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].