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1955 AHMSE Problem 20 asks when $\sqrt{25 - t^2} + 5 =0.$

I know square root of real numbers cannot be negative. So t cannot be real.

But I don't know whether imaginary numbers' square root can be negative or not. I think square roots can never be negative. Also, I don't think we can classify imaginary numbers as positive or negative.

Can square roots of imaginary numbers be negative?

saulspatz
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Ram Keswani
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    You are right that imaginary numbers cannot be classified as positives or negatives. As for your other questions, have you looked into a pre calculus book lately? – imranfat Aug 01 '19 at 16:19
  • Imaginary numbers are quite simple to understand. Just let $i$ be a number such that $i^2=-1$ and operate with them is a normal fashion (for instance: $(3+4i) + (1+2i) = 4+6i$; similarly, $(3+4i) \cdot (1+2i) = 3+4i+6i+8i^2=3-8+10i=-5+10i$ – David Aug 01 '19 at 16:23
  • @calicus Imoved my comment to answer. Positive only is a convention, not a mathematical truth. – herb steinberg Aug 01 '19 at 16:33
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    It is true that $x^2=25⇒x_{1,2}=±5$. But $25$ is just $\sqrt{5}$ – callculus42 Aug 01 '19 at 17:05
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    I agree with callculus, apart from the typo – Adam Rubinson Aug 01 '19 at 17:10

7 Answers7

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Read carefully the first paragraph of what wikipedia has to say about square roots, noting the difference between the definitions of "square root" and "principal square root":

https://en.wikipedia.org/wiki/Square_root

$\sqrt{x} \ $ doesn't mean "square root of x". $\sqrt{x} \ $ means, by definition, $\mathit{the \ principal \ square \ root \ }$ of x, which means, $\mathit{the \ square \ root \ of \ x \ with \ positive \ real \ part }$ . Note that this definition can be applied to complex numbers also, e.g. for the following A-Level question:

enter image description here

if we did not require the principal square root to have a positive real part, then the answer would be: Solutions are 7-3i and -7+3i, whereas given the definition of principal square root, which is standard convention, we must throw away -7+3i, but keep 7-3i as a solution.

4 and −4 are square roots of 16.

However, $\sqrt{16} = 4$, $\mathit{not} \pm 4$.

For further evidence in favour of my understanding of the conventional meaning of $\sqrt{} \ \ $, In Rudin's PMA, Theorem 1.21 states:

For every real $x > 0$ and every integer $n > 0$ there is one and only one $\mathit{positive}$ real y such that $y^{n} = x$. The number y is written $\sqrt[n]{x} \ $ or $ \ x^{1/n}$.

Funnily enough, as pointed out by LegionMammal978 in the comments, Rudin left out the fact that y is positive in his book.

Going back to the original questions,

"I know square root of real numbers cannot be negative. So t cannot be real."

I agree.

"But I don't know whether imaginary numbers' square root can be negative or not."

All complex numbers have two square roots (though they may be repeats). An imaginary number like $9i$ will be have two complex square roots. This is true for complex numbers also- they too have two complex square roots.

I actually think what you meant to ask here was, "can we square root a $\mathit{negative}$ number?" In the complex realm, yes, but we have to be careful if we are calculating the principal square root and are using the $\sqrt{} \ \ $ sign. see:

https://en.wikipedia.org/wiki/Imaginary_unit#Proper_use

for why and when we have to be careful.

-4 has two complex square roots, namely $2i$ and $-2i$. Since the real part of both of these square roots is 0, either both or neither are considered principal square roots, depending on if you define the principal square root as having non-negative real part, or strictly positive real part. To be honest I'm not sure which one is "accepted/standard convention", but if I had to guess, I would say the latter option is correct i.e. both are principal square roots of -4. I'm open to confirming whether or not my suspicions of the "accepted/standard convention" are correct or incorrect based on as long as that someone has sources to back up their claim.

If you ask if a number is negative, then in this context you are implicitly implying the number is real and negative, or integer and negative etc, because only real/integer/etc numbers can be negative. Complex numbers with imaginary part not equal to 0 cannot be negative in and of themselves, however, their $\mathit{ \ real \ / \ imaginary \ parts \ }$ can be negative.

"I think square roots can never be negative."

Square roots of real numbers can be negative. -4 is a square root of 16. But remember: $\sqrt{x}$ means the $\mathit{principal}$ square root of x, not just the square root of x. You would be correct to say, "the principal square root of a real number cannot be negative"

"Also, I don't think we can classify imaginary numbers as positive or negative."

Correct. But like I said you $\mathit{can}$ talk about their $\mathit{ \ real \ / \ imaginary \ parts \ }$

Adam Rubinson
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  • I don't get where the claim of "one and only one real $y$" comes from. Just take $x=4$, $n=2$, and $2^2=(-2)^2=4$, with $y=\pm2$. Did you perhaps mean "one and only one real $y>0$"? – LegionMammal978 Aug 02 '19 at 00:46
  • Yes you are right, and that's what I meant to write. I just copied what Rudin wrote in his book, forgetting Rudin made this error. I'll edit my answer. – Adam Rubinson Aug 02 '19 at 11:01
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Did someone read the original question? Since $$\sqrt{25-t^2}=-5\Rightarrow 25-t^2=25\iff t=0$$ and zero is no solution to the original equation, option (a) is the right answer.

Michael Hoppe
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  • So, $t=2\implies t^2=4$ and therefore $t=-2$ is a solution? – Asaf Karagila Aug 02 '19 at 06:27
  • $t=2\Rightarrow t^2=4\Rightarrow t=-2$ but not reverse. A squared equation might have a larger solution set as the original one. But why are you asking? – Michael Hoppe Aug 02 '19 at 10:16
  • It wasn't a question, it was a rhetorical device. You've squared the equation, therefore increased the pool of possible solutions, and then used one of them without verifying it against the original equation. And indeed putting $t=0$ gives $5=\sqrt{25}=-5$, which is not generally true. – Asaf Karagila Aug 02 '19 at 10:17
  • I don't get your point.The squared equation has only a single solution, namely $t=0$. So I've verified all solutions, namely $t=0$, against the original equation. – Michael Hoppe Aug 02 '19 at 11:40
  • You've added a solution, $t=0$, to an equation without a solution, and then checked it against the squared equation. I am sorry, but with my years doing mathematics, I still don't see how $5=-5$. – Asaf Karagila Aug 02 '19 at 11:55
  • Excuse my dumbness, please, but I wrote: "zero is no solution to the original [i.e., the unsquared] equation." – Michael Hoppe Aug 02 '19 at 12:09
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We get from your equation $$\sqrt{25-t^2}=-5$$ since $$\sqrt{25-t^2}\geq 0$$ for $$25-t^2\geq 0$$ so we get no solution.

Dr. Sonnhard Graubner
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All numbers have two square roots. By convention the positive square root is assumed to be the square root for positive reals. However in your example $t=0$ is O.K., since $\sqrt{25}=\pm 5$.

herb steinberg
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You can use the third binomial formula to see that $t=0$ is no solution.

$$\sqrt{25 - t^2} + 5 =0$$

$$\sqrt{25 - t^2} =- 5 $$

$$\sqrt{5 - t}\cdot \sqrt{5+t} =- 5 $$

$$\sqrt{5}\cdot \sqrt{5} =- 5 $$

$$5 =- 5 \qquad \color{red}{\times}$$

callculus42
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  • "third binomial formula"? What is that supposed to be? Even if I guess it is the difference-of-squares formula, what's the point of using it? You could just substitute $0$ in at the beginning and skip from line 1 to line 4. – rschwieb Aug 01 '19 at 17:27
  • I just wanted to make clear that $t=0$ is no solution. That´s the way I´ve chosen. – callculus42 Aug 01 '19 at 17:30
  • I know, I'm just pointing out you included an unnecessarily complicated step when doing that. Unless you have some qualms about what $25-0^2$ is or something. – rschwieb Aug 01 '19 at 17:31
  • But it doesn´t seem so clear that $t=0$ is no solution if you look at the comments and answers here. – callculus42 Aug 01 '19 at 17:33
  • I think we're both very clear on all the other comments and questions, but it seems as if you didn't read the first comment in this thread at all... ;) I didn't say anything was wrong with your choice of strategy, I'm just saying you added complication that isn't necessary. – rschwieb Aug 01 '19 at 17:34
  • The comments and answers are very different with different results. Yes I haven´t included the first comment in my answer. Why should I? – callculus42 Aug 01 '19 at 17:36
  • OK, well if you aren't going to read it and understand it, I'll try one more time here: your answer could easily read: "It's easy to see $0$ is not a solution. After substituting $0$ (into the first line) $\sqrt{25}=-5\implies 5=-5$ Contradiction, so $0$ is not a solution." Instead you made spurious reference to difference-of squares and added a few unnecessary lines to your solution. My point is you could remove the useless lines. – rschwieb Aug 01 '19 at 17:47
  • @rschwieb Maybe it would be better you would make your complaints to answers which lead to wrong results. Why you don´t do that? – callculus42 Aug 01 '19 at 17:57
  • OK, I get it. You don't want to make easy and significant improvements to your answer. I will avoid helping you improve your answers in the future as well. Have a great day. I see it doesn't matter much since Michael Hoppe has already posted the simplified version of your answer. – rschwieb Aug 01 '19 at 18:04
  • Thanks, a great day for you as well. – callculus42 Aug 01 '19 at 18:06
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Let's begin with $\sqrt{25+t^2}+5=0\implies \sqrt{25+t^2}=-5.$ If we square both sides, we get $25+t^2=25\implies t^2=25-25=0\implies t=0.$ Zero is real so $t$ is real.

poetasis
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  • 'squaring' is not an equivalent term transformation. – callculus42 Aug 01 '19 at 17:02
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    However you figure it, the only value of $t$ that works is zero and zero is real. No other values under the radical will give you $\pm 5$. – poetasis Aug 01 '19 at 17:04
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    This issue here seems to be some people believe $\sqrt{25} = 5$ by definition, whereas others say $\sqrt{25} = \pm 5$, i.e. +5 or -5. Going by Rudin's PMA and wikipedia, the former seems to be the accepted convention/definition. – Adam Rubinson Aug 01 '19 at 17:08
  • @AdamRubinson In general the value of the number $\sqrt{a}$ has to be unique, otherwise we would run into a chaos. – callculus42 Aug 01 '19 at 17:18
  • In formulas I have developed for finding Pythagorean triples, given only one value, I have sometimes found that there are $2$ triples with the desired attribute(s), one for $+$ and one for $-$ out of the square root. – poetasis Aug 01 '19 at 17:27
  • @callculus The mapping $x\mapsto x^2$ is a perfectly well-defined function, and you can square both sides of an equation. The problematic one is taking the square root of both sides. – rschwieb Aug 01 '19 at 17:29
  • @rschwieb Try to solve the following equation by squaring: $$\sqrt{2-x}=x$$ See the answers as well here – callculus42 Aug 01 '19 at 17:54
  • @callculus OK, if you mean "equivalent term transformation" to mean that the solutions of the former are exactly the solutions of the latter, then I agree with you. I wanted to dispute the idea that one shouldn't square both sides, because it is a perfectly valid thing to do. The mistake in the work is not in squaring both sides but in assuming all solutions to the result are automatically all solutions to the original expression. – rschwieb Aug 01 '19 at 18:00
  • @rschwieb I'm really fascinated that you don´t care about answers which lead to wrong results. – callculus42 Aug 01 '19 at 18:03
  • @callculus I'm not sure which posts you're referring to, but my feedback on the posts with obvious flaws comes in the form of upvoting comments pointing out the mistakes, and downvotes on the solution itself. So, it sounds like you're operating under a mistaken assumption. Anyhow I think the question of what you meant is now apparently resolved for future readers and me. – rschwieb Aug 01 '19 at 18:07
  • @rschwieb I think it would be better to point out that you don´t agree with the solution as well-only you have time. At the moment it looks like that both kinds of answer are right. This is causing confusion for other users on MSE. – callculus42 Aug 01 '19 at 18:14
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    @poetasis See this wiki article – rschwieb Aug 01 '19 at 18:48
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Can square roots of imaginary numbers be negative?

No -- the only thing that can conceivably have a negative number as a square root is the square of that negative number, which is positive real rather than imaginary.

As other answers already point out, the negative number can be considered a square root of its positive square, but is not the principal square root that is usually notated with the $\sqrt{\phantom{z}}$ sign.

The square root of a purely imaginary number is a complex number on one of the lines at 45° angles to the real and imaginary axes. For example, $$ \sqrt{2i} = 1 + i $$

hmakholm left over Monica
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