Least Upper Bounds based problem

Question

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).

score 1 · Answer 1 · answered Jun 11 '13 at 23:01

1

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.

answered Jun 11 '13 at 23:01

Michael Hardy

1

(b) doesn't require the least upper bound property. Just that the reals are archimedean (which is usually proven using that they have the least upper bound property, admittedly). The conclusion is true for the rationals as well. – kahen Jun 11 '13 at 23:27
@kahen : The fact that the reals are Archimedean is indeed enough, and it is weaker than the least upper-bound-property, BUT how does one prove the reals are Archimedean in the context of Spivak's book without using the least-upper-bound property? Spivak says the reals are a complete ordered field. "Complete" means they satisfy the least-upper-bound property. Given that, you then have to prove they're Archimedean. And that's what I did above. – Michael Hardy Jun 11 '13 at 23:30
(+1) for verifying the hint. I find it common for people to just assume hints when in fact they require attention. – Jun 11 '13 at 23:39

Ink · Accepted Answer · 2013-06-12T02:23:48.373

1

(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).

edited Jun 12 '13 at 02:23

answered Jun 11 '13 at 23:19

Ink

5,126

How do you know $\ell + 1 > x$ and not $\geq$? – Lemon Jun 11 '13 at 23:27
@sidht By choice of $l$, $l + 1 \ne x$, so $l + 1 > x$. – Ink Jun 11 '13 at 23:30
What do you mean? What if $\ell = 0$ and $x = 1$? – Lemon Jun 12 '13 at 00:47
@sidht I defined $l$ to be the largest integer satisfying $l \le x$, not $l < x$. So if $x = 1$, then $l = 1$. – Ink Jun 12 '13 at 01:01
The hint says $\ell < x$, not $\ell \leq x$ – Lemon Jun 12 '13 at 01:37
@sidht I'm not following the hint. If you see something wrong with my argument, please point it out. – Ink Jun 12 '13 at 02:17
but you say you are following the hint. Please read your answer again. You said Following the hint, let $l$ be.... – Lemon Jun 12 '13 at 02:21
You are correct. I just checked my Spivak's book and it appears the OP made a typo which fixes my confusion with your answer. – Lemon Jun 12 '13 at 02:45

score 0 · Answer 3 · 2013-06-11T23:49:14.063

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!

Least Upper Bounds based problem

3 Answers3