How to prove $a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+.....+a_{n}\sqrt[b_{n}]{c_{n}}$ is irrational?

Asked Feb 15 '16 at 22:13

Active Feb 15 '16 at 22:13

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Let's define the number

$$A=a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+.....+a_{n}\sqrt[b_{n}]{c_{n}}$$

where $a_{1}, a_{2}, ..., a_{n}$ are positive integers and $b_{1}, b_{2}, ..., b_{n}, c_{1}, c_{2}, ..., c_{n}>1$ are positive integers.

For every $1\leq i\leq n$, $c_{i}$ cannot be divided by any $p^{b_{1}}$ where $p$ is a prime number.

How to prove $A$ is irrational?

asked Feb 15 '16 at 22:13

esege

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Can you prove it for n=2? – marty cohen Feb 15 '16 at 22:40
It's not true as it's currently stated. Take for example $n=1$, then $1\cdot \sqrt[2]{4}$ is not irrational – Feb 15 '16 at 22:52
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@vrugtehagel : $4$ can be divided by $p^{b_1}$, where $p=2,b_1=2$. – Watson Feb 15 '16 at 22:55
Excuse me, I thought that was a result that the OP got from given conditions. I didn't realize it was a condition. – Feb 15 '16 at 22:56
Related: https://math.stackexchange.com/questions/440453 – Watson Nov 27 '18 at 15:37
One may use the approach given in this answer, see also this excerpt from Kato, Saito Number theory 2. – Watson Nov 28 '18 at 13:44
(Observe that if $A$ were rational then $$a_{1}\sqrt[b_{1}]{c_{1}} \in K := \Bbb Q(\sqrt[b_{2}]{c_{2}}, ..., \sqrt[b_{n}]{c_{n}}),$$ so that the techniques from my above comment apply.) – Watson Nov 28 '18 at 13:46

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