## College Algebra (11th Edition)

$( x+y)^{2}\mathrm{\ is\ not\ equivalent\ to} \ x^{2} +y^{2}$.
$\displaystyle \begin{array}{|l|l|} \hline ( x+y)^{2} =x^{2} +y^{2} & \mathrm{Assume\ }( x+y)^{2} =x^{2} +y^{2}\\ & \\ \hline \begin{array}{ c l } ( x+y)^{2} & =( x+y)( x+y)\\ & =x( x+y) +y( x+y)\\ & =x^{2} +xy+xy+y^{2}\\ & =x^{2} +2xy+y^{2}\\ & =x^{2} +y^{2} +2xy \end{array} & \mathrm{Then\ expand\ }( x+y)^{2}\\ & \\ \hline x^{2} +y^{2} +2xy=x^{2} +y^{2} & \begin{array}{{>{\displaystyle}l}} \mathrm{Now\ we\ set\ the\ result\ of\ the\ }\\ \mathrm{expansion\ equal\ to} \ x^{2} +y^{2} .\\ \mathrm{At\ this\ point,\ it\ is\ clear\ something\ }\\ \mathrm{is\ wrong.\ Nevertheless,\ we\ continue} . \end{array}\\ & \\ \hline \begin{array}{ r l } y^{2} +2xy & =y^{2}\\ 2xy & =0 \end{array} & \begin{array}{{>{\displaystyle}l}} \mathrm{Subtract} \ x^{2} \ \mathrm{from\ both\ sides}\\ \mathrm{Subtract} \ y^{2} \ \mathrm{from\ both\ sides} \end{array}\\ & \\ \hline \begin{array}{ c c } xy & =0\\ & \end{array} & \begin{array}{{>{\displaystyle}l}} \mathrm{Divide\ both\ sides\ by} \ 2.\ \mathrm{At\ this\ point,}\\ \mathrm{we\ can\ see\ that\ the\ equation\ can\ only\ }\\ \mathrm{be\ true\ if} \ x\ =\ 0\ \mathrm{or\ if} \ y\ =\ 0.\ \mathrm{In\ other\ }\\ \mathrm{words} ,\ ( x+y)^{2} \ =\ x^{2} +y^{2} \ \mathrm{is\ not\ true\ for}\\ \mathrm{all} \ ( x,y) \ \in \mathbb{R} \end{array}\\ & \\ \hline ( x+y)^{2} \neq x^{2} +y^{2} & \begin{array}{{>{\displaystyle}l}} \mathrm{Thus\ we\ can\ conclude\ that} \ ( x+y)^{2}\mathrm{\ is\ }\\ \mathrm{not\ equivalent\ to} \ x^{2} +y^{2} \end{array}\\ \hline \end{array}$