Least Upper Bounds based problem I am having difficulty with this problem from chapter 8 in Spivak's Calculus, any help is appreciated.

(a) Suppose that $y - x > 1$. Prove that there is an integer $k$ such that
  $x < k < y$.  Hint: Let $l$ be the largest integer satisfying $l \le x$, and
  consider $l + 1$.
(b) Suppose $x < y$. Prove that there is a rational number $r$ such that $x <
r < y$.  Hint: If $1/n < y-x$, then $ny - nx > 1$
(c) Suppose that $r < s$ are rational numbers. Prove that there is an irrational
  number between $r$ and $s$. Hint: As a start, you know that there is an
  irrational number between $0$ and $1$.
(d) Suppose that $x < y$. Prove that there is an irrational number between $x$
  and $y$. Hint: It is unnecessary to do any more work; this follows from
  (b) and (c).

 A: The part of this that involves the least-upper-bound property is (b).  You know $y-x>0$, so how do you know that there is some integer $n>0$ such that $1/n<y-x$?
If there is none, then every positive integer $n$ is less than $1/(y-x)$.  That means the set of all positive integers has an upper bound in $\mathbb R$; hence a least upper bound in $\mathbb R$.  Call that least upper bound $c$.  Then $c-1$ is not an upper bound, so there is some integer $n>c-1$.  So $n+1>c$ and $n$ is an integer, and we have a contradiction.
A: (a) Let $l$ be the largest integer satisfying $l \le x$.  Then $l + 1 > x$, and we have $$x < l + 1 \le x + 1 < y$$
(b) Assume without loss of generality that $0 < x < y$.  Choose $n \in \mathbb{N}$ such that $1/n < y - x$, and let $k$ be the largest positive integer such that $k/n \le x$.  Then we have
$$x < \frac{k+1}{n} = \frac{k}{n} + \frac{1}{n} \le x + \frac{1}{n} < y$$
(c) We have 
$$r < r + \frac{s-r}{\sqrt{2}} < r + s - r = s$$
(d) We can find rational numbers $r_1$ and $r_2$ such that
$$x < r_1  < r_2 < y$$
by (b) and we can find an irrational between $r_1$ and $r_2$ by (c).
A: I'll try and help. Sidht has already done (a) for you, but I'd encourage you to show why such a $l$ exists. I won't do (b) as the hint gives it a way. I'll do (c) for you and you should then be able to work out (d) I hope. 
Using the hint, we know there exists an irrational number between $0$ and $1$. Do you really? Lets find one. Almost everyone has seen the proof that $\sqrt{2}$ is irrational and clearly, $$0 < 2 < 4 \implies 0 < \sqrt{2} < 2 \implies 0 < \frac{\sqrt{2}}{2} < 1.$$
Now to the problem, assume that $r < s$ and let $x$ be the irrational we found above between $0$ and $1$. As $0 < x < 1$ and $s - r > 0$ we have,
$$ 0 < x(s-r) < s - r.$$
This implies that,
$$r < x(s-r) + r < s.$$
So $y = x(s-r) + r$ is our irrational between $r$ and $s$. You should check that the product of an irrational and a nonzero rational is irrational and that an irrational plus a rational is still irrational.
Hope this helps!
