Original strip http://spikedmath.com/477.html

The “Shoes and Socks Rule” is all about ‘un-doing’ things. If you want to ‘un-do’ putting on your socks and shoes then you have to do things backward, first take your shoes off then take your socks off, which is just the opposite of how you got them on. Well this much is all clear.

The question is what does this have to do with the following formula:

[1] (XY)^{-1} = Y^{-1} X^{-1}

What it is trying to say is when you do that “Minus-One-Exponent-Thingy” you have to reverse the order. And the “**Minus-One-Exponent-Thingy**”, well that really means “**INVERSE FUNCTION**”. In other words the function that ‘un-does’ the original one (it’s inverse).

The normal way to write this is to do it this way:

[2] (f o g)^{-1}(x) = (g^{-1} o f^{-1})(x)

Another way to write (f o g)(x) is to say f(g(x)) so the above becomes:

[3] f(g(x))^{-1} = g^{-1}(f^{-1}(x))

And the proof is really cool see:

“http://planetmath.org/encyclopedia/InverseOfCompositionOfFunctions.html”

The basis of the proof is that you only need to show that one function undoes a second function and that second function undoes the first function to prove that the two functions are inverses of each other. The rest of the proof relies on the fact that function composition is associative (you can re-group it) and that a function composed with its own inverse gives the ‘identity-function’ (i.e. the do-nothing function) and the identity function composed with anything just gives back that anything since it does nothing.

So here is how that would work: the original function is:

[4] (f o g)(x)

And, the inverse function of that — we want it to turn out to be:

[5] (g^{-1} o f^{-1})(x)

So, to prove this then we think what does a function do? It takes elements from a set (let’s call it set X) to another set (let’s call it set Y). The inverse of the function will take elements of the set Y back to the original set X. Also if you take a function and then apply its inverse the inverse function will UN-DO what the function does. Another way to say this is to say that it leaves things alone.

Or you could say if the combination of a function and another function LEAVES the input set ALONE and the combination of the second function with the first function also leaves the the input set ALONE then the two functions are INVERSES of each other. In symbolic notation this is:

if

[6] (p o q)(x) = x

and

[7] (q o p)(x) = x

then

[8] p(x) and q(x) are inverse functions of each other.

For the proof we are working on we have to show the following two things (here we know f(x)^{1} and f(x) are already inverses of each other and that g^{1}(x) and g(x) are inverses of each other:

[9] ((f o g) o (g^{-1} o f^{-1}))(x) = x

[10] ((g^{-1} o f^{-1}) o (f o g))(x) = x

Here is where the fact that function composition being associative comes in. Notice that if we re-group the parenthesis in [9] the middle two functions turn in to the ‘do-nothing’ function call it “I” for “identity” so we get:

[10] (f o I o f^{-1})(x) = x

And if we notice that composing a function with the “I” or “identity” function just gives us the same function then that “I” in the middle just drops out giving us:

[11] (f o f^{-1})(x) = x

And for sure statement [11] is true because that is just the definition of the action of a function and it’s inverse.

Since you can do the same thing to statement [10] this shows that both statement [9] and statement [10] are true so it is the case that statement [2] is true.

Showing that function composition is associative would be the final thing to show. (Well that and showing that a function composed with the “identity” function is the same as just the function itself.)