Question

Find f:Natural rightarrow(0,1) that is 1-1

Hopefully I already have the answer I just want someone to confirm it. It's been over a year since I took mathematical reasoning so I'm a bit rusty.

Anyway I believe I can answer this by mapping the set of natural numbers (1,2,3,4,5,...) to the set (0.1,0.2,0.3,0.4,0.5,...) which is contained in the set (0,1).

Answer 1

\[f:\mathbb{N} \rightarrow(0,1) \ that \ is 1-1\]
messesd up the formatting in the question... this is how it should look

Answer 2

i don't think that will work

Answer 3

what is \(f(100)?\)

Answer 4

i am sure there are lots of ways to do it though, just have to make sure that they do not get to 1

Answer 5

good point.

yeah, I will think on it a little more. I knew that seemed a little too easy but I hadn't thought it all the way to there.

Answer 6

don't think too hard

Answer 7

hint, any one to one rational function where the numerator is less than the denominator

Answer 8

ah... so mapping (1,2,3,4,...) to (1/1, 1/2, 1/3, 1/4,... should do it.

Answer 9

yeah that will work for sure

Answer 10

or \(\frac{n}{n+1}\) or anything like that

Answer 11

very cool.. and yeah I was overthinking it. thanks so much!

Answer 12

yw