\neq bk$ for any $k\in\Bbb Z$. Then $$ab^{-1}=a^{-1}b\Longrightarrow (ab^{-1})^2=1\Longrightarrow a=\pm b,$$which is a contradiction to our assumption that $a\neq bk$. We have used the commutative property of multiplication here.
An alternative, but essentially equivalent method: if $ab^{-1}=a^{-1}b$, then we have $a^2=b^2$. Now all we need to do is to demonstrate a pair of elements $a,b$ whose squares are not equal; in particular, $3^2=9\neq 49=7^2$, so $\frac 37\neq\frac73$. Since it fails for some pair of elements $a,b$, division cannot be commutative in general.
Asserting that $\frac73\neq\frac37$ is a full proof of the fact that division is not commutative. A simpler proof would be the fact that $\frac21\neq\frac12$.