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Can we simplify $\sqrt{a}*\sqrt{a}$ to $a$ when $a \in \mathbb{R}$ and we do not know whether a is positive or negative? (Since $\sqrt{a}$ by itself is undefined in $\mathbb{R}$ when $a$ is negative)

I was wondering about how we can handle the above problem in $\mathbb{R}$ when I had to do some exercise that involved squaring the squareroot in $\mathbb{C}$

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  • $\begingroup$ If you have written $\sqrt{a}$ and you are working solely with real numbers (no complex numbers), then you do know that $a$ is nonnegative, or otherwise you would not be entitled to write $\sqrt{a}$ in the first place. Similarly, if you see an identity about real numbers that involves $\ln(x)$, you know that $x$ must be greater than zero. As a third example, if $z$ is a complex number and someone writes $z \geq 0$ they must already mean that $z$ is real, because otherwise they could not use $\geq$. $\endgroup$ – Carl Mummert Oct 23 '13 at 12:26
  • $\begingroup$ And what in case of the quadratic formula? You don't really know if what you are taking the square root from is positive or negative, you only know once it is negative it becomes 'undefined'. So we are not entitled to write the quadratic formula. Am I wrong then? @CarlMummert $\endgroup$ – xcrypt Oct 23 '13 at 12:42
  • $\begingroup$ The quadratic formula is intended to be interpreted in the realm of complex numbers. Indeed, if $a,b,c$ are any complex numbers then the quadratic formula gives precisely the two complex roots of $ax^2 +bx +c$, because the derivation goes through perfectly well in $\mathbb{C}$. But you are correct that if we limit ourselves to just real numbers then we cannot use the quadratic formula when the part under the radial was negative - but this is not very interesting. Of course we can't use the complex roots if we have declared that we can only use real numbers! We have to decide what we want. $\endgroup$ – Carl Mummert Oct 23 '13 at 12:51
  • $\begingroup$ @CarlMummert I understand what you just said. But take a look at my comment on Michael's answer. I understand that in order for the expression to be sane, sqrt(a) has to be a real number. Meaning that a has to be nonnegative. But this also means that you still have to figure out whether a is negative or not, else we could all discard the quadratic formula when working with reals. With sqrt(a)^2, basically the question is can we make the entire expression sane in R by temporarily using the complex roots, because it's the entire expression that matters not the individual parts right? $\endgroup$ – xcrypt Oct 23 '13 at 13:01
  • $\begingroup$ Yes, you can just treat everything in the expression as a complex number. The result is that you will compute what is called the "principal value" of the roots in question. For real square roots, the principal square root is the positive one. For complex square roots, the issue is more complicated. The key issue in the question in your comment is that you can just treat everything in the expression as if it is a complex number - you never need to "convert back" at the end. In that case, the answer by Carsten Schultz already expresses the key point. $\endgroup$ – Carl Mummert Oct 23 '13 at 13:29
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Surely you could define a square-root in $\mathbb{C}$ that is either nonnegative real, or has positive imaginary part. Then, I think, $(\sqrt{a})^2$ would equal $a$ because the square-roots would be consistent.
Unfortunately, you lose $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if you do that.

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    $\begingroup$ This is similar to what I was thinking. We can temporarily use complex numbers to simplify $(\sqrt{a})^2$ to $a$, and the first expression which is partly undefined in R would be simplified to a complete defined real number. I guess I asked the question somewhat wrong and it should be more like "is it allowed to temporarily use another number set when working in a set, to achieve results that will later on simplify to the first set?" $\endgroup$ – xcrypt Oct 23 '13 at 11:24
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Of course you can. $\sqrt{x}$ requires $x\ge0$, so we know for fact that $a\ge0$ and thus $\sqrt{a}^2=a$ with $a\ge0$ always stands true...

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  • $\begingroup$ If we know that $\sqrt{a}$ exists, then always $a\ge0$.. $\endgroup$ – costashatz Oct 23 '13 at 9:45
  • $\begingroup$ and I wrote the wrong power by mistake...What I meant in my answer is that if $\sqrt{a}$ is defined, then the product of the two square roots is equal to $a$..but with the edited question @xcrypt says that we can't take it for fact..so..we can't know.. $\endgroup$ – costashatz Oct 23 '13 at 9:48
  • $\begingroup$ That's why I took away the downvote...I saw the power on the wrong side and I didn't see the poster changed his question...so we are good!! $\endgroup$ – Eleven-Eleven Oct 23 '13 at 9:49
  • $\begingroup$ I actually didn't change my question, I just gave extra information. $\endgroup$ – xcrypt Oct 23 '13 at 9:50
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    $\begingroup$ Yes, while you didn't change the question, you altered what you said you knew thereby giving those who didn't know more clarity to answer your question. $\endgroup$ – Eleven-Eleven Oct 23 '13 at 9:54
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Yes, by definition $\sqrt a$ is a number whose square is $a$.

Clarification: Whenever there is something that you can sensibly call $\sqrt a$, the equation $\sqrt a\cdot \sqrt a=a$ will hold. Of course there are contexts, in which $\sqrt a$ does not make sense to begin with.

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    $\begingroup$ -1 "...$\sqrt{a}$ is a positive number whose..." $\endgroup$ – JP McCarthy Oct 23 '13 at 11:08
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    $\begingroup$ @JpMcCarthy and the downvoters: Did Algebra suddenly stop working in fields which are not ordered? $\endgroup$ – Carsten S Oct 23 '13 at 11:29
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    $\begingroup$ What are you on about? In this case $\mathbb{R}$ is ordered and anyway you don't need an (total) order to talk about positive square roots. $\endgroup$ – JP McCarthy Oct 23 '13 at 11:57
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    $\begingroup$ @JpMcCarthy what you cite is a convention that is often used when one works exclusively with real numbers. Elsewhere the symbol $\sqrt a$ is used for any $x$ with $x^2=a$. And yes, this is usually not unique, so care is needed as pointed out by Michael. $\endgroup$ – Carsten S Oct 23 '13 at 12:15
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    $\begingroup$ Just to point out an obvious example: in $\mathbb{Z}/13\mathbb{Z}$, the number $3$ is a quadratic residue (a perfect square), and if we say $y = \sqrt{3}$ in that setting we just mean that $y$ is either $4$ or $9$, which are the two square roots of $3$ modulo $13$. We cannot prove that $\sqrt{3}$ is unique, but we can prove that any candidate for $\sqrt{3}$ will satisfy $\sqrt{3}\sqrt{3} = 3$. $\endgroup$ – Carl Mummert Oct 23 '13 at 12:36

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