Note that I have proved that xR1y is antisymmetric by doing this:
Assume xR1y ∧ yR1x
x=y so the relation for the first case is indeed antisymmetric
Now how can I prove for the symmetric for the first case, and prove whether the second case is antisymmetric or symmetric?
You have already proved that R 1 is anti-symmetric. I have... View the full answer
- I don't understand your part on y−9y+3(x+3y)...How do you get this figure? Please let me know
- Apr 02, 2019 at 2:38am
- y+3x=y-9y+3x+9y=-8y + 3(x+3y)
- Apr 02, 2019 at 2:39am
- so you are subsititute xR1y=x+3y into y+3x so that y+3(x+3y)=y+3x+9y right? where did you get -9y figure?
- Apr 02, 2019 at 2:48am
- We are working with R2 here...I have assumed x R2 y. that means x+3y is an integer multiple of 8.. to show that, the relation is symmetric we need to prove y R2 x....And for that, we need to show that y+3x is an integer multiple of 8
- Apr 02, 2019 at 3:24am
- To show y+3x is an integer multiple of 8, I subtracted 9y and then added 9y...So, y+3x = y-9y+3x+9y which equals -8y+3(x+3y).
- Apr 02, 2019 at 3:26am
- I didnt get what did you mean by substitution.. If we want to show y R2 x, then we need to work with y+3x..
- Apr 02, 2019 at 3:27am
- ok now I get it. Thanks!
- Apr 02, 2019 at 3:49am
- :) You're welcome
- Apr 02, 2019 at 4:10am