irrational number

zhaobin

let $a_1,a_2,\cdots,a_n$ be $n$ positive rational numbers and let $k_1,k_2,\cdots,k_n$ be $n$ positive integers such that $a_i^\frac{1}{k_i}$ is irrational.prove that $\sum_{i=1}^{n} a_i^\frac{1}{k_i}$ is irrational.

**Posted:** Thu Feb 17, 2005 6:59 pm

3X.lich

zhaobin: maybe you want to impose restrictions to $a_i$ . What happens if $a_i=b_i^{k_i}$ for some positive rational number $b_i$ , for every $i=1\ldots n$ ?
Certainly, $\sum_{i=1}^n a_i^{\frac{1}{k_i}}=\sum_{i=1}^n b_i$ , which is a rational number, contradiction.

**Posted:** Thu Feb 17, 2005 7:07 pm

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zhaobin

but I'm afraid you didn't read the problem carefully.I have already imposed restrictions.isn't it? Smile

**Posted:** Thu Feb 17, 2005 7:22 pm

vess

Yes, it is famous. Wink

It has at least appeared in the Problems column of the Elemente der Matematik. I think it was solved by Pierre on this forum (at least in the case of square roots, but the proof can be easily generalized). However, I can't find his post right now, so I'll type the simple solution below.

But, first of all, why do you introduce the different exponents $1/k_1,\ldots,1/k_n$ ? By putting them into a common denominator $k$ and raising each $a_i$ to the power $k/k_i$ , we may clearly assume that $k_1 = \cdots = k_n=k$ (which we do, of course Smile

).

Consider the polynomial
$P(t):= \prod \big( t - a_1^{1/k} - \omega_2 a_2^{1/k} - \cdots - \omega_n a_n^{1/k} \big) \in \mathbb{Q} [ a_1^{1/k},\ldots, ...$
where the product is over all $k^{n-1}$ choices of $(n-1)$ -tuples $(\omega_2,\ldots, \omega_n)$ of $k$ th roots of unity, and $\zeta$ is a fixed primitive $k$ th root of unity. Because, for every $i \in \{2,\ldots,n\}$ and $k$ th root of unity $\omega$ , changing $a_i^{1/k}$ to $\omega a_i^{1/k}$ does not affect this polynomial, $P(t)$ has the form
$(*) \quad P(t) = P_0(t) + a_1^{1/k}P_1(t) + a_1^{2/k}P_2(t) + \cdots + a_1^{(k-1)/k}P_{k-1}(t) \in \mathbb{Q}[a_1^{1/k}][t],$
where $P_0,\ldots, P_{k-1} \in \mathbb{Q}[ t]$ .

Now, let $S=a_1^{1/k}+ \cdots + a_n^{1/k}$ , and suppose to the contrary that $S$ is rational. By (*) and the fact that $a_1^{j/k}$ is irrational for $1 \leq j < k$ , and the fact that $P(S) = 0$ , it must be that $P_0(S) = \cdots = P_{k-1}(S) = 0$ . In particular, it is also true that
$P_0(S) + \zeta a_1^{1/k} P_1(S) + \cdots + \zeta^{k-1} a_1^{(k-1)/k} P_{k-1}(S) = 0,$
i.e.
$\prod \big( S - \zeta a_1^{1/k} - \omega_2 a_2^{1/k} - \cdots - \omega_n a_n^{1/k} \big) = 0,$
where, once again, the product is over all $k^{n-1}$ choices for the $k$ th roots of unity $\omega_2,\ldots,\omega_n$ . Therefore, there are choices of the $\omega_i$ for which
$x_1(1 - \zeta) + x_2(1 - \omega_2) + \cdots + x_n (1 - \omega_n) = 0$
(here, $x_i = a_i^{1/k} \in \mathbb{R}^+$ ), which is clearly a contradiction because each summand has non-negative real part, and the first summand has strictly positive real part (as $\zeta$ is a primitive $k$ th root of unity and $k > 1$ ). Hence, $S$ is irrational, proving the claim. Q.E.D.

Remark. The argument clearly works to show the more general assertion that the set
$\big\{ q^{1/k} \mid q \in \mathbb{Q}, k \in \mathbb{N}, q^{1/k} \notin \mathbb{Q} \big\}$
is linearly independent over $\mathbb{Q}$ .

Hope this helps,
--Vesselin

vess

There is actually a very minor error in this proof which can be immediately fixed but which I am too lazy to correct right now Smile

. Can you find it? (and fix it Wink

)

**Posted:** Fri Feb 18, 2005 4:49 am

zhaobin

$vess$ ,thank you very much.
I'll try to understand your solution. Smile

**Posted:** Mon Feb 21, 2005 5:42 pm

ondrob

Is it only me who thinks that there's something wrong in Vess's solution ?
Today I was trying to solve this for few hours and was very happy, when found your solution, but:

$P_0(S) + \zeta a_1^{1/k} P_1(S) + \cdots + \zeta^{k-1} a_1^{(k-1)/k} P_{k-1}(S) = 0$

I think it's not true. It would mean that $S$ is root of $P(x)$ but it's true only when $\zeta=1$ , isn't it?

Sorry if I'm misunderstooding something.

**Posted:** Tue May 17, 2005 7:40 am

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vess

As I mentioned in post #4, there is a minor mistake in the proof which is immediately fixed -- but it's not in the line you quote. The proof I wrote is correct in the case when $a_1$ is not a $d$ th power for any $d \mid k, d>1$ (so that the numbers $a_1^{1/k}, a_1^{2/k}, \ldots, a_1^{(k-1)/k}$ are $\mathbb{Q}$ -linearly independent). In that case, however, the line By (*) and the fact that $a_i^{1/j}$ is irrational for $1 \leq j < k$ should be read By (*) and the fact that the numbers $a_i^{1/j}, 1 \leq j < k$ are linearly rationally independent. In view of the $\mathbb{Q}$ -linear independence, it follows from $P_0(S) + a_1^{1/k}P_1(S) + \cdots + a_1^{(k-1)/k}P_{k-1}(S) = P(S) = 0$ and $P_0(S), \ldots, P_{k-1}(S) \in \mathbb{Q}$ that $P_0(S) = \cdots = P_{k-1}(S) = 0$ ( $S$ is a root of each of these polynomials), so it must as well be true that $P_0(S) + \zeta a_1^{1/k} P_1(S) + \cdots + \zeta^{k-1} a_1^{(k-1)/k} P_{k-1}(S) = 0$ (each of the summands is $0$ ). From here, the contradiction is easily obtained: the last equality is equivalent to
$\prod \big( S - \zeta a_1^{1/k} - \omega_2 a_2^{1/k} - \cdots - \omega_n a_n^{1/k} \big) = 0,$
but, as we saw, each multiplier has strictly positive real part and is thus non-zero.

But this, when $a_1^{1/k}, \ldots, a_1^{(k-1)/k}$ are linearly independent. And what if this is not the case? Let $d>1$ be the greatest divisor of $k$ for which $a_1$ is a $d$ th power, and let $k=dl$ (by assumption, $l > 1$ , or else $a_1$ would be a $k$ th power). Then, the numbers $a_1^{1/l}, \ldots, a_1^{(l-1)/l}$ are linearly independent over the rationals, and it follows from $P(S) = 0$ that
$a_1^{(j/l)}P_j(S) + a_1^{(j+l)/l}P_{j+l}(S) + \cdots + a_1^{(j+(d-1)l)/l}P_{j+(d-1)l}(S) = 0 \quad \forall \, j \in \mathbb{Z...$
Multiplying this equality by $\zeta^{dj}$ and summing over $j=0,1,\ldots,l-1$ , we get that
$\prod \big( S - \zeta^d a_1^{1/k} - \omega_2 a_2^{1/k} - \cdots - \omega_n a_n^{1/k} \big) = 0,$
which is likewise a contradiction: every multiplier has strictly positive real part and is therefore non-zero.

Hope this helps Smile

, and sorry for the confusion Blush

...

--Vesselin

vess

In fact, it is not difficult to generalize the proof a little to show the stronger result:

Theorem. Let $b_1, \ldots, b_n \in \mathbb{Q}\setminus \{0\}; a_1,\ldots, a_n \in \mathbb{N}$ , and $k_1,\ldots,k_n > 1$ are integers such that the $a_i$ are multiplicatively independent and no $a_i$ is a perfect $d$ th power for any divisor $d \mid k_i, d>1$ , then the degree of the algebraic number
$b_1a_1^{1/k_1} + \cdots + b_na_n^{1/k_n}$
is precisely $k_1k_2 \cdots k_n$ .

Here is an illustration -- a special case that you may consider a nice olympiad problem:

Problem. Let $p_1,\ldots,p_n$ be distinct primes. Prove that the polynomial
$\prod \big( x \pm \sqrt{p_1} \pm \cdots \pm \sqrt{p_n} \big) \in \mathbb{Z}[x]$
is irreducible over $\mathbb{Z}[x]$ . Here, the product is over all $2^n$ choices of the $\pm$ signs.

--Vesselin

**Posted:** Tue May 17, 2005 12:52 pm

ondrob

Thanx very much, now I understand it quite clearly Smile

I was trying something else when trying to prove it, but I was not able to finish it...

**Posted:** Tue May 17, 2005 1:06 pm

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Mrzne tu, ni příští, i chlebem s nedochodím.
---Michal Májek

vess

What was your idea?

**Posted:** Tue May 17, 2005 10:42 pm

ondrob

My idea was almost the same as you, but I didn't do that trick with $\zeta$ ...
I constructed polynomial (In same way that you did) for which:
$P(a_1^{1/k}+a_2^{1/k}\cdots+a_{n-1}^{1/k})=0$
And from it flows that
$P(S-a_n^{1/k})=0$
But from this point I was unable to move. This can be finished in the same way as you did.

**Posted:** Wed May 18, 2005 2:43 am

_________________
Mrzne tu, ni příští, i chlebem s nedochodím.
---Michal Májek

ondrob

Heh, now I noticed your last post Vess.
I thought that the theorem you wrote is true, but I'm unbale to prove such things, because these things from higher math are spanish village for me... I could prove some easy theorems about algebraic numbers using elementary knowledge, but I think that proviing your Theorem needs some bigger "guns" to kill it Wink

Problem you gived seems nice, I'll try to make it.