Let f : R → R and g : R → R be defined by f(x) = x + 1 and g(x) = x − 1. Show that fog = gof = I_R.

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Solution

Given, f : R → R and g : R → R
$\Rightarrow$ fog : R → R and gof : R → R (Also, we know that I_R : R → R)
So, the domains of all fog, gof and I_Rare the same.
$(f o g) (x) = f (g (x)) = f (x - 1) = x - 1 + 1 = x = I_{R} (x) . . . (1) (g o f) (x) = g (f (x)) = g (x + 1) = x + 1 - 1 = x = I_{R} (x) . . . (2) From (1) and (2), (f o g) (x) = (g o f) (x) = I_{R} (x), \forall x \in R Hence, f o g = g o f = I_{R}$

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