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I'm learning calculus and I saw the following statement:

If $|a|+1>0$ then $\sqrt{|a|+1}>0$. By same logic we get $\sqrt{|b|+1}>0$. Then we get $\sqrt{|a|+1}+\sqrt{|b|+1}>0$.

I don't understand why $\sqrt{|a|+1}>0$. It could be a negative number no? For example if $a=3$ then we get $\sqrt{|3|+1}=\pm2$.

kickstart
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    By convention, when a real number has square roots, $\sqrt{;\phantom{x}}$ denotes the non-negative root. – Bernard Mar 19 '19 at 00:24
  • Why? It could have a negative root. – kickstart Mar 19 '19 at 00:25
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    It's just convention - it's just commonly understood to mean the positive root unless there's a $-$ or $\pm$ in front. There's no deep underlying reason to it, as far as I know. – PrincessEev Mar 19 '19 at 00:26
  • It dies have a negative root, but as both roots opposite, the negative root of $a$ is denoted $-\sqrt a$. – Bernard Mar 19 '19 at 00:29
  • Another duplicate is here – Ross Millikan Mar 19 '19 at 00:30
  • This idea is called the principal square root of a nonnegative number: http://mathworld.wolfram.com/PrincipalSquareRoot.html I can't vouch for this, but I think mathematicians agree that there should be only one square root of a number because we tend to definite operations as functions and a function can't return two output values for a single input value. – Michael Rybkin Mar 19 '19 at 00:32

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