Silly question about properties of operations

Question

This is a very frivolous question about basic arithmetic. In a precalculus book I'm going through substraction and division are informally defined as the "opposites" of addition and multiplication, and an "inverse" to a number is defined with respect to both. So the inverse of $a$ with respect to addition is $(-a)$ and with respect to multiplication $(1/a)$. Now while both "original" operations are commutative, neither of the new ones is, but both substraction and divison seem to be 'reversible' with respect to their inverses. So while $(a-b)\neq (b-a)$, it is true that $(a-b)=-(b-a)$. The same with division:$$\frac{a}{b}\neq \frac{b}{a}$$ but $$\frac{a}{b} = \frac{1}{\frac{b}{a}}$$ Nice pattern. Can't make much more of it though. Any advice on how to think about it?

Answer 1

This is a consequence of the existence and unicity of the inverse.

For the addition we have that $a-b$ has an opposite ( the ''inverse'' for this operation) $c$ such that: $$ (a-b) +c=0 $$ and this means: $$ a+(-b)+c=0 $$ and, by the existence of the opposite of $a$ and $b$ we find, adding these terms to both sides: $$ a+(-b)+c+(-a)+b=-a+b \Rightarrow c=b-a $$

In a similar way you can prove the result for the division.

Answer 2

You're thinking the right thoughts.

In general inverse functions or commutative functions shouldn't be commutative.

If $a * b = b*a$ and $a \# b = x$ means $a = x*b$ then if $a\#c = c\#a$ then $a = a\#c*c = c\#a*c = c*c\#a$ so $a*a = c*c\#a*a = c*c$ for all a, and c which, I suppose is possible but very restrictive.

An example where a - b = b -a is $\mathbb Z_2$ where 0+0 = 0, 0+1 = 1+0 = 1; and 1+1 = 0. Note: a+a = 0 for all a. Another is $\mathbb Z_2 \times \mathbb Z_2$ where (a,b) + (c,d) = (a+b, b+d). In this case, again, (a,b)+(a,b) =(0,0) for all (a,b).

Let there be an element $\phi$ such that $a*\phi = \phi*a = a$ for all $a$ and let $!a$ be an element such that $a*!a = \phi$ or in other words $!a = \phi \# a$. Does $!(a\#b) = b\#a$? Well..yes.

$!(a\#b)*(a\#b)= \phi => $

$(a\#b)*a =!(a\#b)*(a\#b)*b = \phi*b = b =>$

$ !(a\#b) = !(a\#b)*a\#a = b\#a$.

However, I am pulling a fast one on you by making a HUGE assumption about a*b. Do you know what it is?