Thread: Order Among the Numbers

1) Prove if x < y, then it is false that y < x; if x > y, then it is false that y > x.

2) Prove if x < y and z is a number, then x-z < y-z.

Originally Posted by noles2188

1) Prove if x < y, then it is false that y < x;

Assume x < y, and assume to the contrary that y < x. since x < y, we have x - y < 0. But also, since y < x, we have x - y > 0. this is a contradiction. (it violates the law of tricotomy)

if x > y, then it is false that y > x.

this is similar to the last one.

2) Prove if x < y and z is a number, then x-z < y-z.

Assume x < y. then that means x - y < 0. But x - y = x + 0 - y = x - z + z - y = x - z - (y - z) < 0

adding y - z to both sides, we obtain the desired result.