Question

Let $f\left(x\right):\{\begin{array}{c}x,xisrational\\ 0,xisirrational\end{array}$

and

$g\left(x\right):\{\begin{array}{c}0,xisrational\\ x,xisirrational\end{array}$

If $f:R\to R$ and $g:R\to R$, then $(f-g)$ is

• A. one-one and into
• B. neither one-one nor onto
• C. many-one and onto
• D. one-one and onto

Answer 1

The correct option is A one-one and into

When x is rational $f\left(x\right)=x,g\left(x\right)=0$

$\therefore (f-g)=x-0=x$.So for each rational no. in $(f-g)$ there is the same rational number in co domain $f\left(x\right)=0,g\left(x\right)=x$ [when x is irrational]

$\therefore (f-g)=0-x=-x$

So for each irrational number x in domain, we have $(-x)$ in co domain.

Conversely for each negative irrational number, there is an element in domain.

But for positive irrational no. there is no corresponding element.

Thus $(f-g)$ is one -one and into.