Allen Career Institute: Discrete Math

Question

Let $f : A \rightarrow B$ be a bijection and let $E,F$ be subjects of $A$, Now, we consider the following statements about the function $f :$ $P : f(E \cup F) = f (E) \cup f(F)$ $Q : f(E \cap F) = f (E) \cap f(F)$ Which of the following is TRUE about $P$ and $Q$ ?

$(1)$	Only $P$ is correct
$(2)$	Only $Q$ is correct
$(3)$	Both $P$ and $Q$ are correct
$(4)$	None of $P$ and $Q$ is correct

 

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I thought $Q$ is true, but answer is both true. Is both true because of bijective function or ans given incorrect?

Comments

If $A\rightarrow B$ is a function, not bijection then $P$ are true,but there are given  $A\rightarrow B$ is bijection so $P$ and $Q$ both are true.

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P can not be true, as far I know

Otherwise check with some example

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The answer has to be 3, as each element in set A has only one unique incidence in set B.

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Can u explain more?
In union reflexivity and transitivity property maintained , but symmetric relation property is not maintained

isnot it?

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It is given that E & F are subsets of A, so lets say elements of E maps to some elements of co-domain and elements of F maps to some elements of co-domain..

So the incidence of elements of E U F will be same as that of incidence of elements of E U incidence of elements of F

let me take an example and explain..

let A = {1,2,3,4,5} and B = {v,w,x,y,z} where 1 -> v, 2-> w, 3-> x, 4->y, 5->z

and E = {1,2,3}, F= {2,5}
f(E U F) = {v,w,x,z} , f(E) = {v,w,x}, f(F) = {w,z}  and f(E) U f(F) = {v,w,x,z},

f(E U F) = f(E) U f(F) = {v,w,x,z}

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@balchandar reddy san

Still doubt in ur ans

Say, for these function

$f\left ( 1,2 \right )=\left \{ \left ( 1,1 \right ) , \left ( 2,2 \right ),\left ( 1,2 \right ),\left ( 2,1 \right )\right \}$

$f\left ( 2,3 \right )=\left \{ \left ( 2,2 \right ) , \left ( 3,3 \right ),\left ( 2,3 \right ),\left ( 3,2 \right )\right \}$

$f\left ( 1,2,3 \right )=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ) ,\left ( 3,3 \right ),\left ( 1,2 \right ),\left ( 2,1 \right ),\left ( 1,3 \right ),\left ( 3,1 \right ),\left ( 2,3 \right ),\left ( 3,2 \right )\right \}$

See here

$f\left ( E\cup F \right )\neq f\left ( E \right )\cup f\left ( F \right )$

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Are trying to say that ordered pair (1,2) in set A maps to 4 ordered pairs in set B, if yes then does it satisfy one-one property?

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I donot understand, what they mean by bijection

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bijective = injective (one-one) + surjective (onto)

injective ( one-one) ---- every element in set A can be mapped to only one unique element in set B

surjective ( onto) ---- all the elements in set B should be mapped by atleast one element from set A  ( co-domain = range)

I hope, this is clear..

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@balchandar reddy san

What is the basic difference between Co-domain and Range?

Please explain using some examples?

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let A = {1,2,3,4,5} and B = {v,w,x,y,z} where 1 -> v, 2-> w, 3-> x, 4-> y, 5-> y

Here, the function is neither one-one nor onto..

why not one-one? because, 4 & 5 maps to same element,

why not onto? because, co-domain: {v,w,x,y,z} and range: {v,w,x,y}

co-domain: all elements of set B

range: elements which have been mapped in set B ( from set A)

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@balchandar reddy san

I got it,thank you so much

Answer 1

Ans-->3

It is always better to solve these types of question by a test case.

Bijection means Function which is both 1 to 1 as well as onto that is each element of the domain must map with a single element of the co-domain.

Say we have a set A = {1,2,3,4} & set B ={W,X,Y,Z} and the mapping be {1-> W, 2-> X, 3->Y,4->Z }

let E = {1,2} and F={2,3}

Now , f{EUF} = f{1,2,3} = {w,x,y} and similarly f{E} U f{F} = {W,X} U {X,Y} = {W,X,Y}

        f{E ∩ F} = f{2} = {x} and similarly f{E}  ∩ f{F} = {W,X}  ∩ {X,Y} = X

So both are true

PS--> if it was given only a function but not a bijection then Q would have never been true..! Then only P should be true,this can be proved in the same way as above

Refer--> https://gateoverflow.in/721/gate2001-2-3.

Comments

But for an equivalence relation the opposite case holds true.. I was thinking of equivalence case at the time of writing the answer so i wrote P instead of Q there..my bad..!

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yes, but that was not bijection

right?

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No

that's why I said that if it is an ordinary function then Q cant be true.