5c) Suppose that $r<s$ are rational numbers. Prove that there is an irrational number between $r$ and $s$. Hint: as a start, you know there is an irrational number between $0$ and $1$.
There is no solution for this problem in the solution manual to Spivak's Calculus.
What irrational number is obviously between $0$ and $1$? This is chapter 8 of Spivak's Calculus, and the only proofs in the book so far touching on this matter are
- a proof that $\sqrt{2}$ is not rational, given before the least upper bound property was introduced:
If $A$ is a set of real numbers, $A \neq \emptyset$, and A bounded above, then $A$ has a least upper bound.
and
- a proof that every positive number has a square root (ie, 2 has a square root), and the proof depends on the least upper bound property.
Therefore, I'm guessing he means $\frac{1}{\sqrt{2}}$ is an irrational between $0$ and $1$.
My Proof of 5c
Since $f(x)=x^2$ is continuous, and since $f(1)=1<2<4=f(2)$, $\exists x \in (1,2)$ such that $f(x)=x^2=2$. We denote such an $x$ as $\sqrt{2}$, and therefore $\sqrt{2} \in (1,2)$. We can also prove that $\sqrt{2}$ is not rational.
Therefore, $\frac{1}{2}<\frac{1}{\sqrt{2}}<1$.
We need the following lemmas:
Lemma 1: the sum or difference of a rational and an irrational number is irrational.
Lemma 2: if $a$ is irrational, then $\frac{1}{a}$ is irrational.
Lemma 3: if $a$ and $b$ are rational, so are $a+b$ and $a-b$.
Therefore, $\frac{1}{\sqrt{2}}$ is irrational.
Now, given rational $r$ and $s$, with $r$ < $s$, let $d=r-s>0$.
There are only two possible cases: $d > \frac{1}{\sqrt{2}}$ or $d < \frac{1}{\sqrt{2}}$. Note that $d = \frac{1}{\sqrt{2}}$ is not possible, because if this is true then $r-s$ is irrational, contradicting lemma 3.
Case 1: $d > \frac{1}{\sqrt{2}}$
Let $n=s+\frac{1}{\sqrt{2}}$.
Then $n$ is irrational and
$$s<n<s+d=r$$
Case 2: $d<\frac{1}{\sqrt{2}}$
Let $i_1 = \frac{1}{\sqrt{2}}$.
Then $0<d<i_1<1$.
If we find an irrational number $m$ between $0$ and $d$, we are in case 1 again with $\frac{1}{\sqrt{2}}$ replaced by $m$.
The number $i_2=i_1-d$ is irrational and $0<i_2<i_1$.
In part b) of this problem we proved
$x<y \implies \exists r \in \mathbb{Q}$, $x<r<y$
Thus, $\exists r_1 \in \mathbb{Q}$ such that $0<i_2<r_1<i_1$.
Subtract $i_2$
Then, $0<r_1-i_2<i_1-i_2=d$
But $r_1-i_2$ is irrational, since it is the difference between a rational and an irrational number.
Thus we have found $m=r_1-i_2=r_1+d-\frac{1}{\sqrt{2}}$ that is irrational and $0<m<d$. Thus we are in case 1, and so if we let $n=s+m$, then $n$ is irrational and $s<n<s+d=r$.