# Bijective function with different domain and co-domain element count

To be bijective is to be both injective and surjective.

Which in other words, have to have a one-on-one match right?

Then how am I supposed to come up with a bijective function if the domain has a even number of naturals and the co-domain has a odd number of naturals?

For example, if the Domain: $\{0,1\}$ | Co-domain: $\{3,4,5\}$ ==> even in this situation, it will not be surjective because not all of the co-domain are hit, and it can't be that an element in the domain hits two or more elements in the codomain because then it won't be injective!

• You can't find a bijective function between $\{0,1\}$ and $\{3,4,5\}$. – Zircht Apr 17 '14 at 5:21
• But, cardinality is a bit of a fun concept. For example, you can have a bijection between a strict subset of an infinite set and itself - for example, $(-\frac{\pi}{2},\frac{\pi}{2}) \to \mathbb{R}$ via $\tan(x)$ is a bijection, but $(-\frac{\pi}{2},\frac{\pi}{2}) \subset \mathbb{R}$ with the inclusion being strict (of course, both $(-\frac{\pi}{2},\frac{\pi}{2}) , \mathbb{R}$ have the same cardinality though). – Batman May 19 '14 at 16:15