Quote:
It's really elementary, so I may be misunderstanding your question:
Multiply top and bottom by the conjugate of the bottom. That makes the denominator a real number, so regular primaryschool division can proceed.

Thanks. You weren't misunderstanding my question. I believe that there's something hidden with maths like it's a mystery or I'm sereptticiously tricking myself into making simple things seem hard so I can feel better about how good I am at maths. Perhaps also that I think that the simple form should be smaller than it's numerator and multiplying the numerator by a number greater than 1 wouldn't have occurred to me.
So,
\(\frac{6+23\sqrt{2}}{2+\sqrt{2}}=\frac{(6+23\sqrt{2})(2\sqrt{2})}{(2+\sqrt{2})(2\sqrt{2})}
=\frac{1246+(466)\sqrt{2}}{2}=\frac{34+40\sqrt{2}}{2}=17+20\sqrt{2}\)
/* editted out the mistakes */
Quote:
Note that, in general, you don't get an algebraic integer (just like dividing an integer by another integer doesn't give an integer, in general, but a rational number).

I take your point but isn't it okay to say that in the ring \(Z[\sqrt{2}]\),\(6+23\sqrt{2}\) has a factor \(2+\sqrt{2}\)?
Thanks very much to both of you.