Norm Question....

**AAM** · Mar 18th 2009, 01:32 PM

Hi guys!

I've just looked at a past year's number theory questions and one is as follows:

Let R = Z[Sqrt(-6)] & K = Q[Sqrt(-6)]. Put a = 3+2Sqrt(-6) & b = 7-6Sqrt(-6). By choosing c close to b/a, or otherwise, find c in R such that:

Normk(b-ac) < Normk(a)

I have no ideas! :-s

Any help would be most appreciated! x

**ThePerfectHacker** · Mar 18th 2009, 07:35 PM

Originally Posted by AAM

Hi guys!

I've just looked at a past year's number theory questions and one is as follows:

Let R = Z[Sqrt(-6)] & K = Q[Sqrt(-6)]. Put a = 3+2Sqrt(-6) & b = 7-6Sqrt(-6). By choosing c close to b/a, or otherwise, find c in R such that:

Normk(b-ac) < Normk(a)

I have no ideas! :-s

Any help would be most appreciated! x

I assume $N(x+y\sqrt{-6}) = x^2 + 6y^2$ . Form $\tfrac{b}{a} = p+q\sqrt{-6}$ where $p,q\in \mathbb{Q}$ by multiplying the numerator and denominator of $\tfrac{b}{a}$ by the conjugate. Next choose $p_0,q_0$ , intergers, so that $p_0,q_0$ are the closest integers to $p,q$ respectively. Set $c = p_0 + q_0 \sqrt{-6}$ and see that $N(b-ac) < N(a)$ .

**AAM** · Mar 19th 2009, 02:30 AM

Oh, fantastic! :-D Nice and simple! Just wasn't quite sure what the question was asking for. :-s But I completely get it now!

Thank you. :-)

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