Let be given with . Prove that there exists an irrational number such that .

*Note: To do this problem, I think we need to assume the existence of an irrational number. We will prove the existence of such a number (the ) in I.3.12, Exercise #12.*

* Proof. * Since the rationals are dense in the reals I.3.12, Exercise #6, we know that for with there exist such that

Now, assume the existence of an irrational number, say (see note preceding the proof about this). Since we know and from the order axioms exactly one of or is positive ( is nonzero since ). Without loss of generality, let . Then, since , we know there exists an integer such that

Also, since , we have ; thus, .

Then, by I.3.12, Exercise #7 we have irrational and hence irrational.

Thus, letting , we have with irrational

Assume z is irrational and z<x

Let x and y be real numbers s.t x<y

– If x<z then z= x+(z-x)

– if x<y then x+h/n<y

– Since h is real, there exists an h= x-z such that x+(z-x) = x+h/n where z-x= h

– Also since the set of positive integers is not bounded above, there are infinite values for z-x/n

if we let x+(z-x)/n = r we have x<r<y

There exist a irrational number between a rational and a irrational number

I think it is not necessary to put those x and y since the proof only argues over s and r, and as you see, the conclusion is that the irrational number is between s and r, these two numbers can be any real, so the proof is already complete.

You need the rational numbers because if you just used x for eg, x can be irrational, and an irrational plus an irrational can be rational so the proof wouldn’t work in that case