If $m_1 , m_2, \cdots m_n$ are natural numbers where at least one of them is not a perfect square, then how do I prove that the sum $$\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$$ is irrational? I'm hoping to prove this using methods from pre-calculus level algebra.

  • $\begingroup$ Are they all distinct? Otherwise they can be rational. $\endgroup$ – AvZ Jan 30 '15 at 8:18
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    $\begingroup$ @AvZ Sure about that? $\endgroup$ – alex.jordan Jan 30 '15 at 8:18
  • $\begingroup$ @user7530 i want an elementry method for solving the problem. $\endgroup$ – user120269 Jan 30 '15 at 8:23
  • $\begingroup$ The accepted answer uses only elementary number theory, and my answer uses only basic algebra for special cases of few terms. I'm not sure what more you're hoping for: I very much doubt a high-school algebra proof of the general case exists. $\endgroup$ – user7530 Jan 30 '15 at 8:52
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    $\begingroup$ See this answer for a more general result, including literature references. $\endgroup$ – Bill Dubuque Jan 30 '15 at 16:34

We cannot do this for arbitary $n$, maybe, but going through the procedure, we can see how far we can push the possibilities using only induction, the base case being $n=1$ is classical.

Assume there is an integer, $N$ such that

$$\sqrt{m_1}+\ldots +\sqrt{m_n}=N\qquad (*)$$

then clearly we may assume $m_n>m_i$ for all other $i<n$. We see

$$\sqrt{m_1}+\ldots +\sqrt{m_{n-1}}=N-\sqrt{m_n}\qquad (**)$$

squaring both sides gives

$$m_1+\ldots +m_{n-1}+2\left(\sum_{1\le i<j\le n-1}\sqrt{m_im_j}\right)=N^2+m_n-2N\sqrt{m_n}$$

rearranging gives

$$N^2+m_n-m_1-\ldots -m_{n-1}=2\left(N\sqrt{m_n}+\sum_{1\le i<j\le n-1}\sqrt{m_im_j}\right)=2S$$

We see that

$$S-N\sqrt{m_n}=\sum_{1\le i<j\le n-1}\sqrt{m_im_j}$$

and squaring again and replacing $m_im_j$ by the symbol $m_{ij}$

$$S^2-2SN\sqrt{m_n}+N^2m_n=\sum_{1\le i<j\le n-1}m_{ij}+2\sum_{1\le i<j\le n-1}\left(\sum_{k\ne i,j}m_k\right)\sqrt{m_{ij}}$$

From both sides subtract


to obtain


$$=\sum_{1\le i<j\le n}m_{ij}+2\sum_{1\le i<j\le n-1}^{n-1}\left\lbrace\left(\sum_{k\ne i,j}m_k\right)-S\right\rbrace\sqrt{m_{ij}}=S'$$

Reducing this gives


Now, we see clearly from $(**)$ and $(*)$ and that


that $S-\sum_{k\ne i,j}m_k$ for $i,j$ fixed and $i,j<n$ gives a positive value.

Then we conclude by rationality of $S'$ that the $m_{ij}$ are all squares--we are doing induction, and there are fewer than $n$ summands of roots of rationals there, indeed there are ${n-1\choose 2}={(n-1)(n-2)\over 2}$ choices, and since

$$2n-(n-1)(n-2)=5n-n^2-2=-\left(n-{5\over 2}\right)^2+{13\over 4}$$

which is positive exactly when

$$\left|n-{5\over 2}\right|\le {\sqrt{13}\over 2}\approx 1.8$$

We see this is the case for $n=2,3,4$, so that the number of summands in $S'$ which are roots, is less than $n$ for $n=2,3,4$ and by induction it follows that they must all be rational.

The proof breaks down for higher $n$, because then the number of summands is just too large, so we have to resort to either an even harder induction (and man, wasn't this already a lot of work!?) or use non-elementary means (this is the best idea).

  • $\begingroup$ The 6'th equation is false. @Adam Hughes $\endgroup$ – user120269 Jan 30 '15 at 13:16
  • $\begingroup$ See this answer for a more general result, including literature references. $\endgroup$ – Bill Dubuque Jan 30 '15 at 16:36
  • $\begingroup$ @billdubuque thanks, I already know a couple slick ways to prove the general result, I was going for the algebra/pre-calc as much as possible because of the op's request, and since I was curious how far such methods could go. :-) $\endgroup$ – Adam Hughes Jan 30 '15 at 17:55
  • $\begingroup$ @mathstudent ah you're right, but I got lucky: it doesn't matter! The formula is true for $n=2,3,4$ which are the key cases, so the rest of the proof works up through $n=4$ as before, and we run into the same problem as before that the number of summands blows up at $n=5$. :-) $\endgroup$ – Adam Hughes Jan 30 '15 at 18:06

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