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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 $\,x + y\;?$

Am I thinking about this correctly or am I mistaken somehow?

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    $\begingroup$ You're going to need to write out the whole thing. You haven't written any inequality. $\endgroup$ – dfeuer Jun 21 '13 at 20:55
  • $\begingroup$ @Jessica To get $\sqrt{x^2+y^2}$ type \$\sqrt{x^2+y^2}\$. $\endgroup$ – Git Gud Jun 21 '13 at 20:56
  • $\begingroup$ @dfeuer $\sqrt{x^2 + y^2} < 1$ $\endgroup$ – Jessica M. Jun 21 '13 at 21:04
  • $\begingroup$ $(x + y)^2$ is not equal to $x^2 + y^2$ – try e.g. $x = 1$ and $y = 1$. $\endgroup$ – Ben Millwood Jun 21 '13 at 21:04
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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): $$\begin{align} \sqrt{x^2 + y^2} \lt 1 & \implies \left(\sqrt{x^2 + y^2}\right)^2 \lt (1)^2\\ \\ & \implies x^2 + y^2 \lt 1\end{align}$$

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  • $\begingroup$ I'm thinking about it but nothing is happening in my head. $\endgroup$ – Jessica M. Jun 21 '13 at 21:00
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    $\begingroup$ amWhy, I'm trying really hard to understand where you came up with that guess for what inequality JessicaM might have intended, and am coming up blank. $\endgroup$ – dfeuer Jun 21 '13 at 21:01
  • $\begingroup$ @dfeuer I delayed my upvote for the same reason, I just can't see it. $\endgroup$ – Git Gud Jun 21 '13 at 21:05
  • $\begingroup$ My guess for the moment is that amWhy has read every popular middle school, high school, and undergraduate math text and remembers all the exercises. $\endgroup$ – dfeuer Jun 21 '13 at 21:07
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    $\begingroup$ Thanks everyone. This is just a simple concept I don't know how i got confused. Thanks for all the help. $\endgroup$ – Jessica M. Jun 21 '13 at 21:37

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