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  1. $2+17 = a^2/b^2$
  2. $19b^2 = a^2$ ($a$ is divisible by 19)
  3. $19b^2 = (19k)^2$
  4. $19b^2 = 361k^2$
  5. $b^2 = 19k$ ($b$ is divisible by 19)

Since both numbers are divisible by 19,it means they have a common factor.

Is this accurate? If not,please elaborate. Thank you in advance.

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    $\begingroup$ How do you arrive at the first line $2+17=a^2/b^2$? Please do not say, by squaring both sides of $\sqrt 2+\sqrt{17}=a/b$ $\endgroup$ – Hagen von Eitzen Nov 9 '18 at 21:10
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    $\begingroup$ Yes, your proof is incorrect. $\sqrt{2+17}\neq \sqrt{17}+\sqrt{2}$ $\endgroup$ – Uri Goren Nov 9 '18 at 21:12
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    $\begingroup$ Looks like you have fallen for the "Freshman's dream." $\endgroup$ – Doug M Nov 9 '18 at 21:13
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    $\begingroup$ Alternatively, start with $\frac{a}{b} = \sqrt{2}+ \sqrt{17}$, square both sides (correctly!) and simplify to get a contradiction where a square root of a nonsquare must be rational. $\endgroup$ – rogerl Nov 9 '18 at 21:17
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    $\begingroup$ What a number of the posts are referring to is the "Freshman's dream", the mistaken belief that $(a+b)^n=a^n+b^n$. You do this when you say that $\sqrt{2}+\sqrt{17}=a/b$, and then square both sides, but this is incorrect. Consider, for example, $(2+3)^2=5^2=25$. The freshman's dream would tell us that $(2+3)^2=2^2+3^2=13$, which is wrong. You need to just evaluate $(\sqrt{2}+\sqrt{3})^2$ by FOILing. $\endgroup$ – Kevin Long Nov 9 '18 at 21:27
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I'm sorry, but your argument is incorrect. From $$ \sqrt{2}+\sqrt{17}=\frac{a}{b} $$ and squaring, you get $$ 2 + 2\sqrt{34} + 17=\frac{a^2}{b^2} $$ and not $19=a^2/b^2$ as you claimed.


You can instead observe that $$ \frac{(\sqrt{17}+\sqrt{2})(\sqrt{17}-\sqrt{2})}{\sqrt{17}-\sqrt{2}}=\frac{a}{b} \tag{*} $$ which is tantamount as saying that $$ \sqrt{17}-\sqrt{2}=\frac{15b}{a} \tag{**} $$ Subtracting (**) from (*) you get $$ 2\sqrt{2}=\frac{a}{b}-\frac{15b}{a} $$ which would imply that $\sqrt{2}$ is rational.

Note that this method will work in all instances of $\sqrt{2}+\sqrt{x}$ and $\sqrt{2}-\sqrt{x}$ (so long as $x\ne2$).

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As already noticed in the comments, assume that $\exists q\in \mathbb{Q}$ such that

$$\sqrt{2} + \sqrt{17}=q \implies (\sqrt{2} + \sqrt{17})^2=q^2$$

$$2+17+2\sqrt{34}=q^2 \implies \sqrt{34}=\frac{q^2-19}2\in \mathbb{Q}$$

which is a contradiction, see for that the related

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    $\begingroup$ You have dropped a $2$. Must be $2\sqrt{34}$ $\endgroup$ – Mason Nov 9 '18 at 21:25
  • $\begingroup$ @Mason Opssss...thanks I've fixed that :) $\endgroup$ – user Nov 9 '18 at 21:26
  • $\begingroup$ And why is it a contradiction that $\sqrt{34}$ is rational? The proof of this is an analog of this $\endgroup$ – Mason Nov 9 '18 at 21:31
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    $\begingroup$ @Mason Yes that's a classical proof, I add that! Thanks $\endgroup$ – user Nov 9 '18 at 21:33

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