k is a NON-INTEGER rational is what you want to say. also, consider the case where x = 0. it is trivial, but if you are dividing by x, you must consider it.

again, say k is a non-integer rational, clearly. and consider the case y = 0.SYMMETRIC:

Suppose xRy.

x = yk-1.

Suppose yRx.

y = xm-1.

Plugging in for x, and solving for y

y = (yk-1)m-1

y = y(km - m) - 1

1 = (km-m) - 1/y

(km-m) = (y + 1)/y, once again, (km-m) is rational, so R is not symmetric.

now this one is a little shaky. how do we know that (x + 1)/z is not an integer? can z = x + 1? and of course, consider the case z = 0 since you divide by itTRANSITIVE:

Suppose xRy.

x = yk-1.

Suppose yRz

y = zn-1

Plugging in for y and solving for x

x = (zn-1)k - 1

x = z(nk - k) - 1

x/z = (nk - k) -1/z

x/z + 1/z = (nk - k)

(x+1)/z = (nk - k), so again, (nk - k) is rational, so R is not transitive

This is the only way that I could come up to disprove the relation. Please correct me if I am wrong on any work and show how I can correct with an explanation. Much thanks.