Compare two sets of natural numbers I am sorry to bother you. I know that this is pretty basic. I have two sets $A = \{a/2, a \in N\}$ and $B = \{2b, b \in N\}$. I have to compare them.  
So far I have:  
For $|A| = |B|$ I need to find bijection. To find bijection, I need to find a morphism, that is injective and surjective.  
I tried to put  
$a/2 = 2b$
a=4b
b = a/4  
I can say, that for every natural number, there is just one natural number, which equals four times that number.  
But can not say, that for every natural number n, there is another natural number equal to n/4 (e.g. 1/4 = 0.25, which is not a natural number).
Can I, therefore, say, that there is no bijection, only injection and $B < A$? And how to write it to look mathematically?
 A: 
Can I, therefore, say, that there is no bijection, only injection and $B < A$?

Nope. You haven’t found a bijection, but that doesn’t mean that there is no bijection.
Here’s a hint. One set is based on $a \in N$ and the other is based on $b \in N$. So rather than setting $a/2 = 2b$ and looking for a bijection there, try simply setting $a = b$. See if you can find a bijection that way.
A: You start with the elements of $A$ being $\alpha=\frac a 2$ where $a \in \mathbb N$ and the elements of $B$ being $\beta=2b$ where $b \in \mathbb N$.  So you can write $a=2\alpha$ and $b=\frac \beta 2$ where both $a$ and $b$ are natural numbers
This would change your attempt at a bijection to $$2\alpha = \frac \beta 2$$ with $\alpha \in A$ and $\beta \in B$, and thus $$\alpha=\frac \beta 4$$ or in the other direction $$\beta=4\alpha$$ which is almost the reverse of what you had.  This works. 
A: Intuitively, these must be in bijection since they are both indexed by $\Bbb N$. In other words, they are "obviously" both the same size as $\Bbb N$.
In this case, it might be easier to show both sets are in bijection with $\Bbb N$. For example, given $n\in \Bbb N$, define $f(n)=n/2$. Then set $g\colon A\to \Bbb N$ using $g(a)=2a$. Now clearly $g\circ f(n)=g(n/2)=n$ and $f\circ g(a)=f(2a)=a$, so $g=f^{-1}$, and $f$ is a bijection.
Next define $F\colon \Bbb N\to B$ using $F(n)=2n$, and $G\colon B\to \Bbb N$ with $G(b)=b/2$. Note that $b/2\in \Bbb N$ since every element of $B$ is even. You can take it from here.
