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Am going through this notes...kindly let me know if there is a mistake on highlighted part. I think it ought to be;

##α^2=5+2\sqrt{6}##

##α^2=5+2\sqrt{6}##

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In summary, the conversation discusses a mistake in notes regarding the value of ##\alpha^2##. The correct value is ##\alpha^2=5+2\sqrt{6}##, as listed at the end. The procedure is also deemed correct. They go on to mention a text providing a proof of the irrationality of ##\sqrt{2}## using Eisenstein's theorem, but the conversation then points out that it is not necessary to use Eisenstein, as its conditions can be applied instead.

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##α^2=5+2\sqrt{6}##

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You are right. It is a mistake in the book and should be ##\alpha^2=5+2\sqrt{6}.## It is the correct value on the list at the end again. The procedure itself is correct.chwala said:Summary:See attached

Am going through this notes...kindly let me know if there is a mistake on highlighted part. I think it ought to be;

##α^2=5+2\sqrt{6}##

View attachment 315473

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Thanks...let me peruse through...fresh_42 said:You are right. It is a mistake in the book and should be ##\alpha^2=5+2\sqrt{6}.## It is the correct value on the list at the end again. The procedure it self is correct.

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Yes, but Eisenstein uses the fact that ##\mathbb{Z}## is a UFD and ##2## is prime. With that, you don't need Eisenstein anymore:WWGD said:

\begin{align*}

\sqrt{2}=\dfrac{m}{n} \Longrightarrow 2n^2=m^2 \Longrightarrow 2\,|\,m \Longrightarrow 4\,|\,m^2\Longrightarrow 2\,|\,n^2

\end{align*}

contradicting the assumption that ##\dfrac{m}{n}## was cancelled.

Hence, you do not use Eisenstein, you use its conditions.

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