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Consider an expression of the form $a + b \sqrt{2}$, where $a,b \in \mathbb{Q}$. How can I extract $b$ (or equivalently $a$) from this expression?

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One can define the conjugate and use it to construct the rational and radical coefficients (rat and rad resp.). Just as PowerExpand assumes bases are positive reals, conj[x] will be correct only if the symbolic variables and functions in an expression x represent rational numbers.

conj[x_] := x /. Sqrt[2] -> -Sqrt[2];
rat[x_] := (x + conj[x])/2;
rad[x_] := (x - conj[x])/(2 Sqrt[2]);

Through[{rat, rad}[a + b Sqrt[2]]]
(*
  {a, b}
*)

Through[{rat, rad}[(a + b Sqrt[2])^2]]
% // Simplify

(*
  { 1/2 ((a - Sqrt[2] b)^2 + (a + Sqrt[2] b)^2),
    (-(a - Sqrt[2] b)^2 + (a + Sqrt[2] b)^2)/(2 Sqrt[2]) }
  {a^2 + 2 b^2, 2 a b}
*)

Simplify@Through[{rat, rad}[(3 - 2 Sqrt[2])/10]]
(*
  {3/10, -(1/5)}
*)

More generally, one can extend the definitions to numbers over an arbitrary quadratic extension of the rationals.

Clear[conj, rat, rad];
conj[x_, sqroot_: Sqrt[2]] := x /. sqroot -> -sqroot;
rat[x_, sqroot_: Sqrt[2]] := (x + conj[x, sqroot])/2;
rad[x_, sqroot_: Sqrt[2]] := (x - conj[x, sqroot])/(2 sqroot);
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  • $\begingroup$ Yes, of course. Very clever solution. $\endgroup$ – Tyson Williams Dec 4 '14 at 17:09
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You can use ToNumberField:

2/3 + 1/4 Sqrt[2]
ToNumberField[%, Sqrt[2]]

which produces

AlgebraicNumber[Sqrt[2], {2/3, 1/4}]
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  • $\begingroup$ Very nice. Thanks! $\endgroup$ – Tyson Williams Dec 4 '14 at 15:39
  • $\begingroup$ What if $a$ and $b$ are symbolic? $\endgroup$ – Tyson Williams Dec 4 '14 at 15:41
  • $\begingroup$ @TysonWilliams: I'm not sure how to handle that case, as I rarely use the abstract algebra functions. However, Daniel Lichtblau might know how to do it, so you could try asking him (or you could ask it as another question). $\endgroup$ – DumpsterDoofus Dec 4 '14 at 15:48
  • $\begingroup$ As a mathematician / theoretical computer scientist, I think of $a$ and $b$ as "symbolically rational" (i.e. some rational numbers by assumption). Of course I also know, but often forget, that (1) symbolic, (2) symbolically rational, and (3) and rational are three different "data types" that can all behave differently in Mathematica. $\endgroup$ – Tyson Williams Dec 4 '14 at 16:18
  • $\begingroup$ BTW: One can use ToNumberField[num][[2, 1]] to extract the rational part ("a") of any quadratic field. No need to even specify the generator! $\endgroup$ – kirma Feb 1 '16 at 21:42

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