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#1
Stuck on this problem, any help is appreciated:

Set A = {1,2,3,4,5,6,7,8} and Set B = {{1,8},{2,7},{3,6},{4,5}}

Define function f: A->B by f(a)=b whenever a ∈ b.

Need to find f(1) anf f(7).

Thanks
0
3 weeks ago
#2
(Original post by SS__)
Stuck on this problem, any help is appreciated:

Set A = {1,2,3,4,5,6,7,8} and Set B = {{1,8},{2,7},{3,6},{4,5}}

Define function f: A->B by f(a)=b whenever a ∈ b.

Need to find f(1) anf f(7).

Thanks
Consider f(1) and look to the definition of f. You're looking to find the element of B (the element in itself is a set), which contains 1.
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#3
(Original post by ghostwalker)
Consider f(1) and look to the definition of f. You're looking to find the element of B (the element in itself is a set), which contains 1.
So f(1) is just {1,8} or no?
0
3 weeks ago
#4
(Original post by SS__)
So f(1) is just {1,8} or no?
Yep.
0
3 weeks ago
#5
(Original post by SS__)
So f(1) is just {1,8} or no?
Yep.
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#6
(Original post by ghostwalker)
Yep.
And would f(7) be {2,7}?
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3 weeks ago
#7
(Original post by SS__)
And would f(7) be {2,7}?
Yep, again.
0
#8
(Original post by ghostwalker)
Yep, again.
Thanks for that. Any hints on how I’d go about showing this function is surjective/injective/bijective?
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3 weeks ago
#9
(Original post by SS__)
Thanks for that. Any hints on how I’d go about showing this function is surjective/injective/bijective?
Standard methods.

Surjective: Does every element of B have an element of A that maps onto it.

Injective: Does f(x)=f(y) imply x=y?

Bijective: Follows from the truth or falsehood of the previous two.
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#10
(Original post by ghostwalker)
Standard methods.

Surjective: Does every element of B have an element of A that maps onto it.

Injective: Does f(x)=f(y) imply x=y?

Bijective: Follows from the truth or falsehood of the previous two.
Would I be right in saying the example I gave is a function that is surjective, not injective therefore not bijective?
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3 weeks ago
#11
(Original post by SS__)
Would I be right in saying the example I gave is a function that is surjective, not injective therefore not bijective?
That's correct.

Every element of B has at least one element of A that maps onto it, under f, so surjective.

f(1)=f(8) for example, so not injective.

And hence not bijective.
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