Inequality with square root and squaring each side of the inequality

My book says if I take $\sqrt{x^2 + y^2} \lt 1,\;$ and it says if I "square each side of the inequality" the result will give the inequality $\;x^2 + y^2\lt 1,\;$ but I don't understand the concept.

If you find the square root of $5^2$ isn't that $5$?

Then isn't the square root of $\,x^2 + y^2\,$ equal to $\,x + y\;?$

• You're going to need to write out the whole thing. You haven't written any inequality. Jun 21 '13 at 20:55
• @Jessica To get $\sqrt{x^2+y^2}$ type \$\sqrt{x^2+y^2}\$. Jun 21 '13 at 20:56
• @dfeuer $\sqrt{x^2 + y^2} < 1$ Jun 21 '13 at 21:04
• $(x + y)^2$ is not equal to $x^2 + y^2$ – try e.g. $x = 1$ and $y = 1$. Jun 21 '13 at 21:04

Please understand: While it is true that $\;\sqrt{(x + y)^2} = x+y,\,$ note that \begin{align} \sqrt{x^2 + y^2} & \color{blue}{\;\large \bf \neq\;}x + y \\ \\ \text{so}\quad \left(\sqrt{x^2 + y^2}\right)^2 & \neq (x + y)^2 \\ \\ \text{because}\quad x^2 + y^2 & \neq x^2 + 2xy + y^2 \end{align}
Regarding your inequality (for why I'm addressing this, please note the OP's comment below, and the fact that the original post before editing, included a statement about being perplexed as to why the following is true): \begin{align} \sqrt{x^2 + y^2} \lt 1 & \implies \left(\sqrt{x^2 + y^2}\right)^2 \lt (1)^2\\ \\ & \implies x^2 + y^2 \lt 1\end{align}