# Thread: Explain how these proofs work

1. ## Re: Explain how these proofs work

Originally Posted by Plato
I personally do not like the terms domain/codomain. I have used initial set and final set.

In $f: \mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ the set $\mathbb{Z}\times\mathbb{Z}$ is the initial set and $\mathbb{Z}$ is the finial set.
Ordered pairs of integers are mapped to an integer: $(m,n)\mapsto k$ so $f=\{((m,n),k) : (m,n)\in(\mathbb{Z}\times\mathbb{Z})~\&~k\in \mathbb{Z}\}$.
so to prove it should I come up with an example set? such that m,n =1 so f(1,1) = 2(1)+1 = 3 = k

that is, since f(1, 1) = 3 and 3 is an integer, 3 = k (because k is some integer)?

2. ## Re: Explain how these proofs work

Originally Posted by michaelgg13
so to prove it should I come up with an example set? such that m,n =1 so f(1,1) = 2(1)+1 = 3 = k
that is, since f(1, 1) = 3 and 3 is an integer, 3 = k (because k is some integer)?
I have actually lost the theme of this thread. I even missed - sign on one of your posts.
Here is what I read as the OP:
Originally Posted by michaelgg13
Let f : Z x Z --> Z state whether the following function is onto or not, then prove or disprove.
f(m, n) = m2 + n4
$f: (\mathbb{Z}\times\mathbb{Z})\to\mathbb{Z}$ defined by the rule $(m,n)\mapsto m^2+n^4$.

If again I have not missed something, it is clearly $f(m,n)\ge0$ so can it be onto?

3. ## Re: Explain how these proofs work

Originally Posted by Plato
I have actually lost the theme of this thread. I even missed - sign on one of your posts.
Here is what I read as the OP:

$f: (\mathbb{Z}\times\mathbb{Z})\to\mathbb{Z}$ defined by the rule $(m,n)\mapsto m^2+n^4$.

If again I have not missed something, it is clearly $f(m,n)\ge0$ so can it be onto?
It can't be cause they don't map to anything, correct?

In post #14 I mentioned
Originally Posted by michaelgg13
Don't give me the answer, but for example, explain how you would incorporate this domain/codomain into proving/disproving f(m,n) = 2m+n
And then I took a shot at it by stating this
Originally Posted by michaelgg13
so to prove it should I come up with an example set? such that m,n =1 so f(1,1) = 2(1)+1 = 3 = k

that is, since f(1, 1) = 3 and 3 is an integer, 3 = k (because k is some integer)?
Sorry for getting you mixed up. When I said OP I wasn't trying to refer to the question as a whole, but to understanding the domain/codomain, which you explained to me very well.

4. ## Re: Explain how these proofs work

Suppose we want to prove $f: \Bbb Z \times \Bbb Z \to \Bbb Z$ given by $f(m,n) = 2m+n$ is onto.

Then we must show that for ANY integer $k$, there is at least one pair $(m,n)$ with $f(m,n) = k$.

We can pick $m,n$ any way we want, as long as they are both integers. Suppose we pick: $m = 0, n = k$?

I can tell you that $f$ is NOT injective. See if you can find two different pairs $(m_1,n_1),(m_2,n_2)$ such that:

$f(m_1,n_1) = f(m_2,n_2)$. I suggest finding two such pairs with both images of $f$ equal to 3.

5. ## Re: Explain how these proofs work

Originally Posted by michaelgg13
Sorry for getting you mixed up. When I said OP I wasn't trying to refer to the question as a whole, but to understanding the domain/codomain, which you explained to me very well.
@michaelgg13, May I suggest that you should start a new thread if you change the question.
That is, stick with the PO in any thread..

6. ## Re: Explain how these proofs work

Let's look at the last function:

$f(n) = n^2 + 1$.

Can we find an $n$ such that $f(n) = -1$? Suppose that (just for the time being) one existed. Then:

$n^2 + 1 = -1$
$n^2 = -2$.

If $n = 0$, then $n^2 = 0 \neq -2$. So $n$ is not 0.

If $n > 0$, then $n^2 > 0$ so $n^2 \neq -2$, since $-2 < 0$.

If $n < 0$ then $n = -k$ for some $k > 0$, and thus $n^2 = (-k)(-k) = (-1)(k)(-1)(k) = (-1)(-1)k^2 = k^2 > 0$, so again $n^2 \neq -2$.

So there is NO integer $n$ with $f(n) = -2$, so $f$ CANNOT be onto, since the value -2 is not in the range of $f$.

Can $f$ be injective? Again, no, because $f(-1) = (-1)^2 + 1 = 1 + 1 + 2 = 1^2 + 1 = f(1)$, so $f$ maps -1 and 1 to the same value. That is:

$f(-1) = f(1)$ but $-1 \neq 1$, and an injective function must satisfy:

$f(n) = f(m) \implies m = n$ for ALL $m,n$ and this simply isn't true for n= -1, and m = 1.

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