Question

exhibit a bijection between N and the set of all even integers which are strictly less than 2013.

Answer 1

is N the set of natural numbers here?

Answer 2

r u there @aceecabalquinto ?

Answer 3

yes

Answer 4

sorry about that lol

Answer 5

Let
\[ \large \begin{align*}
f:\mathbb{N}&\to\{x\in\mathbb{Z}:x\text{ is even and }x<2013\}\\
n&\mapsto -n+2013
\end{align*} \]

Answer 6

i'll let u prove it is a bijection.
it is easy.

Answer 7

why is it -n+2012? is "n" any number?

Answer 8

2013*

Answer 9

f(1)=-1+2013=2012

Answer 10

i made a mistake.
wait please.

Answer 11

ok

Answer 12

it should be
\[ \large f:n\mapsto -2n+2014 \]

Answer 13

how come?

Answer 14

the idea was this
\[ \large 1\mapsto2012\qquad2\mapsto2010\qquad3\mapsto2008 \]

Answer 15

oh ok. that makes sense. i'm just gonna try out the bijection right now

Answer 16

go ahead.

Answer 17

ok so i got the injective part alright. i'm kinda confused with the surjection part