# about "Boolean Relationships on Venn Diagrams"....

#### Hannibal666

Joined Jan 25, 2016
12
Hi, may any one help me in that subset of that chapter ...
Im studing about benn diagrams, in my way I stoped here :

may any one explain to me that "explanation" cause Im beginner,
how we

"we get the illustration above right which corresponds to the inclusive OR function of A, B....."

how can be that ?

"Note that everything outside of double hatched AB is AB-not..."

what is everthing ? is it mean the elements of A and B and (AB)' and the univers?

or just element of (AB)'

may I have an example please? and a simple explanation,,

and that is the link below :

#### Marley

Joined Apr 4, 2016
502
Inclusive OR is the "proper" way of saying OR and (as you correctly say) is the 45degree hatched area on the top right.
There is a function: exclusive OR which is A OR B but not both together. This would be two moon shaped areas without the central overlap. Normally denote with plus sign inside circle.

Everything outside double hatched AB is everything else! I would say this as "NOT(AB)". The rectangular box IS the rest of the universe.

Basically I think you get it. Good for questioning though. Your examples are the rock-bottom fundamentals of logic. Understand this and everything else is easy!

#### WBahn

Joined Mar 31, 2012
30,257
Think of how we use the words OR and AND in everyday usage (though we use OR two different ways allowing context to indicate which).

Would you agree that if we throw a dart at the paper that the region that describes if the dart hits A OR B is the red region at the top right?

Would you agree that if we throw a dart at the paper that the region that describes if the dart hits A AND B is the red/blue region at the bottom right?

#### Hannibal666

Joined Jan 25, 2016
12
I didnt understand

may you give me an example by numbers?

#### WBahn

Joined Mar 31, 2012
30,257
I didnt understand

may you give me an example by numbers?
Say that A is the set of numbers from 10 through 20 (inclusive) and B is the set of numbers from 15 through 25 (inclusive).

You have a hat that contains tickets with numbers from 1 to 50 (inclusive).

You pick the following tickets from the hat: {3, 7, 12, 17, 19, 22, 24, 39, 42, 49}

How many winning tickets do you hold if a winning ticket must be from A OR B?

How many winning tickets do you hold if a winning ticket must be from A AND B?

#### Hannibal666

Joined Jan 25, 2016
12
thanks,,,

may I know why four possibilities? why not five ..

why not five,,, cause the second example is :

it can be the fifth Where we can tell : set B is contained within set A....etc
is that right?

#### shteii01

Joined Feb 19, 2010
4,644
thanks,,,

may I know why four possibilities? why not five ..
View attachment 105523

why not five,,, cause the second example is :

View attachment 105524

it can be the fifth Where we can tell : set B is contained within set A....etc
is that right?
Third pic covers the fifth possibility. It can be looked at as B contains A, but that is already covered in second pic. The other way to explain third pic is A contains B which is the fifth possibility you are asking about. So the third pic covers two possibilities.

#### Hannibal666

Joined Jan 25, 2016
12
Third pic covers the fifth possibility. It can be looked at as B contains A, but that is already covered in second pic. The other way to explain third pic is A contains B which is the fifth possibility you are asking about. So the third pic covers two possibilities.
the third example cover and explain an perfect overlaping...!
I mean
if A={1,2} set B={1,2,3,4,5,6,7,8} that is the second example
why we can tell
B={1,2} Set A={1,2,3,4,5,6,7,8}? and it will be the fifth....

#### WBahn

Joined Mar 31, 2012
30,257
Notice that the diagram does not claim that these are the ONLY possibilities. They are simply the four possibilities of interest.

1) Two sets have no overlap.
2) One set is a proper subset of the other set.
3) The two sets are identical.
4) The two sets partially overlap.

The case of B being within A and the case of A being within B are equivalent from the perspective of that discussion.