Let R1={(1,2),(2,3),(3,4)} and
R2={(1,1),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4)} be relations from {1,2,3} to {1,2,3,4}.

For the ordered pair (a,b), write the ordered pairs in increasing order of a and then b, separated by commas without any spaces. For example, (1,2),(1,4),(2,3),(3,4),(3,5). Enter 0 if the set is empty.

a) Find R1∩R2.

R1∩R2={ }

b) Find R1−R2.

R1−R2={ }

Question

Let R1={(1,2),(2,3),(3,4)} and 
R2={(1,1),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(3,4)} be relations from {1,2,3} to {1,2,3,4}. 

For the ordered pair (a,b), write the ordered pairs in increasing order of a and then b, separated by commas without any spaces. For example, (1,2),(1,4),(2,3),(3,4),(3,5). Enter 0 if the set is empty.

a) Find R1∩R2.

R1∩R2={  }

b) Find R1−R2.

R1−R2={  }

vortish · Answer

@Jhannybean could use some help with this one

anonymous · Answer

For $R_1\cap R_2$, is the elements $(x,y)$ in both sets.

anonymous · Answer

So go through each $(x,y)$ in $R_1$, and if it is in $R_2$, then it is in $R_1\cap R_2$ as well.

vortish · Answer

im lost with that wio it may make sense to you but i am lost with that translation

anonymous · Answer

For example, the first element in $R_1$ is $(1,2)$.

anonymous · Answer

Is $(1,2)$ in $R_2$?

vortish · Answer

yes

anonymous · Answer

Then $(1,2)$ is in $R_1\cap R_2$, since it is in both.

anonymous · Answer

Do that for each pair.

vortish · Answer

so for a we should have (1,2)(2,3)(3,4)

anonymous · Answer

Yes

vortish · Answer

and for b we remove those from r2

vortish · Answer

@wio and for b do remove those from r2

anonymous · Answer

Yes

anonymous · Answer

For b, it is every pair in $R_1$ that is not in $R_2$.

vortish · Answer

so its the same as a

anonymous · Answer

Nope

anonymous · Answer

Since $(1,2)$ is in $R_2$, it is not in $R_1-R_2$.

anonymous · Answer

We remove the ones in $R_2$ from $R_2$.

vortish · Answer

then that would leave us with 0

anonymous · Answer

Yep, with nothing

anonymous · Answer

It will be an empty set

vortish · Answer

ok thanks I appreciated it