Foggy Dirichlets

Algebra Level 3

The functions $f(x)$ and $g(x)$ are defined $\mathbb {R^+ \to R}$ such that $f(x)=\begin{cases} 1-\sqrt{x}\quad \text{x is rational} \\ \quad x^2\quad\quad\text{x is irrational}\end{cases}\\g(x)=\begin{cases} \quad x\quad\quad~~~ \text{x is rational} \\ 1-x\quad\quad\text{x is irrational}\end{cases}$

The composite function $f \circ g(x)$ is

many-one, into one-one, onto many-one, onto one-one, into

2 solutions

Pranjal Jain
Apr 12, 2015

Lets try to define $fog(x)$ .

If $x$ is rational, $g(x)$ will be rational, thus, $fog(x)$ will be $1-\sqrt{x}$ .

If $x$ is irrational, $g(x)$ will be irrational, thus, $fog(x)$ will be $(1-x)^2$

$fog(x)=\begin{cases} 1-\sqrt{x}\quad \text{x is rational} \\ (1-x)^2\quad\text{x is irrational} \end{cases}$

$fog(x)$ can never return perfect squares greater than $1$ , for example, $4,9$ . Thus, its an $into$ function.

$fog(\dfrac{1}{\sqrt{2}}+1)=fog(-\dfrac{1}{\sqrt{2}}+1)$ , thus, its a $many-one$ function

The point $-\sqrt{2}+1$ does not lie in the domain of the function $fog$ . Hence this fact can't be used to prove the many one fact of of the function $fog$ .

Prakhar Gupta - 6 years, 2 months ago

Thanks for reporting. I think its fine now.

Pranjal Jain - 6 years, 2 months ago

Yes, it is perfectly OK now.

Prakhar Gupta - 6 years, 1 month ago

@Prakhar Gupta – Anyway, the function $f\circ g$ is ill-defined.

Maxence Seymat - 1 year, 5 months ago

Karan Shekhawat
Apr 14, 2015

One can kill this , by using fact that g(x) has single point continuty that is at $x=0.5$

Hance many-one and into...

Hence Killed

Foggy Dirichlets

2 solutions

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