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Old 2019-07-01, 21:05   #4
wildrabbitt
 
Jul 2014

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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 primary-school 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,


\(\frac{6+23\sqrt{2}}{2+\sqrt{2}}=\frac{(6+23\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}
=\frac{12-46+(46-6)\sqrt{2}}{2}=\frac{-34+40\sqrt{2}}{2}=-17+20\sqrt{2}\)


/* 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.

Last fiddled with by wildrabbitt on 2019-07-01 at 21:39
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