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Yesterday, I had the pleasure of teaching some maths to a high-school student. She wondered why the following doesn't work:

$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$.

I explained it as follows (slightly less formal)

  • For your hypothesis to hold, it should hold given an arbitrary set of operations performed on your equation.
  • For example, it should hold if we square the equation, and after that take the square root, i.e. (note that I applied her logic in the second line; I know it's not OK to do maths like that) $\sqrt{a+b}=\sqrt{a}+\sqrt{b}\\ a+b=a+b+2\sqrt{ab}\\ \sqrt{a+b}=\sqrt{a}+\sqrt{b}+\sqrt{2\sqrt{ab}}$

  • We now arrive at a contradiction, which means that your hypothesis is false.

However, she then went on to ask 'But why then is it false? You only proved that it's false!'. As far as I'm concerned, my little proof is a perfect why explanation as far as mathematicians are concerned, but I had a hard time convincing her - the only thing I could think of is to say that the square root operator is not a linear operator, but I don't really think that adds much (besides, I really don't want to be explaining and proving linearity to a high school student).

So, my question: is there anything 'more' as to why the above doesn't work, or was I justified in trying to convince her that this is really all there is to it?

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    $\begingroup$ How about $\sqrt{2} \neq 2$? $\endgroup$ Sep 17, 2015 at 12:02
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    $\begingroup$ Before one comes to terms with accepting mathematical proofs as they are (and perhaps even after), when one comes across a statement and ask whether it's true or not, it's common to not find a simple arithmetical proof satisfying. I believe she wants more of a "philosophical" reason why it's false, a "moral" to the story. Using $a = b = 1$ to demonstrate that it isn't always the case, as well as, say, $a = 4, b = 0$ to show that it is, in fact, true some times might help. $\endgroup$
    – Arthur
    Sep 17, 2015 at 12:06
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    $\begingroup$ Can you not draw a diagram with squares of sides (a+b), a and b and illustrate graphically that the conjecture is false? It seems that that is more the level on which the original question is being asked. $\endgroup$
    – Sean Reid
    Sep 17, 2015 at 13:47
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    $\begingroup$ This might be a good question for matheducators.stackexchange.com. matheducators.stackexchange.com/questions/926/… seems to address part of this. $\endgroup$ Sep 17, 2015 at 16:33
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    $\begingroup$ @user133281: Using the rule $\sqrt{x+y} = \sqrt{x} + \sqrt{y}$, of course! $\endgroup$
    – user14972
    Sep 17, 2015 at 22:33

15 Answers 15

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I was one of those students that was fanatical about the "why," so I can give my own viewpoint on this. When I asked "why" on something like this, its because I was looking for a general principle I could apply. I wasn't just concerned with the problem at hand, but rather the situation at hand raised a red flag because my intuition wasn't lining up with what I was told. I was concerned with the legion of potential misconceptions I might be holding onto because there was some underlying thing I was missing. I needed to understand, in my own terms, what the "real deal" was.

That "real deal" is different for each individual. Several other answers have already pointed out that there won't be a one-size-fits-all answer for this, because each student thinks about math in a different way. Some are visual learners, needing pictures. Personally, I viewed a great deal of this from the perspective of semantics vs syntax: What meanings are true versus what syntactic manipulations are valid. So naturally when I view the issue your student had, I view it from my own perspective.

From my perspective, the issue is that it looks like there is a valid syntactic manipulation, $\sqrt{a+b}=\sqrt a + \sqrt b$. If I am told that that manipulation is invalid, I want to understand why. The words I would now use to describe why is that what I am trying to do here is "distribute" the square root over the addition operation, and you're not allowed to do that. If I were to remove the square root, and replace it with a more generic function, $f(a+b) = f(a) + f(b)$, it starts to become more clear that such an assumption is not always valid. In fact, it starts to look like a truly special case (especially when you consider that the above assumption doesn't even hold water for simple cases like $f(x) = x + 1$). The ability to distribute one function over another is not the norm, its the special case that occurs in a few situations (like distributing multiplication over addition).

This ends up reframing the problem away from "why can't I distribute a square root over addition" to "what special properties are needed to allow the distributed property?" It moves the specialness away from square roots, and puts specialness on things like addition and multiplication. It starts to lead one to appreciate why addition and multiplication are so useful: they have so many nice convenient properties other functions don't have!

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  • $\begingroup$ +1 I was planning to write an answer like this but now there's no need. The "why" in this particular question is indeed best answered by thinking about distributing a function over addition, which beginners want to invoke often - for squaring, exponentiating, ... Of course it works only rarely. I like to point out that the important place where it does is what they actually know as the "distributive law" for the function $f(x) = cx$, Of course you don't go on to discuss to formal linear operators ... $\endgroup$ Sep 18, 2015 at 16:38
  • $\begingroup$ Actually, $\sqrt{a·b} = \sqrt a · \sqrt b$ works (where both are defined) ... so she used just the wrong operator here. $\endgroup$ Sep 19, 2015 at 23:54
  • $\begingroup$ @PaŭloEbermann Good point. That also makes it even worse: some operators allow you to do that manipulation, others don't! =D $\endgroup$
    – Cort Ammon
    Sep 21, 2015 at 16:49
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You did not arrive at a contradiction, and technically, you did not even have a theorem in the first place.

To be strict, $\sqrt{a+b} = \sqrt a + \sqrt b$ is not a mathematical theorem. A mathematical theorem is

For all real numbers $a,b$, the equality $$\sqrt{a+b} = \sqrt a + \sqrt b$$ is true.

This theorem is false because the statement

There exist two real numbers $a,b$ such that $$\sqrt{a+b} = \sqrt a + \sqrt b$$ is not true.

which is the negation of the theorem, is true. This statement is true because you can set $a=b=1$ and prove that it is true.


Really, what you did was not a contradiction. You simply showed that if $\sqrt{a+b} = \sqrt a + \sqrt b$ is true, then $\sqrt{a+b} = \sqrt a + \sqrt b + \sqrt{2\sqrt{ab}}$ is also true. You did not prove that this other statement is false.

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  • $\begingroup$ Well I started out with an example with numbers, and stated that since I can show a counterexample, her theorem was false. But she really wants a why it's not true, rather than a proof that it's not true. As far as I'm concerned, that's one and the same thing. Do you agree? $\endgroup$
    – Sanchises
    Sep 17, 2015 at 12:05
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    $\begingroup$ @sanchises Yeah, mostly. Try asking her "why is the statement 'all people have three feet'" false. If she says "because I have two feet",say that that is just proof that it is not true, not why it is not true :P $\endgroup$
    – 5xum
    Sep 17, 2015 at 12:07
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    $\begingroup$ Not to be too much of a pedant, but I'd say that "For all real numbers ... is true." isn't a theorem because the claim isn't true. It's a proposition or sentence, but it's not a provable one (with respect to the proof system, etc., at hand), so it's not a theorem. The question is really whether "forall a, b, ...." is provable or not. If it is, then it's a theorem. The sentence $\forall a, b\,\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ is false, and that can be shown by demonstrating that $\exists a, b\,\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$. $\endgroup$ Sep 17, 2015 at 16:02
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    $\begingroup$ @5xum That's exactly what I did (well, I used 'all strawberries are red', which is true until you find a green one) . Then she said, "Well then. Can you explain why some strawberries are green?" $\endgroup$
    – Sanchises
    Sep 17, 2015 at 19:51
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    $\begingroup$ @kasperd I never said that the equality is false for all pairs of real numbers. I only said that the statement 'This equality is true for all pairs of real numbers' is false. So no, there is no factual error in my answer. There is no false statement that I claim is true. $\endgroup$
    – 5xum
    Sep 18, 2015 at 11:03
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I've found that people often respond well to visual proofs as satisfactory answers to "why" a theorem is true. In this case you could use a diagram that looks something like the following:

enter image description here

This is meant to be a square with side length $\sqrt{A}+\sqrt{B}$, but which clearly does not have area $A+B$, and therefore the side length cannot also be expressed as $\sqrt{A+B}$.

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    $\begingroup$ And note - this diagram shows exactly when the result is true - it's when A or B is zero. So with this picture, we know when it's true as well as when it's not true. $\endgroup$
    – Joel
    Sep 20, 2015 at 11:39
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The way I see it, in your argument you didn't reach a contradiction: you proved that if $\sqrt {a+b}=\sqrt {a}+\sqrt {b} $, then $$\sqrt {a}+\sqrt {b}=\sqrt {a}+\sqrt {b}+\sqrt {2\sqrt {ab}}. $$ That is $\sqrt {2\sqrt {ab}}=0$, so $ab=0$.

The conclusion is that $ \sqrt {a+b}= \sqrt {a}+\sqrt {b}$ holds precisely when at least one of $a,b $ is zero.

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    $\begingroup$ The conclusion is not that a or b must be zero for it to hold but that it might hold for that case, but surely doesn't for all other cases. $\endgroup$ Sep 17, 2015 at 14:31
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    $\begingroup$ The argument shows that the equality implies $ab=0$. If $ab=0$, the equality holds. So it is an "if and only if", and that's why I used the word "precisely". $\endgroup$ Sep 17, 2015 at 15:46
  • $\begingroup$ @MartinArgerami however you still need to verify that it holds if $ab = 0$ as the chain is broken when we square in the first step $\endgroup$ Sep 17, 2015 at 17:39
  • $\begingroup$ @thanosQR: I have no idea what "chain" you are talking about. If $ab=0$, then at least one of them is zero. If $a=0$, $\sqrt{a+b}=\sqrt{b}=\sqrt{a}+\sqrt{b}$, and similarly when $b=0$. $\endgroup$ Sep 17, 2015 at 19:25
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    $\begingroup$ @MartinArgerami He's talking about the fact that you only proved one direction in your answer, namely: If $\sqrt a + \sqrt b = \sqrt{a+b}$ holds, then $a = 0$ or $b = 0$, you lost the equivalence when you squared. $\endgroup$ Sep 17, 2015 at 23:48
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This isn't a technical issue; it's a human one. So it needs a human approach.

Theorem

All people are blonde.

If she disagrees, ask her why.

Help her see that a counterexample to the proposed theorem shows both that it's false and "why" it's false (which are really the same thing).


She may be wondering why her intuition is wrong; I suspect everyone has wondered that (theorem! ;). This is particularly true of distribution.

Explain that just because something looks like it could be true, doesn't mean that it is true. Distribution of multiplication over addition ($a(b + c) = ab + ac$) is very special; most operators don't have this property.

This is why a careful study of math is important; intuition can lead us astray.

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Assuming it wasn't about the definition Problem the other answers have brought up:


Explain it with the "Pizza problem":

You can either take one pizza with radius a, or two pizzas with radius a/2; Which should you take?

It explains the core concept in a nice visceral way, which some people need before they can accept the logic behind something.

(Alternatively use squares instead of circles, it makes the answer even more obvious.)

From there you can then expand the concept to squareroots.

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    $\begingroup$ -1: this does not answer the question. $\endgroup$
    – Deusovi
    Sep 17, 2015 at 17:32
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    $\begingroup$ I agree that a geometrical or visual proof may help understand why the hypothesis is not true. $\endgroup$
    – lodebari
    Sep 17, 2015 at 17:47
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    $\begingroup$ I think this might be a good way to go. I think this answer would get a lot more upvotes if you included a visual example. I'd use square pizza's though, since that's easier to relate to square roots. $\endgroup$
    – Sanchises
    Sep 17, 2015 at 19:53
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The fact is that for a very large number of interesting functions $f$, including the square root, $f(x+y)$ is not generally equal to $f(x) + f(y)$. When a function $f$ does satisfy $f(x+y) = f(x) + f(y)$ for all $x$ and $y$, it's a remarkable property of that function.

For the square root in particular, you can appeal to the Pythagorean Theorem. Since you've already used $a$ and $b$, let's label the sides of a right triangle $p,q,r$ instead, with $r$ as the hypotenuse. Then $p^2 + q^2 = r^2$.

Draw the usual graphic representation of this with a square on each side of the triangle. Now suppose the areas of the squares on the two legs are $a$ and $b$; in other words, given $a$ and $b$ let's draw a right triangle whose legs are $p = \sqrt a$ and $q = \sqrt b$. The area of the square on the hypotenuse is $r^2 = p^2 + q^2 = a + b$, so $r = \sqrt{a + b}$. Now, looking at this picture, how does $\sqrt{a + b}$ (the length of the hypotenuse) compare to $\sqrt a + \sqrt b$ (the lengths of the two legs)?


Another way to show "why" $\sqrt{a + b}$ is not $\sqrt a + \sqrt b$ (or perhaps rather how these two expressions come to be unequal) is to imagine we start with two squares of areas $a$ and $b$; if the squares are not equal, let $b$ be the smaller one. Now put square $b$ alongside square $a$ so they share a vertex and part of the side of the larger square. Notice that the longest side of the resulting figure (consisting of collinear sides of the two squares) has length $\sqrt a + \sqrt b$.

Now deform $b$ into a border around two sides of square $a$ so that square $a$ plus the border is a slightly larger square. Notice how we had to "flatten" the smaller square to make this work, so instead of a figure with a side $\sqrt a + \sqrt b$, we get a square whose side is larger than $\sqrt a$, but not nearly that much larger. That is, an amount of "space" in the plane with area $b$ is a lot "fatter" when we draw it as a single square than it is when we "stretch" it over two sides of a larger square.

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How about a counterexample to disprove a universally quantified statement? For instance $\sqrt{9+16}\ne \sqrt{9}+\sqrt{16}$.

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  • $\begingroup$ -1 What makes you think I didn't try a simple numerical answer first? $\endgroup$
    – Sanchises
    Sep 17, 2015 at 20:37
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    $\begingroup$ @sanchises And when that didn't help you thought; I know - I'l make the argument more abstract. That's bound to make everything more understandable. $\endgroup$
    – Taemyr
    Sep 18, 2015 at 13:20
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Because any continuous function from real numbers to real numbers where $$f(a+b) = f(a) + f(b)$$ has a graph that is a straight line through the origin, and the graph of $y = \sqrt x$ is curved.

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  • $\begingroup$ Love it. This kind of "why" explaination will make her shut up! $\endgroup$
    – Jakob
    Sep 18, 2015 at 9:12
  • $\begingroup$ No -- there are functions which satisfy $f(a+b)=f(a)+f(b)$ for all real $a,b$, but whose graph is not a straight line. See en.wikipedia.org/wiki/Cauchy's_functional_equation. $\endgroup$ Sep 19, 2015 at 19:07
  • $\begingroup$ @FedericoPoloni There are no continuous additive functions $\mathbb{R} \to \mathbb{R}$ that are not a multiplication with a constant. $\endgroup$
    – user10904
    Sep 20, 2015 at 10:06
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I'm not sure if I'm just stating the obvious here, but it has only been implicitly touched by the other answers, so I'll explicitly write it down here.

For me the answer to "Why is this wrong?" is "Because this is breaking the rules that we agreed on!":

In mathematics, a "theorem" or a "statement" is stated in a specific language and to decide if we want to call it true or not is judged by certain rules. If it doesn't comply with the rules, we call it wrong. In your example the statement is not compliant with the rules of the underlying logic as you already showed.

I know that in real world applications (and situations) this abstract reason feels unsatisfactory for some people and that's perfectly fine. It just shows that you didn't present your argument in a way that convinces the other person to agree with your view of (not) complying with the rules.

For your specific question this means:

  1. There is nothing more to "Why?" than what you already understood. But that doesn't mean you can't say anything more to answer the question.

  2. I absolutely do not think that you should stop at the statement "this really is all there is to it" because even if this might be true from your point of view, she obviously is not convinced that this is the case and in the end she didn't understand what she wanted to know in the first place.


I find the second part of your question (2. above) quite hard to answer and as it was already hinted at in the comments, this should most likely be answered not by mere mathematicians but by someone with a background in education/didactics. My arguments above focused on the mathematical aspect of your question. (Not repeating formal issues already covered by other answers.)

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  • $\begingroup$ Exactly right. It is always because the rules are not made clear to the student (and usually not even clear to the teachers) that mathematical proofs and disproofs cannot be understood. The rules must be laid down right at the beginning and never changed. Then the game begins! $\endgroup$
    – user21820
    Sep 18, 2015 at 12:33
  • $\begingroup$ In other words, one shouldn't be asking why certain mathematical structures are the way they are before asking why we chose the logical rules we chose. Logic always comes first, and should follow naturally from the semantic interpretation we want the symbols to have. Once logic is fully grasped everything becomes an order of magnitude easier. $\endgroup$
    – user21820
    Sep 18, 2015 at 12:36
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Others have pointed out the weakness, I would like to make some supplementary remarks.

It is clear that not all $a,b \in \mathbb{R}$ are such that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$; $a := -1$ and $b:=2$ will do. This may lead us to guess that the "problem" seems to lie at the thing that some real numbers cannot be eaten by the square root function. So now you may lead him or her to formulate a conjecture: Is it true that, if $a,b \in \mathbb{R}$ such that $\sqrt{a+b}$, $\sqrt{a}$, $\sqrt{b}$ are meaningful, then $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$? And then use a simple counterexample, say $a := b := 1$, to show the student that this conjecture is also not true.

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It's true that false statements aren't required to have a fundamental logical reason for being false - they are false because a counter-example exists. In this case though, I feel the why of it is really asking this: why is the square root operator not linear, unlike some other operators that you have seen in class. The simple answer is that most operators are not linear, and you have to examine each one to determine if it is or not. Which you go about by attempting to prove the sort of identity that she assumed was true.

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Let $f(a,b)=\sqrt{a+b}$. Then $f$ is defined for $a+b\ge0$.

Let $g(a,b)=\sqrt{a}+\sqrt{b}$. Then $g$ is defined for $a\ge0,b\ge0$.

Then the domain of $f$ is a strict superset of the domain of $g$, so the functions are fundamentally different.

This shows that the original identity does not make sense.

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Here's another explanation approach:

Suppose that $$\sqrt{a+b}=\sqrt a + \sqrt b$$

If this is true, then we can square both sides of the equation:

$$(\sqrt {a+b})^2=a+b=(\sqrt a+\sqrt b)^2=a+b+2\sqrt {ab}$$

This resulting equality can then be manipulated as follows:

$$a+b=a+b+2\sqrt{ab}\\ 0=2\sqrt {ab}\\ 0=ab$$

By the rules of real numbers, which do not have zero-divisors, if $ab=0$ then either $a=0$ or $b=0$. Looking back at the original supposition, it will hold given either condition. So the answer is that the supposition is correct as long as at least one of $a,b$ is $0$, or in more explanatory language, "you can do that, as long as either $a$ or $b$ is zero."

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I have a different perspective on this question that I want to share. The answer to your questions is that there is nothing more to why that hypothesis is false and you are completely justified in trying to convince her of this.

To begin with, it is important to understand that there are different connotations of the 'Why?' question. One of the connotations is "Why are you saying that this is true/false?", and another is "What causes this to be true/false?". The first question can be answered in many ways, while the second question can only be answered by a causal explanation.

For instance, consider the proposition

A ball thrown into the air always falls at a angle of $45^{\circ}$ to the ground.

One can say that this proposition is false by showing a counter example, but to answer what causes this to be false, one has to show that this contradicts law of gravitation, which we surely know to be true.

For propositions that are true, showing that no counter examples exist is not a causal explanation, though it is a valid proof of the truth of the proposition. A causal explanation of why the proposition is true should invoke one or more laws that subsume the proposition, i.e., the proposition is an instance of the operation of the law. This is same as what Cort Ammon mentioned in his answer. For example, consider the proposition

For $a, b \in \mathbb{R}$ and $n\in \mathbb{Z}$ $$(ab)^n = a^n b^n$$

This can be shown to be true because the operation of multiplication of real numbers is commutative. In fact, this holds for any commutative operation on a set. And therefore one can abstract this property from the operation of multiplication of real numbers and state it to be the cause of the above identity.

Likewise, for a proposition that is false, a counter example is not a causal explanation, though it is a valid proof of the falsity of the proposition. A causal explanation must show that the proposition contradicts either axioms, definitions or established theorems. To be more clear, a proposition $X$ is false because another proposition $Y$ is true, where $X$ contradicts $Y$. This is what Piwi mentioned in his answer.
For example, consider the hypothesis in your question

For $a,b\in\mathbb{R}$ $$\sqrt{a+b} = \sqrt{a} + \sqrt{b} $$

Following your steps, we can show that the above hypothesis contradicts that $\sqrt{ab}\neq 0$ for some $a,b \in \mathbb{R}$. Therefore, we can say that the hypothesis is not true because $\sqrt{ab} \neq 0$ for some $a,b\in\mathbb{R}$.

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