Question

Let f : R \(\to\) R be the Signum Function defined as \(f(x) =   \begin{cases}     1,     & \quad \text x>0 \\     0,  & \quad x=0 \\  -1, &\quad x<0 \end{cases}\)

and \(g: R \to R\) be the Greatest Integer Function given by g (x)= [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0,1]?

Answer 1

It is given that,f : R\(\to\)R is defined as \(f(x) =   \begin{cases}     1,     & \quad \text x>0 \\     0,  & \quad x=0 \\  -1, &\quad x<0 \end{cases}\)
Also, g : R \(\to\) R is defined as g (x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let x ∈ (0, 1].
Then, we have:
[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1. 
\(\therefore\) \(f(x) = f(g(x))=f([x])  \begin{cases}     f(1),       & \quad \text{if } x=1 \\     f(0),  & \quad \text{if } x\in (0,1)   \end{cases}\)\(=   \begin{cases}     1,       & \quad \text{if } x=1 \\     0,  & \quad \text{if } x\in (0,1)    \end{cases}\)
gof (x) = g (f(x))= g (1) [x>0]
=[1]=1.
Thus, when x ∈ (0, 1), we have fog (x) = 0 and gof (x) = 1.

Hence, fog and gof do not coincide in (0, 1].