Question

Prove:
The set \( T=\{ n \in \mathbb{Z} : n=2(mod6) \} \) is denumerable

Answer 1

Is the following proof correct?

Answer 2

Define a function F on the integers by setting F(z)= 6z+20. F is one to one because if F(u)=F(v) then 6u+20 = 6v+20 so u=v. Every element t of T has the form 6k+2 for some integer k and t=F(k-3) so F maps onto T. Therefore T is equivalent to \( |mathbb{Z} \) which is denumerable

Answer 3

@KingGeorge can you check this when you have the time. Now that i reread it i think its correct but i just want ur opinion on this

Answer 4

I believe your proof is correct.

Answer 5

ya but y cldnt we just have \(\huge f(z) = \frac {z-2} {6} \) ?

Answer 6

I am just wondering if we can choose nething?

Answer 7

You can't quite choose anything. The example you just posted is one you probably should avoid since it doesn't output an integer for all \(z\in\mathbb{Z}\).

Answer 8

For example like what?

Answer 9

Let \(z=3\). Then \(f(3)=\frac{3-2}{6}=\frac{1}{6}\)

Answer 10

ohhhh that is cool

Answer 11

Unless of course you defined the function as follows \[f\;:T\longrightarrow\mathbb{Z}\]\[\quad \;\;\,z\longmapsto\frac{z-2}{6}\]This function would be fine to use.

Answer 12

ohhhhh okkkkk i seeeeee

Answer 13

Thanks you are awesome

Answer 14

You're welcome.