Note that I have proved that xR1y is antisymmetric by doing this:

xR1y⇔x+3y=8

Assume xR1y ∧ yR1x

By definition:

x+3y=8 (k1)

y+3x=8 (k2)

y+3x=x+3y

__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?

#### Top Answer

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
- daltopn
- Apr 02, 2019 at 2:38am

- y+3x=y-9y+3x+9y=-8y + 3(x+3y)
- nabaneetdas
- 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?
- daltopn
- 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
- nabaneetdas
- 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).
- nabaneetdas
- 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..
- nabaneetdas
- Apr 02, 2019 at 3:27am

- ok now I get it. Thanks!
- daltopn
- Apr 02, 2019 at 3:49am

- :) You're welcome
- nabaneetdas
- Apr 02, 2019 at 4:10am