Disproving surjectivity of $f : \Bbb Z \times \Bbb Z \rightarrow \Bbb Z$, $f(u,v) = 3u + 6v$

Question

A function $f : \Bbb Z \times \Bbb Z \rightarrow \Bbb Z$ is defined as $f(u,v) = 3u + 6v.$ Is the function surjective? Prove it.

I had the following proof.

Proof
Pick $x = 2$, then $3u + 6v = 2 \Rightarrow 3(u + 2v) = 2$
Let $y = u + 2v$ $\exists y \in \Bbb Z \times \Bbb Z$.
Thus $3y = 2 \Rightarrow y = \frac{2}{3}$.
This is a contradiction because $\frac{2}{3} \not\in \Bbb Z \times \Bbb Z.$
The function is therefore not surjective.

I am a novice at this whole LaTex thing and relatively new to proofs and these surjective proofs are killing me I can't seem to get anything right. Any help is greatly appreciated.

Answer 1

You've got the right idea, but the second half is a bit jumbled, mostly because you introduced a lot of unnecessary variables. I'd phrase it this way:

Suppose that there are $u,v\in\mathbb{Z}$ such that $3u+6v=2$. Then $3(u+2v)=2$, so $u+2v=2/3$. This is a contradiction, since $2/3$ is not an integer and the sum of two integers is also an integer.

I'd say that captures the same idea of your proof, but using half the number of variables.

Answer 2

You have the right idea. But you did not write it properly.

Pick $x=2$, then $3u+6v=2$ implies $3(u+2v)=2$.

So far it is fine. But really you plan to show that $2$ is not in the image/range of the function so you should write in the beginning that "Suppose $f$ is surjective. Let $u$ and $v$ be such that $f(u,v)=2$. Then, ...".

Let $y=u+2v$ $\exists y\in\Bbb Z\times\Bbb Z$.

This does not make sense, since $y$ is supposed to be an integer. You should write: "Then, with $y=u+2v\in\Bbb Z$, we have $3y=2$."

So, $y=2/3\notin\Bbb Z$ a contradiction.

Yeah.

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With that said, here is a rephrasing of your proof (sticking to your notation and writing as much):

Suppose $f$ is surjective. Then, let $(u,v)\in\Bbb Z\times\Bbb Z$ be such that $f(u,v)=2$. That means $3u+6v=2$, which implies $3(u+2v)=2$ or $u+2v=2/3$. But $u+2v\in\Bbb Z$ - a contradiction. Therefore, $f$ is not surjective.

Hope this helps. :)

Answer 3

You should do it without reference to $2/3$.

The idea is right: suppose $2=f(u,v)$, for some $u,v\in\mathbb{Z}$. Then $$ 2=3u+6v=3(u+2v) $$ but this implies that $3$ divides $2$. Contradiction.

Note that saying that $2/3\notin\mathbb{Z}$ is exactly the same as saying that $3$ doesn't divide $2$, so it's not necessary to go beyond integers.

Answer 4

The image of your map is the ideal $$I:=<3,6>\ \subset \ \mathbb{Z}.$$ Because $6 = 2 \cdot 3$ you have $I=<3>$, the set of integer multiples of $3$. Because $2$ is not an integer multiple of $3$ the map is not surjective.