# The sum of two squares is zero

1,768

## Solution 1

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 $b = 0$.

## Solution 2

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

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### B. David

Math Student. Universidad Complutense de Madrid

Updated on October 09, 2020

### Comments

• B. David about 3 years

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?

• Stella Biderman about 6 years
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.
• MCCCS about 6 years
if $a=0$, then how can $a^2$ be bigger than 0?
• B. David about 6 years
@StellaBiderman Now the title is correct
• Stella Biderman about 6 years
@Daniel I've added a few tags and rewritten your question to abide by the community guidelines for titles that I mentioned.
• B. David about 6 years
Thank you @StellaBiderman