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Determine whether the function $f: Z×Z→Z$ is onto if

a) $f(m,n)=m+n$.

b) $f(m,n)=m^2+n^2$

c) $f(m,n)=m$.

d) $f(m,n)=|n|$.

e) $f(m,n)=m−n$.

I was able to answer these questions properly, I just wanted to make sure that I was using the proper notation to answer these questions.

Let $z = f(m,n)$ For the functions to be surjective, $\forall(m,n), (m,n) \in \mathbb{Z}^2, \exists z, z \in \mathbb{Z} (f(m,n)=z)$; and could I make it more palatable by writing it as $\forall(m,n) \exists z, (f(m,n)=z)$, where $(m,n) \in \mathbb{Z}^2$ and $z \in \mathbb{Z}$? I have a feeling both say the same thing, and thus are equivalent.

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Rather than writing $\forall x, x\in A$ it is more common to just write $\forall x\in A$. This is more readable, at least once you get used to it (and neither is completely correct from a formal perspective anyway). – Tobias Kildetoft Mar 15 '13 at 15:17
up vote 1 down vote accepted

More notation is not necessarily better. I wouldn’t clutter it up with so much formal notation at all: I’d simply say that for each $\langle m,n\rangle\in\Bbb Z^2$ there is a $z\in\Bbb Z$ such that $f(m,n)=z$. Except that that isn’t what you actually want to say. You have the quantifiers backwards:

$f$ is surjective if for each $z\in\Bbb Z$ there is a pair $\langle m,n\rangle\in\Bbb Z^2$ such that $f(m,n)=z$.

If you insist on using the formal quantifiers, this is

$$\forall z\in\Bbb Z\,\exists\langle m,n\rangle\in\Bbb Z^2\big(f(m,n)=z\big)\;.$$

(And if you’re being very formal, you really ought to write $f\big(\langle m,n\rangle\big)$, not $f(m,n)$: technically $f:\Bbb Z^2\to\Bbb Z$ takes a single argument, and that argument is an ordered pair of integers.)

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Thank you for that. For some reason, using the symbols is more intuitive for me. – Mack Mar 15 '13 at 16:58
@Mack: You’re welcome. – Brian M. Scott Mar 15 '13 at 17:03

(a) is surjective

Proof: For ∀x ∈ Z,∃ (x-m,m) ∈ Z^2 such that f(x-m,m) = x-m +m = x Hence f is surjective

(e) Is surjective

Proof: For ∀x ∈ Z,∃ (x+m,m) ∈ Z^2 such that f(x+m,m) = x+m -m = x Hence f is surjective

(c)Is surjective

∀x ∈ Z,∃ (x,y) ∈ Z^2 such that f(x,y) = x Hence f is surjective

b) and d) are not surjective

Counter example: For -2∈ Z there exist no (m,n) ∈ Z^2 such that f(m,n) = -2 Therefore f is not surjective

One counter example is enough to show that f is not surjective

I hope you get the answer

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