I need help with this simple proof. I think I am over analyzing.?

Question

Prove that, for x,y E Reals
If x

I understand this simple concept. It's the proff that is giving me trouble. I would appreciate any help if just some suggestions to start the proof.

Answer 1

I'm not sure exactly what you are allowed to assume (most of these questions are asked because they want you to prove it from first principles, and this depends on what are defined to be "first principles").
But here goes... don't be freaked out by the length (I'm just writing it all out to - hopefully - make things easier to see). The actual proof is pretty short. I'll leave you to be the judge as to whether you were overanalyzing.

The fact that x,y are real is irrelevant... any partial order will do. By definition, a partial order has the following properties (I'll write <= for "less than or equal"):

(1) a <= a (reflexivity)
(2) IF a <= b and b <= a THEN a = b (antisymmetry)
(3) IF a <=b and b <= c, THEN a <= c (transitivity)

Then saying "a < b" is really saying "a <= b, but the two are NOT equal". I'm going to assume that you don't have to prove that the reals (with the normal <= we know) have these properties, but I guess you could.

One (very) useful observation is that if x < y, then x <= y (useful because you can turn things like a < b into a <= b and then use the reflexive/antisymmetric/transitive properties).

There are a number of ways to go about proving this, but one way is to show that
(1) x <= z is true (hint: the observation above), but that
(2) x = z is NOT true (hint: by contradiction)

and that will show x < z. It could probably be made easier using the property of the reals that exactly one of the following is true: either x < y or x = y or x > y (this has a name, but I forget it at the moment). Another approach would be to show that transitivity holds for < the same way it does for <= and then splitting y <= z into two cases (y = z or y < z). But this seemed (to me) to be the most straightforward way.

I'll write the proof below, just in case you need it, but if you just wanted a hint, stop reading here.

First a small lemma: If a <= b, then it is NOT the case that a > b.
To see this, assume (for contradiction) that a <= b and a > b. Since a > b then (by the observation) a >= b, so by antisymmetry a = b. But this is a contradiction, since a > b means that a and b can't be equal.

Now to show the proof:
Assume x < y and y <= z. Since x < y, then (by the observation) x <= y. So by transitivity x <= y and y <=z implies x <= z. Thus if we can show that x and z are not the same, we are done.

Assume (for contradiction) that x = z. Then (by the observation) z <= x. So by transitivity y <= z and z <= x, so y <= x. We also know that x < y, but this is a contradiction (due to the lemma above).

That's it!

Answer 2

if x is less than y and z is equal or greater than y. z will of coarse be greater than x because x is smaller than y which is ether equal or less than z.
example: 3=x which is less than 4=y and z is an unknown number that could be equal to 4=y or greater as in 5 which would make it greater than 3=x simple because z is greater or equal to y.
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Answer 3

if x
if x = 5, y=6 and z=6, x
since x can never equal y and y can equal z, x can never equal z.

you can essentially substitue z for y since y can equal z.

Answer 4

we have

x-z = (x-y) +(y-z)

as x-y < 0 and y-z < equal to 0

we have

x-z =( x-y) + (y-z) < 0 + (y-z) < 0
so
x

Answer 5

z>or=y and y>x then z>x