I had the ability to answer these inquiries properly, I simply wanted to make sure that I was utilizing the correct notation to answer these concerns.
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Let $z = f(m,n)$ For the functions to be surjective, $forall(m,n), (m,n) in rianowens.tvbbZ^2, exists z, z in rianowens.tvbbZ (f(m,n)=z)$; and could I make it even more palatable by writing it as $forall(m,n) exists z, (f(m,n)=z)$, wright here $(m,n) in rianowens.tvbbZ^2$ and also $z in rianowens.tvbbZ$? I have a feeling both say the very same thing, and hence are tantamount.
edited Jun 12 "20 at 10:38
asked Mar 15 "13 at 15:10
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More notation is not necessarily better. I wouldn’t clutter it up with so a lot formal notation at all: I’d simply say that for each $langle m,n angleinBbb Z^2$ there is a $zinBbb Z$ such that $f(m,n)=z$. Except that that isn’t what you actually desire to say. You have actually the quantifiers backwards:
$f$ is surjective if for each $zinBbb Z$ tbelow is a pair $langle m,n angleinBbb Z^2$ such that $f(m,n)=z$.
If you urge on using the formal quantifiers, this is
$$forall zinBbb Z,existslangle m,n angleinBbb Z^2ig(f(m,n)=zig);.$$
(And if you’re being very formal, you really must write $fig(langle m,n angleig)$, not $f(m,n)$: technically $f:Bbb Z^2 oBbb Z$ takes a single discussion, and also that discussion is an ordered pair of integers.)
answered Mar 15 "13 at 15:18
Brian M. ScottBrian M. Scott
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(a) is surjective
Proof: For ∀x ∈ Z,∃ (x-m,m) ∈ Z^2 such that f(x-m,m) = x-m +m = xHence f is surjective
(e) Is surjective
Proof: For ∀x ∈ Z,∃ (x+m,m) ∈ Z^2 such that f(x+m,m) = x+m -m = xHence f is surjective
∀x ∈ Z,∃ (x,y) ∈ Z^2 such that f(x,y) = xHence f is surjective
b) and also d) are not surjective
Counter example: For -2∈ Z tright here exist no (m,n) ∈ Z^2 such that f(m,n) = -2 Because of this f is not surjective
One respond to example is sufficient to present that f is not surjective
I hope you get the answer
edited Mar 15 "13 at 15:37
answered Mar 15 "13 at 15:30
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