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2019-07-01, 21:05   #4
wildrabbitt

Jul 2014

3×149 Posts

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