23
$\begingroup$

I had this problem yesterday. I tried to explain to the kid this: $$\sqrt{(-3)^2} = 3,$$ and he immediately said: "My teacher told us that we can cancel the square with the square root, so it's $$\sqrt{(-3)^2} = -3."$$

He has a lot of problems with maths, and I don't know how I can explain to him this as easily as possible. He still thinks that I lied him.

Thank you.

$\endgroup$
9
  • 3
    $\begingroup$ When I was 14 I might have been amused by the explanation that the operation of squaring and then taking the square root is like a Roach Motel for negative numbers--they go in but they never come out. And then you could show how the images of the negative numbers get "trapped" above zero. But I don't know if today's 14-year-olds have ever heard of a Roach Motel. $\endgroup$ – David K Mar 28 '15 at 19:30
  • 2
    $\begingroup$ Tell him to hang onto his hat. In a couple of years at school (or minutes at a search engine), he'll learn that (complex) exponentiation is multi-valued. This tends to put confusion over the sign of a square root, and over when/whether powers and roots simply cancel, into some perspective, because this is only the beginning. One realizes that many?/most?/all? elementary definitions and rules we learn eventually need some tweaking. At some point, the square-roots-cancel-squares rule is bound to break; it may not be too hard to accept that that threshold is reached with the problem at hand. $\endgroup$ – Blue Mar 28 '15 at 19:36
  • 4
    $\begingroup$ A lot of adults in America would have trouble with this one. $\endgroup$ – Robert Soupe Mar 29 '15 at 0:11
  • 13
    $\begingroup$ Ask him to CALCULATE instead of taking rules for granted. Ask him what is $(-3)^2$ and what is $\sqrt{9}$. $\endgroup$ – N. S. Mar 29 '15 at 1:05
  • 2
    $\begingroup$ similar as you might explain it to a 13 year old. $\endgroup$ – Seyhmus Güngören Mar 30 '15 at 21:27

15 Answers 15

18
$\begingroup$

After having given him the square root function definition, it may be helpful to show the following: $$ \sqrt{(-3)^2}=\sqrt{9}=\sqrt{3^2}=3\neq -3. $$ Although well-definedness may be a bit of a heavy topic for a 14-year old, it should be very easy to explain why you cannot have ambiguous definitions in mathematics.

$\endgroup$
4
  • $\begingroup$ I plus 1 you! :) $\endgroup$ – randomgirl Mar 28 '15 at 17:55
  • 2
    $\begingroup$ @randomgirl Haha thanks. The voting tendencies on MSE must be one of the most bizarre and random things to exist. I recall a precalc problem the other day where the question was upvoted several times while two quality answers were downvoted into oblivion but then pulled back up within the hour. Some weird stuff. $\endgroup$ – Daniel W. Farlow Mar 28 '15 at 17:56
  • 2
    $\begingroup$ I will never downvote anything. I see it as kinda negative. I either upvote or not do anything at all (or I guess I might leave a comment). $\endgroup$ – randomgirl Mar 28 '15 at 17:58
  • 9
    $\begingroup$ @randomgirl Please consider downvoting when something is wrong. This helps users and answerers (and the OP) know that an answer may be wrong. It's really a matter of quality management, but you are right on one thing--don't downvote to target someone. :) $\endgroup$ – Daniel W. Farlow Mar 28 '15 at 17:59
8
+250
$\begingroup$

This is about the difference between a function on one variable and an equation with one variable. A function should only have one result for each argument, whereas an equation can have more than one solution, or no solutions at all.

Let's consider a very simple function, $f(x) = x + 1$. For example, $f(3) = 4$, $f(4) = 5$. If $x$ doesn't change, neither should $f(x)$. You can run the function a hundred times with the same value and it should give the same result. Solve the equation $f(x) = 8$. The words "trivially simple" come to mind. This function has a one-to-one correspondence between inputs and outputs: each number $x$ is matched to a unique $f(x)$.

Now consider a slightly more complicated function, $f(x) = |x| + 1$. For example, $f(-3) = 4$. As with the first function, if $x$ doesn't change, neither should $f(x)$. But now an equation involving this function can have two solutions. If $f(x) = 8$, then $x$ can be $7$, but it can also be $-7$.

For the first function, we can define a look-up function by arithmetic means: $g(y) = y - 1$. But for the second function, if we demand a look-up function that uses arithmetic means rather than memory, we're going to be frustrated. The definition $g(y) = y - 1$ delivers the right result if $x$ was positive, or $0$. The question is whether or not this function $g(y)$ is good enough for our purposes.

That's what the principal square root function is: a function that tells us what value was input into another function if that value was positive, or $0$. The square function is defined as $x^2 = x \times x$. If $y = x^2$, the function $\sqrt{y}$ tells us what $x$ was if we know that $x$ was positive, or $0$.

Calculating $\sqrt{y}$ is not the same as solving the equation $x^2 - y = 0$. At least for this equation there are only two possible solutions: $\sqrt{y}$ and $-\sqrt{y}$. Therefore, $\sqrt{(-3)^2} = 3$, but the equation $x^2 - 9 = 0$ has two solutions, $x = 3$ and $x = -3$. We expect that if $y$ doesn't change, neither does $\sqrt{y}$, whether we run the function at noon, at midnight or any time of the day.

Canceling the square in $\sqrt{(-3)^2}$ to obtain $-3$ rather than $3$ creates a memory function rather than an arithmetic function. A memory function can be an arithmetic function only if the function being undone has a one-to-one correspondence between inputs and outputs.

$\endgroup$
6
$\begingroup$

Give him the definition of the square root function : for $x\in\mathbf{R}_{+}$, the number $\sqrt{x}$ is by definition the unique number $y\in\mathbf{R}_{+}$ such that $y^2 = x$.

$\endgroup$
4
  • $\begingroup$ In general (that is, when $x$ is not necessarily positive anymore), $\sqrt{x^2}$ is equal to the absolute value $|x|$ of $x$. $\endgroup$ – Olórin Mar 28 '15 at 17:58
  • $\begingroup$ Actually yes, because I could never ever come up with this. My bet with myself is won, thx to you. $\endgroup$ – Olórin Mar 28 '15 at 18:15
  • $\begingroup$ I'm sure you could come up with it yourself, just as I could come up with my answer myself. But it is precedence that is important here, just like in art. You have to be the first person throwing paint on a canvas to be original. $\endgroup$ – Yuval Filmus Mar 28 '15 at 18:18
  • $\begingroup$ @crash We're just having fun, don't take it so seriously. $\endgroup$ – Yuval Filmus Mar 28 '15 at 18:18
5
$\begingroup$

He (and his teacher, if the latter did indeed say that) may be right. After all what does the expression "the negative square root" mean. Many people use it, and it makes sense, and the negative square root of $(-3)^2$ is indeed $-3$.

So I believe a proper explanation should take into account that the terms "negative square root" and "square root" do not mean the same, and that it is a matter of convention in mathematics that "square root" is reserved to mean "positive square root" unless specified or implied otherwise from the context. Then it might be easier to comprehend (even for a 7-year old, not sure why age matters here) that $-3$ cannot possibly be the "positive square root" of anything.

Then it could also be explained that each positive integer has two square roots, one positive and one negative, but as we officially only call the positive one "the square root", when we start with $-3$ we end up with positive $3$, that is $\sqrt{(-3)^2}=\sqrt{9}=$ the positive of the two numbers $-3$ and $3$, that is $3$.

Even if you explain the above, you may have a harder time to explain why the above convention is adopted. (And if you work with cube and bigger roots and complex numbers that explanation may be even more difficult.) But the main point is to explain the convention and the terminology, since $-3$ is, after all, a square root of $9$, the negative one.

He is supposed to trust his teacher. So, it is also important to explain that his teacher is right, but that his teacher must have meant that cancellation holds only when we start with a positive number, e.g. $\sqrt{3^2}=3$ and $\sqrt{4^2}=4$. Then explain that his teacher did not mean that the rule applies when we start with a negative number, and in that case we end up with the opposite positive number, e.g. $\sqrt{(-3)^2}=3$ and $\sqrt{(-4)^2}=4$. It may seem a startling revelation, so please do no be disappointed if it takes a few days for it to be appreciated and accepted, after all it is simply a convention, and may need some deliberation why it is ok to accept it (and the reason is because everybody else did, and since we would like to assign a unique meaning to the unqualified term "the square root").

I wonder what the effect of the above lecture would be on a 14-year old ... I might have misses the point of your question. But, to paraphrase Euclid, "there is no 14-year-old road to square roots".

$\endgroup$
6
  • $\begingroup$ Nice answer, Mirko! $\endgroup$ – Relure Mar 28 '15 at 19:14
  • $\begingroup$ Um, "the negative square root of $\sqrt{(-3)^2}$" is actually $\sqrt{\sqrt{(-3)^2}} \approx -1.732$. $\endgroup$ – Robert Soupe Mar 29 '15 at 0:34
  • $\begingroup$ did you mean $-\sqrt{\sqrt{(-3)^2}} \approx -1.732\ \ $ , not $\sqrt{\sqrt{(-3)^2}}$ (the latter being the positive square root)? Thank you for the correction $\endgroup$ – Mirko Mar 31 '15 at 21:10
  • $\begingroup$ "He is supposed to trust his teacher". You seem to think 14-year-olds are little children. You don't have to twist reality for a 14-year-old. He or she can already understand that people can sometimes be wrong or can say something misleading even if they know better. Of course the teacher probably didn't say anything wrong about basic stuff like square roots, but that's not the point. You should discuss the math, not who-said-what. If the teacher did say something wrong, so what? People are often wrong and teachers are people. This shouldn't be shocking or surprising to a 14-year-old. $\endgroup$ – isarandi Apr 1 '15 at 15:57
  • $\begingroup$ @isarandi I would not retract my claim that students are supposed to trust teachers. There is a reason why students are students and teachers are teachers. Also, I did not say that the teacher said something wrong. While students indeed need to be critical and think on their own, I see no point to immediately jump to the conclusion that the teacher was wrong, this would be too easy an explanation. If a teacher is wrong there ought to be a way to clarify the misunderstanding, without jeopardizing the student teacher trust relationship, which is important for the student to be able to learn. $\endgroup$ – Mirko Apr 1 '15 at 17:10
4
$\begingroup$

For a student struggling with math, first I would get the student comfortable with problems like $\sqrt{25}=5$ (square root of a specific positive number is positive).

Also for these initial problems the student should understand that we mean the positive square root (i.e. we would not say that $\sqrt{25}$ is $\pm 5$; nor that it can be $-5$ if we want). This point legitimately causes some confusion since we often say 'square root of $25$' when we read $\sqrt{25}$, and indeed $-5$ is a square root of $25$. (We probably really should say 'the nonnegative square root' or the 'principal square root', but most of us don't).

Finally after the student was comfortable with these ideas, we can address the problem at hand, doing the inside first: $\sqrt{(-3)^2}=\sqrt{9}=3$

$\endgroup$
4
$\begingroup$

If he is struggling with math, he may find a graphic argument more helpful. Draw a graph of $f(x) = x^2$ and show him that every positive element on the $y$ axis has two (symmetric) preimages. Which one we choose to invert the function is indeed (mostly) irrelevant so it is a matter of convention: we like continuous functions, so we want to pick one side of the $y$-axis and stick with it, plus we really like positive stuff.

As others pointed out, this choice has the nice upside that $\sqrt{x^2} = \left| x \right|$. Still, it is worth stressing that this is just a choice and, while which one we make is indifferent, we must agree on one.

$\endgroup$
1
  • $\begingroup$ Could whoever downvoted this explain why they did so, allowing me to improve my answer? $\endgroup$ – A.P. Apr 5 '15 at 12:26
4
$\begingroup$

Explain how the squaring erases the information about the sign of $-3$. The square root operation only sees a 9 coming in, it can't know any more that it was made by squaring a $-3$ or a $+3$. So by convention it is defined to always give the positive answer. We could also define it to always give the negative one (although it would be clumsy), but what we can't do is recover the information that was lost when squaring.

$\endgroup$
4
$\begingroup$

I think the kid has misunderstood something his teacher said about cancelling. A math student of any age is bound to misunderstand his teacher at some point or other. When I was in high school, long, long ago, Mr. Jones was fond of saying that such and such equation has no real solutions, equations like, say, $x^2 + 9 = 0$. I thought he meant such an equation has no solutions at all. It wasn't until long after college that I learned about imaginary numbers.

The equation $x^4 - 81 = 0$ has four solutions, anyone who has studied the fundamental theorem of algebra can tell you. If we want to limit ourselves to real solutions, there are still two solutions left. But when we punch up $\root 4 \of {81}$ on a calculator, we want just one answer, and we want that answer to be the same each time, e.g., if it says the answer is $3$ one time and $-3$ another time, we'd think the calculator has a malfunction of some sort.

And so it is with the square root. We want the calculator to say $\sqrt 9 = 3$ every time, no matter how it is that we got the $9$ in the first place, whether by $(-3)^2$ or $3^2$ or any other operation that could possibly give $9$, such as $56 - 47$.

$\endgroup$
3
$\begingroup$

By definition, the square root of $x$ (if it exists) is the (unique) non-negative number $y \geq 0$ such that $x = y^2$. The square root only exists if $x \geq 0$, and in this case it always exists and is unique.

If $y \geq 0$ then it is true that $\sqrt{y^2} = y$. More generally, $$ \sqrt{y^2} = |y|. $$

For a more complete understanding, your high-schooler will have to study some complex analysis. The square root function is multi-valued, and the usual positive square root function is a specific branch of the function. Another possibility is using the non-positive square root instead of the non-negative square root. These are the only two possibilities which are continuous.

$\endgroup$
12
  • $\begingroup$ "Thou shall never fear to give the very same answer that an answer that was already given." $\endgroup$ – Olórin Mar 28 '15 at 17:47
  • $\begingroup$ @user10000100_u No two stones are alike, especially when one of them is 10 times as large. $\endgroup$ – Yuval Filmus Mar 28 '15 at 17:49
  • $\begingroup$ "In the realm of proverbs no proverb will tell you that $1\not=1$, and you know why." $\endgroup$ – Olórin Mar 28 '15 at 17:50
  • $\begingroup$ @user10000100_u Can I get in on this? "As lightning clears the air of impalpable vapours, so an incisive paradox frees the human intelligence from the lethargic influence of latent and unsuspected assumptions. Paradox is the slayer of Prejudice." (J. J. Sylvester) $\endgroup$ – Daniel W. Farlow Mar 28 '15 at 17:52
  • 1
    $\begingroup$ @user10000100_u Surely you meant "carcass"--I can't see why vultures would want to rest on a car case :P. In terms of pathetic points, in regards to my own answer, I hit my rep cap a few hours ago :/ So don't include me in that!! $\endgroup$ – Daniel W. Farlow Mar 28 '15 at 18:01
3
$\begingroup$

It's not quite right to say that the square and the square root cancel each other (in fact, this example demonstrates that nicely!)

For this problem, we have $\sqrt{(-3)^2} = \sqrt{9} = 3$. But why are these the right two steps? Because we evaluate functions from the inside out. Thus, the first step is to square the $-3$. The second step is a matter of convention: the square root of a positive number is taken to be the positive root (one could choose the negative root, but that would be a different function...)

$\endgroup$
0
3
$\begingroup$

There is an unspoken agreement that $\sqrt{x}$ means the principal square root of $x$. The principal square root of a positive real number is a positive real number, and the principal square root of a negative real number is a positive imaginary number, though of course a negative imaginary number is just as valid a square root of a negative real number.

To put it in terms of computer programming, $\sqrt{(-3)^2}$ means sqrt((-3)^2). By the rules of operator precedence, (-3)^2 is evaluated first, giving 9, just the same as if we had told the computer sqrt(3^2) instead.

So by the same token, with $i$ being the principal square root of $-1$, we have $\sqrt{(-3i)^2} = 3i$, not $-3i$.

If you choose to tell your kid this, he might ask about the principal square root of an imaginary number. I believe that would be a complex number...

$\endgroup$
5
  • $\begingroup$ what do you mean by a "positive" or "negative" imaginary number? As far as I know the terms positive and negative are only defined for real numbers. I guess I see what you mean, but I would probably not adopt your terminology in this context. $\endgroup$ – Mirko Apr 1 '15 at 17:19
  • $\begingroup$ @Mirko I saw this objection coming, but I didn't think of how to address it until now. If $\Re(z) = 0$ and $\Im(z) \neq 0$ but $\Re(-zi) > 0$ then I call $z$ a "positive imaginary" number but if $\Re(-zi) < 0$ I call it a "negative imaginary" number. I could honestly respect someone if they'd rather call $z$ such that $\Re(-zi) < 0$ a "positive imaginary," but I can only sigh at someone who insists both that someone and I are wrong. $\endgroup$ – Robert Soupe Apr 2 '15 at 2:22
  • 1
    $\begingroup$ When I first read your answer I somehow thought you were talking about positive or negative complex numbers, then halfway through writing my comment I realized you were only talking about positive or negative imaginary numbers, the meaning of which is quite obvious (positive or negative $y$ in $(0,y)$ when one identifies an imaginary number with a point in the plane).Your definition above says the same,I believe. The problem for me is that unless one uses the term "pure imaginary number" i tend to think of "imaginary number" as being the same as a "complex number",even if I know this is wrong. $\endgroup$ – Mirko Apr 2 '15 at 3:16
  • $\begingroup$ You're probably not alone with that problem, I think I've caught myself a couple of times thinking along similar lines. Maybe I should have said "purely imaginary number." Or maybe I should have not mentioned imaginary and complex numbers at all. $\endgroup$ – Robert Soupe Apr 2 '15 at 3:36
  • $\begingroup$ complex numbers complicate things ... I wonder if one extends the discussion to cube roots then whether $\sqrt[3]{-1}$ which is $-1$ in the reals, whether this is also the principal cube root of $-1$ in the complex numbers (well, can't be principal as it does not generate the other two). So perhaps one strays away from the original question if complex numbers are involved, but on the other hand for square roots your answer involving complex numbers presents an interesting point of view. $\endgroup$ – Mirko Apr 2 '15 at 3:47
2
$\begingroup$

Many issues I had at a similar age were rectified after being taught by a teacher about order of operations.

First Parentheses, Division, Multiplication, Addition and finally subtraction. It might be useful to introduce the kid to a certain "order" to which you must comply if you want to simplify a formula which might include exponentiation and taking square roots. I was told by my teacher to memorise the first letters of each operation (in Sinhalese) and make a poem of sort to know what operation I should first compute.

After learning this students generally get a good understanding about why parentheses are placed and what operations should be done first when simplifying which mostly gets rid of questions like this.

Then couple that with what others have said here about defining the square root operation (let's call it that instead of a function here) and you might get somewhere.

$\endgroup$
5
  • $\begingroup$ Math is not about manipulating formulas but solving problems. Imagining math as rules to be memorized for transforming one formula to another is the quickest way to making people hate the whole thing. The most important thing is to understand what you are doing. A function is a function even if you use different notations. You can even represent a formula by connected boxes with inputs and outputs if you want (addition with two inputs and one output, square root with one input and one output, etc). What matters is the abstract idea behind it all. $\endgroup$ – isarandi Apr 1 '15 at 16:08
  • $\begingroup$ @isarandi: Key word there - abstract. You can "understand" what happens to numbers when they are manipulated. But when it comes to abstract mathematics you must learn to confine yourself to the definition and only those things that have been proven. The failure to conform to such conventions is what motivates students to do things like "cancelling off" the power and the square root operations. "Memorise" here is a bad word. I propose the student must "understand" how and why the order of operations are as they are. Once he/she grasps that the rest should flow - at least for now $\endgroup$ – Ishfaaq Apr 2 '15 at 0:22
  • $\begingroup$ But for this you don't even need to know the order of operations. It's not even a question of notation. To understand why the square root cannot perfectly reverse the squaring operation, you don't have to manipulate symbols at all. It's simply that squaring maps multiple inputs to the same output so the original input can't be recovered from the output alone. The principle is the same for why division by zero is undefined. Because multiplication by zero maps many (all) things to the same output, so no reversal is possible. $\endgroup$ – isarandi Apr 2 '15 at 0:53
  • $\begingroup$ And actually it's not even a big "mistake" to think that it can be canceled off. We could define the square operation such that it "remembers" and passes along the sign of its input. In computing, we even have negative and positive zero, and negative and positive infinities. There is also the complex exponential function $e^{ix}$ which is basically just a sine wave that remembers which way it is currently going (as opposed to a normal sine wave that takes the same value in two distinct positions with opposite derivatives). You could in fact make your definitions so that the sqrt cancels. $\endgroup$ – isarandi Apr 2 '15 at 1:01
  • $\begingroup$ @isarandi: I am sure we could debate this forever. My next argument might be a choice between the propriety of introducing a 14 year old informally to injective functions and those that are not, and maybe the potential terrors of seeing varying definitions of the same thing in different places - a result of deviating from convention (there is a reason why teaching kids is so hard and streamlined). But this was my humble suggestion - I probably don't even have the credentials to debate this with you. But it helped me and I'm confident it will help those who are willing to try it. $\endgroup$ – Ishfaaq Apr 3 '15 at 4:48
1
$\begingroup$

It may be worth pointing out that one reason to define (at his level) the square root of a non-negative real $\sqrt{x}$ as the positive real whose square is $x$ is that it avoids paradoxes such as: $9 = 9$, which we can write as $3^2 = (-3)^2$, which we can write as $\sqrt{3^2} = \sqrt{(-3)^2}$, "therefore" $3 = -3$.

But in general, the problem doesn't seem mathematical. Like many $14$-year-olds, he doesn't like to be told he was wrong. Eventually, I would guess, he'll come around, but he won't freely admit it.

$\endgroup$
0
$\begingroup$

Clearly that he is looking for some rule to make his life easier. And when he saw power rapped by a square he pooled the best thing he remembered from class:

My teacher told us that we can cancel the square with the square root...

So the solution will be to give him another rule to clear things up. The most simple thing I can think of its:

Power of even number will always result in positive solution:

$x^y > 0$ if $y$ is even

$\endgroup$
-2
$\begingroup$

I think perhaps the boy is right and You've not told him the whole truth. $\sqrt{(-3)^2}$ does actually equal -3 but it also equals +3. In the same way as $\sqrt{9}$ = 3 or -3.

I wonder if he has a grasp of negative numbers yet? If not then I think that would be the place to start. If he does then start by showing him that $3^2 = 9$ and $(-3)^2 = 9$, then you can tell him that $\sqrt{9}$ takes us back from the square to either of the original numbers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.