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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.

gd1035
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user569685
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    How do you arrive at the first line $2+17=a^2/b^2$? Please do not say, by squaring both sides of $\sqrt 2+\sqrt{17}=a/b$ – Hagen von Eitzen Nov 09 '18 at 21:10
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    Yes, your proof is incorrect. $\sqrt{2+17}\neq \sqrt{17}+\sqrt{2}$ – Uri Goren Nov 09 '18 at 21:12
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    Looks like you have fallen for the "Freshman's dream." – Doug M Nov 09 '18 at 21:13
  • Any advice? :'l – user569685 Nov 09 '18 at 21:14
  • Think about the rational root theorem as applied to $(x-\sqrt{2}-\sqrt{17})(x-\sqrt{2}+\sqrt{17})(x+\sqrt{2}-\sqrt{17})(x+\sqrt{2}+\sqrt{17})=0$ – K B Dave Nov 09 '18 at 21:16
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    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. – rogerl Nov 09 '18 at 21:17
  • What am i messing up with the squaring of both of the sides? – user569685 Nov 09 '18 at 21:25
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    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. – Kevin Long Nov 09 '18 at 21:27
  • @user569685. It's not the case that $(a+b)^2= a^2+b^2$. Though it kind of looks like it might be that. It's a common error. You can see it quickly though... Let $a=b=1$. It is however the case $(a+b)^2=a^2+2ab+b^2$ and that should help. – Mason Nov 09 '18 at 21:28

2 Answers2

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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

user
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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 $x\ne2$).

egreg
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