# Example 1.1 in Baby Rudin Generalised

In example 1.1 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition, the author shows that there is no rational number $p$ such that $p^2 = 2$; moreover, the set $A$ of all rational numbers $q$ such that $q^2 < 2$ has no largest element in the set of rationals and the set of all rational numbers $r$ such that $r^2 > 2$ has no smallest element.

Now we would like to state the following.

Let $n$ be a positive integer greater than $1$. Then there is no rational number $p$ such that $p^n = 2$; moreover the set $A$ of all rationals $q$ such that $q^n < 2$ has no largest element, and the set $B$ of all rationals $r$ such that $r^n > 2$ has no smallest element.

How to prove this statement, using the same idea as Rudin has used?

How can we state and prove an even more generalised version of the above statement, with $2$ replaced by an arbitrary (positive) integer---or perhaps even by an arbitrary rational number?

• Well there is no $q\in \mathbb{Q}$ such that $q^2=p$ where $p$ is any prime. That is, in other words, $\sqrt{p}$ is irrational for all primes $p$. That could get you started. – Nap D. Lover Mar 20 '16 at 8:01
• You better not get too arbitrary in what you use to replace $2$. For example, you better not replace it with $4$.... – Barry Cipra Mar 20 '16 at 13:14

Partial Answer: For the first question, Its the same idea. Use proof by contradiction,

assume $\frac ab=p$ where $a,b\in \mathbb Z^+$ and $GCD(a,b)=1$

$a^n=2b^n$ implies that $a$ must be even since $b$ doesnt divide $a$, so let $a=2k$

$2^{n-1}k^n=b^n$, but this means that $2$ divides $b^n$ and thus $b$. Contradicting that $GCD(a,b)=1$

For the second statement, its a bit complicated. Originally the reason why Rudin's argument works is because the iteration $p\rightarrow \frac{2p+2}{p+2}$ would converge to $\sqrt 2$.

So we want an expression of $p$ such that the value of $p$ would converge to $2^{1/n}$. We can do this using Newton's method to find the root of $x^n=2$.

To solve $f(x)=x^n-2=0$, begin with any number $x_0$, by the iteration

$$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$$

So for some rational number $p$, we can associate with $q$ using:

$$q=p-\frac{p^n-2}{np^{n-1}}$$

Then $q=\frac{(n-1)p^n+2}{np^{n-1}}$

$$q^n-2=\frac{[(n-1)p^n+2]^n-2n^np^{n^2-n}}{n^np^{n^2-n}}$$

This looks very awful to simplify, but by applying $a^n - b^n=(a-b)(\sum_{k=0}^{n-1}a^kb^{n-1-k})$

$$q^n-2=\frac{[(n-1)p^n+2 -2^{1/n}np^{n-1}][\sum_{k=0}^{n-1}((n-1)p^n+2)^k(2^{1/n}np^{n-1})^{n-k-1}]}{n^np^{n^n-n}}$$

Here is where I got stuck, it is possible to show that $p^n-2$ is a factor of the numerator. But it turns out Newton's method will give us a result that the numerator would contain a factor of $(p^n-2)^2$ (cant prove it but I checked several cases for $n$), which makes it much difficult to show that $q^n-2$ is negative iff $p^n\in A$. Maybe we would need Halley's method to find the iteration of $q$, but that would be much more complicated.