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

Question

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.

Answer 1

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

• Proving $\sqrt 3$ is irrational.
• Prove that $\sqrt 5$ is irrational

Answer 2

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.

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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 $x\ne2$).