@kory
Your overall aim is not just to show that the function \(\displaystyle f:\mathbb{N} \rightarrow\) 4\(\displaystyle \mathbb{Z}^+\) is 1-1 but also that it is onto.

Informally, 'onto' means that any element of 4\(\displaystyle \mathbb{Z}^+\) is \(\displaystyle f(n)\) for some \(\displaystyle n \in \mathbb{N}\).

Informally, '1-1' means that any element of 4\(\displaystyle \mathbb{Z}^+\) is \(\displaystyle f(n)\) for at most one value of \(\displaystyle n \in \mathbb{N}\).

Even though it is obvious that f is both, in proofs you must always start with the

**definitions**. So I look up the definition of 1-1 and I look up the definition of onto (even though I 'know' what they mean), write out the definitions as they apply to this question and simply prove those statments (no matter what looks obvious or not to me). Lots of statements in mathematics are obvious and many of them are not true! So you always stick rigidly to the definitions. Here is a proof.

**Prove that \(\displaystyle f:\mathbb{N} \rightarrow 4\mathbb{Z}^+\) is a bijection (1-1 and onto)**

where \(\displaystyle f(n)=4n, n \in \mathbb{N}\)
**Onto**: \(\displaystyle \quad y \in 4\mathbb{Z}^+ \rightarrow \exists n \in \mathbb{N}: f(n)=y \quad\) (from definition)

__Proof__
\(\displaystyle 4\mathbb{Z}^+ = \{4n:n \in \mathbb{N}\}\)

\(\displaystyle \therefore y \in 4\mathbb{Z}^+ \rightarrow \exists n \in \mathbb{N}: 4n=y\)

\(\displaystyle \therefore \exists n \in \mathbb{N}: f(n)=y\)

Q.E.D.

**1-1**:\(\displaystyle \quad f(n_1)=f(n_2) \rightarrow n_1=n_2 \quad\) (from definition)

__Proof__
\(\displaystyle f(n_1)=f(n_2)\)

\(\displaystyle \leftrightarrow 4n_1=4n_2\)

\(\displaystyle \leftrightarrow n_1=n_2\)

Q.E.D.

**f is a bijection**
__ºººººººººººººººººººººººººººº__

(1) Next I have to prove that the function is a one-to-one correspondence between the set \(\displaystyle N\) and the set \(\displaystyle 4Z^+\)

(2) Suppose \(\displaystyle y\) is a particular but arbitrarily chosen positive integer.

(3) We must show that \(\displaystyle x\) in the natural numbers, \(\displaystyle x \in N\) such that \(\displaystyle f(n)=y\)

(4) \(\displaystyle 4x+2=y\)

(5) \(\displaystyle 4x=y+2\) by addition

(6) \(\displaystyle 4x\)/\(\displaystyle 4\) = \(\displaystyle y+2\)/\(\displaystyle 4\) by division

(7) \(\displaystyle x=y+1\)/\(\displaystyle 2\)

(8) Since \(\displaystyle y=2k+2\)

(9) then \(\displaystyle x=(2k+2)+1)\)/\(\displaystyle 2\) = \(\displaystyle 2k+2\)/\(\displaystyle 2 = k+1\)

*A few comments*
(1) you are only trying to prove it is 1-1 (but also needs to be onto)

(2) y should be an element of 4\(\displaystyle \mathbb{Z}^+\) not \(\displaystyle \mathbb{Z}^+\)

(3) ...that

*there exists x*...f(x)=y (you have x and n)

(4) the function f(x)=4x not 4x+2

(5) this line does not follow from the previous one

(6)(7) when you divide y+2 by 4, you get \(\displaystyle \frac{y}{4} + \frac{2}{4} \text{ or } \frac{y+2}{4}\)

(8) we know that y is of the form \(\displaystyle 4k\)

(9) the brackets don't match and the division by 2 is done incorrectly.

In summary, this is a good example of a formal proof and there are some good points to learn from it.

It all helps. You'll pick up a few things from this for next time. It all takes time and practice!