Solve 6th sum nd plz tell how is any mapping done.like what is the rule for this mapping

### Asked by lovemaan5500 | 1st Feb, 2019, 09:39: PM

Expert Answer:

###
A = {1, 2, 3, 4}
B = {2, 5, 6, 7}
f : A → B mapping from A to B
f ={(2, 5), (3, 6), (4, 7)}
f is an injective mapping.
As for every element a A there is an unique element b B
We define a mapping g:A→B given by g = {(2, 2)(2, 5)(3, 6)(4, 7)}
g is not an injective mapping. As the element 2 A is not uniquely mapped
Since (2, 2) and (2, 5) both belong to the mapping g, g is not injective.
Define a mapping h : A →B given by h = {(2, 2), (5, 3), (7, 4)}
h is a mapping from B to A since the every ordered puts {2, 5, 7} B to elements in {2, 3, 4} A

A = {1, 2, 3, 4}

B = {2, 5, 6, 7}

f : A → B mapping from A to B

f ={(2, 5), (3, 6), (4, 7)}

f is an injective mapping.

As for every element a A there is an unique element b B

We define a mapping g:A→B given by g = {(2, 2)(2, 5)(3, 6)(4, 7)}

g is not an injective mapping. As the element 2 A is not uniquely mapped

Since (2, 2) and (2, 5) both belong to the mapping g, g is not injective.

Define a mapping h : A →B given by h = {(2, 2), (5, 3), (7, 4)}

h is a mapping from B to A since the every ordered puts {2, 5, 7} B to elements in {2, 3, 4} A

### Answered by Sneha shidid | 4th Feb, 2019, 02:08: PM

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