# Prove that $\sqrt{2} + \sqrt{17}$ is irrational. Is my proof correct?

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.

• 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 9 '18 at 21:10
• Yes, your proof is incorrect. $\sqrt{2+17}\neq \sqrt{17}+\sqrt{2}$ – Uri Goren Nov 9 '18 at 21:12
• Looks like you have fallen for the "Freshman's dream." – Doug M Nov 9 '18 at 21:13
• 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 9 '18 at 21:17
• 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 9 '18 at 21:27

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 $$x\ne2$$).

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

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