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.

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edited Jun 12 "20 at 10:38 Community♦
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asked Mar 15 "13 at 15:10 MackMack
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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^2ig(f(m,n)=zig);.\$\$

(And if you’re being very formal, you really must write \$fig(langle m,n angleig)\$, 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.)

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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

(c)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

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edited Mar 15 "13 at 15:37
answered Mar 15 "13 at 15:30 kalpeshmpopatkalpeshmpopat
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