Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given: $\sqrt{110-n} = n$

It follows that:

$110 - n = n^2$

$n^2 + n - 110 = 0$

$(n + 11)(n - 10)=0$

If $n = 10$, $\sqrt{110 - 10} = \sqrt{100} = 10$, it checks out.

If $n = -11$, $\sqrt{110 - (-11)} = \sqrt{121} = 11$, it doesn't

However!

I thought that $\sqrt{x}$ gives you two possible numbers, $+\sqrt{x}$ and $-\sqrt{x}$, so in the example above, $\sqrt{121} = -11$ is a valid root.

Can someone explain what's going on here? Why aren't they both valid?

share|improve this question
    
Note that your sentence: "I thought that $\sqrt x$ gives you two possible numbers $+\sqrt x$ and $-\sqrt x $", is not correct. $\sqrt x $ has 2 solutions $(+y)$ and (or) $(-y)$ where $y^2 = x$. –  Emmad Kareem Sep 12 '12 at 4:44

3 Answers 3

up vote 11 down vote accepted

When you squared, you introduced a second solution. Consider a much simpler problem: $\sqrt 4=n$. This has the unique solution $n=2$. If you square both sides, however, you get $4=n^2$, which has two solutions, $n=2$ and $n=-2$. The second is not a solution of $\sqrt4=n$, because the square root symbol by convention means the non-negative square root: when you use it, you’re implicitly choosing one of the solutions to the squared equation. Similarly, there are two integers $n$ with the property that $110-n=n^2$, $10$ and $-11$, but when you write $n=\sqrt{110-n}$, you’re forcing $n$ to be non-negative and thereby eliminating one of the two solutions to $110-n=n^2$.

share|improve this answer
    
The convention of it meaning a non-negative was exactly what I needed to hear, thank you Brian (I will accept when stackexchange lets me) –  Henry Sep 12 '12 at 3:38
    
@Henry: Actually you correctly wrote yourself "It follws that:" and not "This is equivalent to:" when you squared. –  Hagen von Eitzen Sep 12 '12 at 5:18

It just has to do with the fact that $(-n)^2 = (n)^2$ does not imply that $-n = n$.

The $\sqrt{n}$ can only be positive by definition.

share|improve this answer
    
+1, interesting first sentence. –  Emmad Kareem Sep 12 '12 at 4:18

You can pinpoint the error by working backwards, finding the first equation which doesn't hold for the value $\rm\:n = -11.\:$ You'll find that all equations except the first hold true for $\rm\:n= -11,\:$ so your inference from the first to the second equation must be erroneous. The problem is that you applied to the first equation a non $1$-to-$1$ operation (squaring), which generally enlarges its solution set, introducing extraneous solutions.

This method of isolating errors works quite generally. It is especially useful when debugging proofs on abstract objects, since the error may become simpler to locate after specializing to more concrete objects. Above, evaluating all equations at $\rm\,n=-11\,$ has the effect of specializing the arithmetic of polynomial and algebraic functions (in $\rm\,n)\,$ to the simpler arithmetic of integers, making the error much easier to locate. See here for further discussion.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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