Question

Group theory. is the fxn injective? is the fxn surjective?
f(x)=x if x is rational
f(x)=2x if x is irrational

Answer 1

Let a and b be two reals such that :
\[f(a)=f(b)\]
1- If a and b are rational then we get :
\[f(a)=a\text{ and }f(b)=b\]
And then :
\[a=b.\]
2- If a and b are irrational then we get :
\[f(a)=2a\text{ and }f(b)=2b\]
And then :
\[2a=2b.\]
and , so :
\[a=b\].
3- If a is rational and b is irrational then :
\[f(a)=a\text{ and } f(b)=2b\]
so :
\[a=2b\]
and then :
\[b=\frac{a}{2}\in\mathbb Q\]
but this last case is impossible, because b is irrational.

Conclusion : f is an injection !