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According to this answer, when $\sqrt{m}$+$\sqrt{n}$ is rational, then $\sqrt{m}$ and $\sqrt{n}$ are rational.

But $\sqrt{100+\sqrt{6156}}+\sqrt{100-\sqrt{6156}} = 18 \in \mathbb{Q}$.

So that would imply $\sqrt{100+\sqrt{6156}}$ and $\sqrt{100-\sqrt{6156}}$ both rational. However, the former is $9 + \sqrt{19}$ and the latter is $9 - \sqrt{19}$, both irrational.

Where is the mistake here?

Thanks.

Bill Dubuque
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2 Answers2

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Your interpretation of "when $\sqrt{m}+\sqrt{n}$ is rational, then $\sqrt{m}$ and $\sqrt{n}$ are rational" is too broad. For example:

$$\sqrt{2}+\sqrt{(3-\sqrt{2})^2}$$

is rational, because it equals $3$. But of course $\sqrt{2}$ is not rational.

The statement "when $\sqrt{m}+\sqrt{n}$ is rational, then $\sqrt{m}$ and $\sqrt{n}$ are rational" might be true when $m,n$ are integers. But the example here has $m=100+\sqrt{6156}$, not an integer.

2'5 9'2
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    Not necessary for m,n to be integers or even rational. As long as $\sqrt{m}+\sqrt{n}$ and $m-n$ are rational, $\sqrt{m}$ and $\sqrt{n}$ will be rational too. – Aryan Dec 25 '20 at 03:39
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    @AryanHemmati That's right, I intentionally phrased it without saying that $m,n$ need to be integers. Also I tried to not over-complicate things by giving the full account. The point is, OP did not realize there were restrictions on $m,n$. – 2'5 9'2 Dec 25 '20 at 05:06
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The problem is that the proof of what you linked is actually equivalent to $m-n$ being rational which isn't correct here

Aryan
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