Let R be the set of real numbers and Q the set of rational numbers. Is the function f:R to R a bijection where f(t)= 2-t^2 if t is in Q and f(t)=t^2-2 if t is R\Q (complement set)

Question

helder_edwin · Answer

the function is not injective, so it is not bijective.

anonymous · Answer

can you explain why it's not injective. Is it because there is more than one value for t in f(t)? like 1 and -1 give the same answer?

helder_edwin · Answer

no. it is from the definition: since
$$\large  f(1)=2-1^2=1 $$
$$\large  f(-1)=2-(-1)^2=2-1=1 $$
but $1\neq-1$ then the function is not injective.

anonymous · Answer

Would it be surjective?