Tuesday, November 11, 2008

Section 1.1

(If you would prefer, a PDF of this work is available here.)

"Then (1) implies m^{2}=2n^{2}."
\inline $p=\frac{m}{n} \Rightarrow p^{2}=\frac{m^{2}}{n^{2}}=2 \Rightarrow m^{2}=2n^{2}.

"...and so \inline m^{2} is divisible by \inline 4."
\inline m even implies that \inline m=2q for some \inline q\in \mathbb{N}. Then \inline m^{2}=4q^{2} \Rightarrow \frac{m^{2}}{4}=q^{2} for some \inline q\in \mathbb{N}.

"\inline $p-\frac{p^{2}-2}{p+2}=\frac{2p+2}{(p+2)^{2}}$."
\inline p=\frac{p(p+2)}{p+2} implies that
\begin{align*} p-\frac{p^{2}-2}{p+2}&=\frac{p(p+2)}{p+2}-\frac{p^{2}-2}{p+2} \notag\\ &=\frac{p(p+2)-(p^{2}-2)}{p+2} \notag\\ &=\frac{4p^{2}+8p+4-2p^{2}-8p-8}{(p+2)^{2}} \notag\\ &=\frac{2p^{2}-4}{(p+2)^{2}} \notag\\ &=\frac{2(p^{2}-2)}{(p+2)^{2}}. \notag\\ \end{align*}

"Then \inline q^{2}-2=\frac{2(p^{2}-2)}{(p+2)^{2}}."
\inline \begin{align*} q^{2}&=\frac{(2p+2)^{2}}{(p+2)^{2}}\\ &=\frac{4p^{2}+8p+4}{(p+2)^{2}},\\ \end{align*}
so therefore
\begin{align*} q^{2}-2&=\frac{4p^{2}+8p+4}{(p+2)^{2}}-\frac{2(p+2)^{2}}{(p+2)^{2}} \notag\\ &=\frac{4p^{2}+8p+4}{(p+2)^{2}}-\frac{2p^{2}+8p+8}{(p+2)^{2}} \notag\\ &=\frac{4p^{2}+8p+4-2p^{2}-8p-8}{(p+2)^{2}} \notag\\ &=\frac{2p^{2}-4}{(p+2)^{2}} \notag\\ &=\frac{2(p^{2}-2}{(p+2)^{2}}. \notag\\ \end{align*}
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