1
$\begingroup$

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!

Please help me find one if its even possible.

$\endgroup$
  • 1
    $\begingroup$ You can't find a bijective function between $\{0,1\}$ and $\{3,4,5\}$. $\endgroup$ – Zircht Apr 17 '14 at 5:21
  • $\begingroup$ Also note you can't have one element sent to two elements because it fails to be a function then. not that it fails injectivity. Injectivity is that for any element in the codomain there's a unique element in the domain that maps to it, not that every element in the domain maps to a unique element (which is characteristic in general of a function) $\endgroup$ – DanZimm Apr 17 '14 at 5:28
3
$\begingroup$

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?

You can't, as commenters said. There is no bijective function between two finite sets with different number of elements. There is no bijective function between a finite and an infinite set, either.

This observation extends to a definition of cardinality, where two sets have the same cardinality if there is a bijection between them. In view of the above, cardinality captures the concept of "number of elements", extending this concept for arbitrary sets.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Batman May 19 '14 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.