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Good-morning Math Exchange (and good evening to some!)

I have a very basic question that is confusing me.

At school I was told that $\sqrt {a^2} = \pm a$

However, does this mean that $\sqrt {a^2} = +2$ *and*$-2$ or does it mean: $\sqrt {a^2} = +2$ *OR*$-2$

Is it wrong to say 'and'? What are the implications of choosing 'and'/'or'

Any help would be greatly appreciated. Thanks in advance and enjoy the rest of your day :)

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For most purposes in mathematics, if $a$ is real then $\sqrt{a^2}=|a|$. It is unfortunate that (some) school systems use a different convention. –  André Nicolas Feb 15 at 7:05
    
Tag linear-algebra changed to the more general abstract-algebra tag, makes more sense, hope you don't mind. –  Joachim Feb 15 at 9:36

4 Answers 4

But the truth being that the squarte root function is always associated with absolute value function. That is

$\sqrt {a^2} = \vert a \vert $

This is very very important to remember and be careful while using the $ \sqrt{} $ .

Note that if it is $ {(a^{2}})^{\frac{1}{2}} $ then we get answer as $\pm a $ but note that $\sqrt {a^2} = \vert a \vert $

Absoulte value is associated as OR

$\vert a \vert$ = $a$ OR $-a$

And is not used as a variable cant have two values at the same time !!

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Why is it not 'AND'? –  Daniel Feinstein Feb 15 at 7:16
    
I think its not needed as square root function is associated with absolute value. –  sammath Feb 15 at 7:18

$\sqrt{\mathrm{a}^{2}}=|a|$= +a OR -a.Remember here OR is used.AND is not used as a function cannot have two values at same time.

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The usual convention is that the square root sign gives the positive square root. When the negative/both is/are required, use the minus/plusminus sign.

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Even though quite a bit has been said already, i wanted to add something.

The numbers which you normally use in school (-1, $\frac{2}{3}$, $ \pi$, etcetera) are called the real numbers. The set of real numbers is denoted by $\mathbb{R}$.

Now the square root of any number $b$ is normally considered to be any number $x$ that satisfies $x^2 = b$, or equivalently $x^2 - b = 0$.

As you pointed out, there are normally two solutions to this, so two values for $x$ will do the trick. However, working in $\mathbb{R}$ this situation is remedied by adopting the convention that the square root of $b$ will be the positive number $x$ that satisfies $x^2 - b=0$. So indeed, when $b= a^2$, we get $$ \sqrt{a^2} = |a|. $$ So with this convention, the solutions to $x^2 - b=0$ become $x=\sqrt{b}$ and $x = -\sqrt{b}$. It is very important to note that this is merely a convention.

Even more: there are more sets of numbers we could work in, where this trick will not work! If we pass from the real numbers $\mathbb{R}$ to the so called complex numbers, denoted $\mathbb{C}$ (check wikipedia), we lose this! In this set of numbers, the notion of a positive number does not make sense, and it is in fact impossible to define a square root function in a nice way on the whole of $\mathbb{C}$ (if you want to know more about this, ask google).

In general there are many more things that i call "sets of numbers" now, in mathematics they are called "fields". In all of them, the square root notion makes sense, as in solving the solution to $x^2 - b=0$. However, the nice $\sqrt{{}}$ function as we have it in $\mathbb{R}$ is rarely found in other fields.

Hope this context was interesting to you.

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