If then .
If then .
If then .
So it is onto.
My book shows that the set of integers, is denumerable by expressing the elements of the set as infinite sequence
, and then shows that the function
is bijective, where
is defined as
The picture shows that there is one-one correspondence, such as this:
.
.
.
Now I want to prove that bijective, but in the expression is making it very clumsy.
To prove the one-to-one isn't too bad, since seems to look clean, but now when I get to the part where I need to prove it an onto function, I got a huge mess, so I am wondering whether it be alright to split the proof in two parts such that
and define and and by so doing, I plan to prove that
there exists an odd positive integer such that where
and there also exists an even positive integer such that where
Questions:
1. Is there a simpler way than to prove the onto-function in two parts?
2. Is there a better way with handling expression ?