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I was reading chapter 29 page 296 of the book A Book Of Abstract Algebra (second edition) written by Charles C. Pinter, and the equation below was given: $$\sqrt{3} = a + b\sqrt{2}$$ where $a, b \in \mathbb {Q}$.

They were solving for $\sqrt{2} $ in terms of $a$ and $b$. The instruction given was to square both sides and solve for $\sqrt{2}$ What I did was $$\big( \sqrt{3} \ \big)^2 = \big(a + b\sqrt{2}\ \big )^2 \implies 3 = a^2 + 2b^2 + 2ab\sqrt{2}.$$ Looking at this, I don't find anyway to get to their final answer of $\sqrt{2} = a$ when solving for $\sqrt{2}$.

Xander Henderson
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EA Lehn
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    What's the question? If all you want to do is to find some solution to the initial equation, take $a=\sqrt 3$, $b=0$. You must have conditions on $a,b$, no? If, say, you required that $a,b\in \mathbb Q$ then there are no solutions. So...what did you mean? – lulu Jan 07 '20 at 23:40
  • Can you please check the page/chapter? In my book, page 296 relates the chapter 31. Where is the equation precisely? – user376343 Jan 07 '20 at 23:58
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    @user376343, I found the excerpt in my answer in the second edition. I googled to it online at https://books.google.com/books?id=ZAo_AwAAQBAJ&q=296#v=snippet&q=296&f=false – Barry Cipra Jan 08 '20 at 00:02

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Quoting from page 296 of Pinter:

Observe that $\sqrt3$ cannot be in $\mathbb{Q}(\sqrt2)$; for if it were, we would have $\sqrt3=a+b\sqrt2$ for rational $a$ and $b$; squaring both sides and solving for $\sqrt2$ would give us $\sqrt2=$ a rational number, which is impossible.

Note, the "a" in "$\sqrt2=$ a rational number" is the article a, not the variable $a$, i.e., "the square root of $2$ would be a rational number."

Barry Cipra
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    Good catch! $\quad$ – lulu Jan 07 '20 at 23:58
  • @BarryCipra Thanks for the clarification – EA Lehn Jan 08 '20 at 00:02
  • For disambiguation, it might have been better to write something like $\sqrt{2} = (\text{a rational number})$, or "...would imply that $\sqrt{2}$ is equal to a rational number". It may also be worth noting the subtle fact that the article "a" is not italicized, so it is not a variable (though this is, frankly, quite subtle). – Xander Henderson Jan 08 '20 at 15:09