$\sum_{i=1}^n {(a_i\sqrt{b_i})} \ne 0$

Question

In a surd $a\sqrt{b}$ ($b \in \mathbb{Z^+}$) the value of $b$ can assumed to be a square-free integer ($b = p_1p_2\dots p_k$, where $p_i$ are distinct primes), since otherwise a multiple prime factor can be (completely or partially) taken out of the square root sign.

I am struggling to find an elementary proof of the following:

If $\{a_i\}_{i=1}^n$ are non-zero integers, $\{b_i\}_{i=1}^n$ are distinct positive square-free integers, then $\sum_{i=1}^n{a_i}\sqrt{b_i} \ne 0$.

This is how it can be proven for n=2.

Suppose $a_1\sqrt{b_1} + a_2\sqrt{b_2} = 0$. By squaring the equation, $\sqrt{b_1b_2}$ can be expressed as $$ \sqrt{b_1b_2} = - \frac{a_1^2b_1+a_2^2b_2}{2a_1a_2} $$

RHS of the equation is a rational number. However, since $b_1$ and $b_2$ are distinct and void of multiple prime factors, there should be a prime factor, containing in one of those, but not in another. This factor appears in $b_1b_2$ only once, therefore $b_1b_2$ is not a full square, so LHS $\sqrt{b_1b_2}$ is irrational.

Tried induction — no luck.

Your statement needs more detail. If $a_1=-a_2$ and $b_1=b_2$, then $a_1\sqrt{b_1}+a_2\sqrt{b_2}=0$. — herb steinberg, Feb 02 '19 at 02:37
@herbsteinberg The question states that "$\left{b_i\right}_{i=1}^n$ are distinct positive integers". Thus, $b_1 \neq b_2$. — John Omielan, Feb 02 '19 at 02:39
People usually say what you wrote about $b$ in a slightly different way: "$b$ is not divisible by the square of any prime" — Zubin Mukerjee, Feb 02 '19 at 02:39
This is not quite the same, but if you consider the more general situation of the right side not being a rational value, which I suspect is also true, then perhaps the somewhat similar question of Proving that if $x_1,\dots,x_n$ are rational numbers and $\sqrt{x_1}+\dots\sqrt{x_n}$ is rational, then each $\sqrt{x_i}$ is rational as well may help, although it deals with only positive quantities. — John Omielan, Feb 02 '19 at 03:31
@John Omielan. Indeed. Assuming $b_n = 1$, we get that the the sum of other surds must not be integer. Then, if there are rationals instead of integers, just multiply by square root of common denominator, and take multiple factors out, if any. Hopefully, no one will notice that, or I will be penalised for a duplicate question:) BTW, that post doesn't seem to have an answer either. — cyanide, Feb 02 '19 at 06:05
@cyanide Note that as the $b_i$ are square-free integers, no $b_i$ may be $1$ since, by the definition I'm aware of, $1$ is not square-free. — John Omielan, Feb 02 '19 at 06:07
Why 1 is not square-free? it doesn't have a square of a prime as a factor. Actually the referred post has a good proof, based on $\mathbb{Q(\sqrt{n}})$ being a filed. See, if I can apply the idea to this case. Not quite elementary, but better than nothing. — cyanide, Feb 02 '19 at 06:14
@cyanide I'm sorry, but you are right that $1$ is square-free. I was thinking of something else. — John Omielan, Feb 02 '19 at 06:38
@cyanide The linked question does have an answer, viz. my answer there shows how to eliminate the (rudimentary) field theory in my linked proof to obtain a more elementary proof. If this is not clear then comment there and I can elaborate. We can do the same elimination in this answer to handle your question. — Bill Dubuque, Feb 02 '19 at 16:28

score 0 · Answer 1 · answered Feb 02 '19 at 17:03

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Here is a different approach to the $n = 2$ proof:

If $a_1\sqrt{b_1} + a_2\sqrt{b_2} = 0$, then, multiplying through by $\sqrt{b_2}$, we have $a_1\sqrt{b_1b_2} = -a_2b_2$

Proceed as with your proof.

Can you see how this is more easily generalizable to $n > 2$?

answered Feb 02 '19 at 17:03

Paul Sinclair

43,643

1

I tried to develop your idea. For three terms $a_1\sqrt{b_1b_2} + a_2\sqrt{b_2b_3} = -a_3b_3$ there is no contradiction straight away while multiplication by a surd doesn't seem to help since it again produces two surds. Well, taking squares leads to a contradiction, however the case of four terms would require taking a square twice. Still I can't see how it can be generalised for an induction proof. Unless you have a different idea... – cyanide Feb 03 '19 at 20:59

$\sum_{i=1}^n {(a_i\sqrt{b_i})} \ne 0$

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