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So, my friend show me prove that $1=-1$ by using this way:

$$1=\sqrt{1}=\sqrt{(-1)\times(-1)}=\sqrt{-1}\times\sqrt{-1}=i\times i=i^2=-1$$

At first sight, I stated "No, $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ is valid only for $a,b\in\mathbb{R}$ and $a,b\geq0$"

But, I remember that $\sqrt{-4}=\sqrt{4}\times\sqrt{-1}=2i$ which is true (I guess).

Was my statement true? But, $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ is also valid if one of a or b is negative real number. Why is it not valid for a and b both negative? If my statement was wrong, what is wrong with that proof?

ions me
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user379677
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    @RobertIsrael: Yes and no -- the OP here presents a particular misunderstanding (namely, confusion between $a,b\ge 0$ being a necessary or a sufficient condition) that the previous question or its answers do not handle. – hmakholm left over Monica Dec 07 '16 at 01:17
  • @user379677 Reading the comment $\sqrt{-4}=2i$ kind of hurts...You got to watch these conclusions with some care – imranfat Dec 07 '16 at 01:26
  • @imranfat I'm sorry. Because I still think that $\sqrt{-1}=i$. That's why I thought it's true. – user379677 Dec 07 '16 at 01:29
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    @user379677 Yes, I can see that and I don't blame you. I see it too in a lot of high school textbooks. For "low level" stuff, this definition works fine, but going higher up with complex variables, that definition has some drawbacks. Saying "The squareroots of $-4$ are $2i$ and $-2i$" is better (I still don't like it), but the radical symbol as an operator on negative numbers, please avoid... – imranfat Dec 07 '16 at 01:31
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    @imranfat Ah... I remember that. I've learned that for all complex number there are always max 2 squareroot of it. Is that why I can't easily state $\sqrt{-4}=2i$? – user379677 Dec 07 '16 at 01:35
  • There is a very interesting answer at this almost identically titled post that I wish could be merged here somehow, but it seems unlikely. – rschwieb Jun 01 '20 at 13:53
  • @imranfat I do not understand what is wrong with "The squareroots of −4 are 2i and −2i". Could you elaborate? – Some Guy Jan 08 '21 at 01:10
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    @SomeGuy In essence it is generally understood that $2i$ and $-2i$ are the solutions to the equation $x^2=-4$. So we can say that the square roots of $-4$ are $2i$ and $-2i$ but it is incorrect to say $\sqrt{-4}=2i$. The reason why one has to be careful with the word "roots" is because it is "deficient" For example: The cube root of $1$ is not just $1$, is it? There are two more (complex) solutions, but we CAN say that $1$ is a solutions to $x^3=1$ – imranfat Jan 08 '21 at 03:01
  • @IonSme Please don't make cosmetic edits to old questions. That moves them to the active question queue which distracts people who look at what's really current. – Ethan Bolker Apr 21 '23 at 19:32

4 Answers4

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As you know, the rule $\sqrt{ab}=\sqrt a \sqrt b$ holds for some but not all combinations of $a$ and $b$. Explaining and remembering exactly which those combinations are is usually more trouble than it's worth, so usually the rule we remember is just

It is a sufficient condition for $\sqrt{ab}$ to equal $\sqrt a\sqrt b$ that $a$ and $b$ are both non-negative reals.

As you have noticed, this condition is not necessary, but that does not keep the rule from being useful.

For the purpose of rejecting your friend's fake proof, even the above version is more than you need; all you need to say is

The rule $\sqrt{ab}=\sqrt a\sqrt b$ does not always hold when we extend the $\sqrt{\phantom a}$ function to complex numbers.

It is not your task to prove that the rule fails in the particular case $a=b=-1$ (thought doing so is a simple matter of computation); it is the guy who wants to prove something who has the responsibility for only using rules he knows apply in the context he's using them in. After you've pointed out that the rule has been stretched beyond the domain we know it to work for, it's up to him to figure out whether he can come up with an argument that it should be valid here.

hmakholm left over Monica
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  • So, there is another condition that $\sqrt{ab}=\sqrt{a}\sqrt{b}$? But, how can I prove that he's wrong by only said the sufficient condition for it? Because what I know about sufficient condition is that if the sufficient condition doesn't hold, then the statement $\sqrt{ab}=\sqrt{a}\sqrt{b}$ doesn't mean to be wrong. – user379677 Dec 07 '16 at 01:26
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    x @user379677: You don't even need to state a sufficient condition in order to point out that the proof is wrong. It is enough to say: This rewriting is not valid at that place, and it is then up to your friend to produce a rule that would apply in that situation and make it valid. Since he can't, you win. – hmakholm left over Monica Dec 07 '16 at 01:30
  • Ah... now I get it... thank you... ^^ – user379677 Dec 07 '16 at 01:40
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    Haha, I once saw a game where the winner didn't win in the normal sense, but forced the opponent to break the rules in some way. This reminds me of that. – Simply Beautiful Art Dec 07 '16 at 01:43
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    @SimpleArt: The idea of "winning" here can be taken quite seriously and leads to things such as game semantics for the meaning of proofs. – hmakholm left over Monica Dec 07 '16 at 01:50
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An alternate way to understand what is happening here is to note that $$ 1=e^{0i} $$ When we take the square root, we have $$ 1=\sqrt{e^{0i}} = \sqrt{e^{-\pi i}\times e^{\pi i}} $$ Notice that $e^{-\pi i}=e^{\pi i}=-1$.

Now, since we are working in polar form, we can evaluate the square roots consistently, arriving at $$ 1=e^{-\pi i/2}\times e^{\pi i/2} = -i\times i = 1 $$

Essentially, the problem lies in the "branch cut" that occurs with the square root operation - you must be careful with the evaluation.

To put it another way, $1=e^{2n\pi i}$ for all integer $n$, and the square root function has to respect its specific value (of $n$), as it can take multiple different values depending on that $n$. To get $1=-1$ as in the question, one must simultaneously use $1=e^{0i}$ and $1=e^{2\pi i}$.

Glen O
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    If I use $1=e^{2\pi i}$ so I split $\sqrt{e^{2\pi i}}$ become $\sqrt{e^{\pi i}\times e^{\pi i}}$ which when I evaluate the square root become $e^{2 \pi i/2}\times e^{\pi i/2}$. It equals $i\times i=-1$. What's wrong with it? – user379677 Dec 15 '16 at 11:32
  • @user379677 - two mistakes. First, $\sqrt{e^{2\pi i}} = e^{\pi i} = -1$ (this is why I said you have to keep an eye on the branch), so that agrees with $i\times i=-1$. Second, you've accidentally written $e^{2\pi i/2}$ in your product. – Glen O Dec 15 '16 at 14:59
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    Essentially, you can't go "well, $1=e^{2\pi i}$, so $1=\sqrt{e^{2\pi i}}$", because those two statements aren't the same. – Glen O Dec 15 '16 at 15:00
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The particular step in question is whether or not it is the case that

$$\sqrt{(-1)\times(-1)}\stackrel?=\sqrt{-1}\times\sqrt{-1}$$

In particular, it is questioned whether or not we can have $\sqrt{(-1)^2}=(\sqrt{-1})^2$, and here, the answer is no due to how we interpret the operations.

This is because the $x^2$ operation cancels the $\sqrt x$ operation, but it does not work the other way around, since

$$y=\sqrt x\implies y^2=x,\underbrace{y>0}_{\text{we lose something here}}$$

This means we lose a possible value in the process, since we only take one of the possible values that could be the solution. On the other hand, regardless of which value a square root is denoted, the squaring operation will take both and make the end result the same.

Simply Beautiful Art
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  • Ah... so my friend just like switch the order of operation and didn't realize that switching it true only if $a,b$ are real nonnegative? – user379677 Dec 07 '16 at 01:32
  • @user379677 No, it has to do with that the operations aren't perfect inverses. For example, take a number. Add one to it, then subtract one. You got the number you started with, right? Now take a number. *Square* it, the square root it. You only get the number you started with if it wasn't negative? What about if you start with a complex number instead? The point here is that$$\sqrt{\square}=...\require{cancel}\cancel{\imply}\square=(...)^2$$ – Simply Beautiful Art Dec 07 '16 at 01:35
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    So square root is not inverse of power by 2 or square root is inverse of power by 2 only for some number? – user379677 Dec 07 '16 at 01:38
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    @user379677 It is taken, usually by definition, to be the principal inverse. The 'total' inverse would have to be given by a $\sqrt\square=\pm...$. Over the complex numbers, the principal value has argument $\in[0,\pi)$, i.e. the top half of the complex plane, not including the negative reals. – Simply Beautiful Art Dec 07 '16 at 01:42
  • I think I get it. Thank you... ^^ – user379677 Dec 07 '16 at 01:44
  • @user379677 Glad to help. :) – Simply Beautiful Art Dec 07 '16 at 01:46
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If both are negative, $$\sqrt{-a} \times \sqrt{-b} = \sqrt{-1}\times \sqrt{-1} \times \sqrt{a} \times \sqrt{b}$$ $$=i^2 \times \sqrt{ab} = -\sqrt{ab}$$ (the rule is not applicable here)

If one of them is negative, $$\sqrt{-a}\times \sqrt{b} = \sqrt{-1} \times \sqrt{ab} = \sqrt{-ab}$$ (the rule is applicable here)

If you have non-negative real numbers, this rule is easily applicable. If negative numbers come, then you have complex numbers involved. For them, this rule behaves differently. As you have seen above, if both of them are negative, the result is $-\sqrt{ab}$ and not just $\sqrt{ab}$. That's why $1$ is not equal to $\sqrt{(-1)}\times\sqrt{(-1)}$(two complex numbers multiplying) becuase now you have complex numbers involved and as we said, for them rules are different!

So, when you bring negative inside the square root, complex numbers arise. $\sqrt{(-1)}\times\sqrt{(-1)}$ is $-1$. But, $\sqrt{(-1) \times (-1)}$ is simply $1$. $$\sqrt{(-1)}\times\sqrt{(-1)} \neq \sqrt{(-1) \times (-1)}$$

Simran
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  • Could you elaborate your second step? When you say $\sqrt{-a}\times \sqrt{b} = \sqrt{-1} \times \sqrt{ab}$, do you mean $\sqrt{-a}=\sqrt{-1}\sqrt{a}$? – IamKnull Sep 14 '19 at 05:55
  • Yes, it means that. Then you have $\sqrt{-1}\sqrt{a}\sqrt{b} = \sqrt{-1}\sqrt{ab} = \sqrt{-ab}$ – Simran Sep 14 '19 at 22:30
  • But that's what you are proving. If you are using $\sqrt{-1\times ab}=\sqrt{-1}\sqrt{ab}$ then you are already using the result that given equality is true if one of them is negative. – IamKnull Sep 15 '19 at 03:19
  • That's what I did for the first part as well. $\sqrt{-a} = \sqrt{-1}\sqrt{a}$. If you multiply $\sqrt{-a}$ and $\sqrt{-b}$ together, you will get $- \sqrt{ab}$. You simply can't multiply the inside numbers ($-a$ and $-b$). But for, $\sqrt{-1}$ and $\sqrt{a}$, you can do that because $\sqrt{-a} \times \sqrt{b} = \sqrt{-ab}$. I'm not proving that, it's just to make things formal. As, $\sqrt{-a} = \sqrt{-1}\sqrt{a}$, I can state that. – Simran Sep 15 '19 at 03:50
  • Your conclusion is, of course, correct but you can't write $\sqrt{-a}=\sqrt{-1}\sqrt{a}$ because as I've mentioned this assumes the result is already true before actually proving it. – IamKnull Sep 15 '19 at 03:59
  • Lets say I'm given $\sqrt{-a} \times \sqrt{-b}$. If I say $\sqrt{ab}$ is the answer then I am wrong as breaking them, I will get $\sqrt{a}\times \sqrt{b}$ and not the original values. And now, if I break $\sqrt{-ab}$, I will get $\sqrt{-a}\times \sqrt{b}$, which contains my original values. It means, the step that I used, $\sqrt{-a} = \sqrt{-1}\sqrt{a}$ is correct! I can break them down!!!! – Simran Sep 15 '19 at 03:59