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Prove: If $a,b \in \mathbb{R}: a^2+b^2= 0\Longleftrightarrow a=0$ and $b=0$

My work:

If $a = 0$, then $a^2 >0$, since $b^2 ≥0$, it follows that $a^2 +b^2 >0$

How can I continue my proof?

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  • $\begingroup$ This problem seems to be missing something. Did you mean to ask about $a^2+b^2=0$? Also, it's generally considered best to put the full question in the body of the post, even if you might need to repeat something in the title. Think of the title like the title of a book. $\endgroup$ – Stella Biderman Oct 8 '17 at 13:32
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    $\begingroup$ if $a=0$, then how can $a^2$ be bigger than 0? $\endgroup$ – MCCCS Oct 8 '17 at 13:35
  • $\begingroup$ @StellaBiderman Now the title is correct $\endgroup$ – B. David Oct 8 '17 at 13:41
  • $\begingroup$ @Daniel I've added a few tags and rewritten your question to abide by the community guidelines for titles that I mentioned. $\endgroup$ – Stella Biderman Oct 8 '17 at 13:52
  • $\begingroup$ Thank you @StellaBiderman $\endgroup$ – B. David Oct 8 '17 at 13:54
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Clearly if $ a = 0 $ and $ b = 0 $ then $ a^2 + b^2 = 0 $.

Conversely, suppose that $ a^2 + b^2 = 0 $ . If $ a \neq 0 $ then $ a^2 > 0 $. But we always have $ b^2 \geq 0 $ so we deduce that $ a^2 + b^2 > 0 $ a contradiction. Likewise we cannot have $ b \neq 0 $. So we must have $ a = 0 $ and $ b = 0 $.

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$$\begin{array}{c|cc}a^2+b^2&\color{green}{a=0}&a>0\\\hline\color{green}{b=0}&\color{green}{=0}&>0\\b>0&>0&>0\end{array}$$

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Suppose $a^2+b^2=0$ and one or both of $a,b$ not zero, then one or both of $a^2,b^2$ would be not zero, but larger than zero and $a^2+b^2$ would not be zero.

If $a=b=0$ then $a^2+b^2=0+0=0$

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