Question 197775
Consider the formula n(A U B) = n(A) + n(B) - n(A ∩ B). 
(a) Show that this relation holds for A = {1, 2, 3, 4} and B= {2, 4, 5, 6, 7, 8} 
<pre><font size = 4 color = "indigo"><b>
A has 4 elements, so n(A)=4
B has 6 elements, so n(B)=6

A U B = all elements which are in either A or B or both =
{1,2,3,4,5,6,7,8)
A U B has 8 elements so n(A U B) = 8

A &#8745; B = all elements which A and B have in common = {2,4}
A &#8745; B has 2 elements, so n(A &#8745; B) = 2

n(A U B) = n(A) + n(B) - n(A &#8745; B).

    8    =  4   +  6   -     2
    8    =      10     -     2
    8    =             8

So it's true.    

(b) Make up your own two sets A and B, each consisting 
of at least six elements. Using these two sets, show 
that the relationship above holds. 

I'll let you do that one by yourself


(c ) Use a Venn diagram and explain why the relation 
holds for any two sets A and B.

{{{drawing(300,187.5,-4,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),locate(3.4,2.5,B),
circle(sqrt(2),sqrt(2),2)
 )}}}


The set (circle) A consists of two parts:

{{{drawing(150,187.5,-4,-.2,-1,4, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}and this part{{{drawing(75,100,-1,1,0.2,2.8, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}

The set (circle) B consists of two parts:

{{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),locate(3.4,2.5,B),
circle(sqrt(2),sqrt(2),2)
 )}}} and this part{{{drawing(150,187.5,0,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}

Now this part {{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}} is common to both sets, so it's A &#8745; B

So if we add the number of elements in A to the number
of elements in B, we would have this

n(A) + n(B)

{{{drawing(150,187.5,-4,-.2,-1,4, 

circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(75,100,-1,1,0.2,2.8, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(150,187.5,0,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  

But as we see, this amounts to adding this part {{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}

TWICE!!!  This part that we have added twice is A &#8745; B.
But we only want to add it ONCE, not TWICE!!!,
So we must subtract the number of elements in A &#8745; B ONCE, so
it will not be added TWICE, but only ONCE  

So we subtract it from n(A)+n(B), and we get 

n(A)+n(B)-n(A &#8745; B)

and we have this:

{{{drawing(150,187.5,-4,-.2,-1,4, 

circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(75,100,-1,1,0.2,2.8, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(150,187.5,0,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  -  {{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}

And the set A &#8745; B that we subtracted away, cancels with
the extra A &#8745; B,

{{{drawing(150,187.5,-4,-.2,-1,4, 

circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(75,100,-1,1,0.2,2.8, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{cross(drawing(75,95,-1,1,0.8,2.2, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 ))}}}  +  {{{drawing(150,187.5,0,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  -  {{{cross(drawing(75,95,-1,1,0.8,2.2, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 ))}}}

and we are left with this:

{{{drawing(150,187.5,-4,-.2,-1,4, 

circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(75,100,-1,1,0.2,2.8, 
 
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}}  +  {{{drawing(150,187.5,0,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
circle(sqrt(2),sqrt(2),2)
 )}}} 
 
which, when we put them back together is this:

{{{drawing(300,187.5,-4,4,-1,4, 
 locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),locate(3.4,2.5,B),
circle(sqrt(2),sqrt(2),2)
 )}}}

which is n(A U B).  That's why the formula works

n(A U B) = n(A) + n(B) - n(A &#8745; B),

the n(A &#8745; B) gets counted once as part of n(A),
and gets counted again in part of n(B), when we add n(A) + n(B),
and so n(A &#8745; B) must be subtracted once to take 
away the extra time it is counted.

Edwin</pre>