Interesting!

Algebra Level 1

$x=\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { ...... } } } } } }$

Find $x$ . You must use the positive root as your answer.

The answer is 2.

4 solutions

Akash Patalwanshi
May 27, 2016

$\large x= \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}$

$\large → x^2 = 2 + x$

$\large → x = 2, -1$

Considering the positive value of x as we know, the square root of any number not result in negative number. So we get $\large \boxed{x= 2}$

Can we treat $\boxed {x=2}$ as a principal solution even if the two solutions don't have same absolute value?

Rakshit Pandey - 5 years ago

Nice observation.

akash patalwanshi - 5 years ago

I believe that it should be mentioned in the question that only positive root must be used as correct answer .

@Kok Hao Kindly fix this detail in your question.

Rakshit Pandey - 5 years ago

@Rakshit Pandey – Ok, I did it. Thanks for your suggestion!

Kok Hao - 5 years ago

@Kok Hao – The square root of a positive is the positive real root.

The reason why the algebraic solution has multiple root is that you're making the assumption of:

If the limit exists and is finite, and we manipulate it as such, we can find a necessary condition that it must satisfy.

However, with the algebraic approach, you have these three issues:
1. Why is the limit finite? Note that $\infty ^2 = \infty + 2$ and so it's algebraically possible for the limit to be infinite.
2. When we square an equation, we risk introducing extraneous roots. How do we know for certain that 2 is not an extraneous root?
3. We only have a necessary condition. Why it (part of) it sufficient?

Instead, you should be using the calculus approach, where you treat the expression as the limit of:

$\sqrt{2}, \sqrt{ 2 + \sqrt{2} } , \sqrt{2 + \sqrt{2 + \sqrt{2} } } , \ldots$

You can use a epsilon delta proof to show that the limit exists and is equal to 2.

Calvin Lin Staff - 5 years ago

@Calvin Lin – Thanks @Calvin Lin , this makes it quite clear.

Rakshit Pandey - 5 years ago

@Calvin Lin – This is way too hard for me!

Firstly, I don't know what you are saying. Secondly, can you teach me, please?

Kok Hao - 5 years ago

@Kok Hao – Read the wikis! They are a great source of knowledge. If you don't understand the epsilon delta wiki, then read up on the more basic concepts to build up your knowledge.

Calvin Lin Staff - 5 years ago

@Rakshit Pandey – It is usual. Since there is difference between "Solution to $x^n = a$ and $n$ th root of $a$ . In order to keep the square root a function so that we should get one and only one output for any valid input, we consider the positive output. For example, thats why, $\sqrt{4} = 2$ and not $-2$ and not $±2$ . Because We want the square root to be a function, so we want a single answer, and we agree to give as an answer the nonnegative solution to $x^2 =4$ .

akash patalwanshi - 5 years ago

@Akash Patalwanshi – I'm sorry but I don't agree with you on this. I understand that Square Root Function exists only in the $R^+ \rightarrow R^+$ mapping but I don't see what that has got to do in this situation. The issue here is that you formed a quadratic equation, which gives two real solutions. One is positive and one is negative. There's no logical reason to prefer positive solution over the negative one. There are no physical aspects to this problem so we cannot argue that negative solution don't make sense.

Quoting from your comment, "Because We want the square root to be a function, so we want a single answer, and $\ldots$ " can you please explain Why do we want the square root to be a function here? What goes wrong if the square root in the problem is not considered to be a function but just an operation? I don't think it matters. Both answers are equally acceptable in my opinion.

Rakshit Pandey - 5 years ago

Well, we observe that the expression only has positive numbers, and it only adds terms. Therefore, the answer will be a positive value. So yes, we can say that the value will definitely be positive

Hung Woei Neoh - 5 years ago

I would like to hear your views on this:

1 + 2 + 3 + 4 + ⋯

This particular case consists of a series that has only positive natural numbers as its terms, and the terms are always added to each other. The result is $\frac{-1}{12}$ .

We cannot treat our intuition as a valid basis to justify correctness of a solution to a mathematical problem.

Rakshit Pandey - 5 years ago

@Rakshit Pandey – This is too high level for me.......

Anyway, there is one more reason for what I said that I forgot to mention:

When we use the radical $\sqrt{}$ sign, we refer to the principal square root, which asks for the positive square root.

Negative numbers do not have real square roots, and we certainly do not get negative numbers from a square root

This is why, in the quadratic formula, we put a $\pm$ in front of the discriminant, to note that there are two values

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – If you read the previous comments that I wrote, you'll find that I already discussed the Principal Square Root issue with @akash patalwanshi . We cannot treat the positive solution as principal root in this case because the positive and the negative root don't have the same absolute value. The concept of Principal Root is applicable when the two roots of the quadratic equation have same absolute values but opposite signs.

I was reading about mathematical fallacies a few moments ago. As it turns out, this solution is mathematically false. Ironically, I used exactly same solution to find the answer for this problem.

Here's why this solution is wrong: Squaring both sides of an Equation

When we square both sides of an equation, we are unknowingly introducing new roots, which is $-1$ in this particular case.

Rakshit Pandey - 5 years ago

@Rakshit Pandey – But then, the roots we obtain, from the quadratic formula, is stated as:

$x=\dfrac{1 \pm \sqrt{9}}{2}$

And from what I read here and here , we always take the positive value for such nested radicals

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – I don't know what to think now. I'm totally confused.

Do you know anyone who can clear this up? Please tag them in this comment thread.

Rakshit Pandey - 5 years ago

@Rakshit Pandey – I'm not sure......ehhh, let's see......

@Calvin Lin @Pi Han Goh @Sandeep Bhardwaj @Ashish Siva @Chew-Seong Cheong

Err......sorry to disturb you guys, but can you help us with this? Thanks in advance

Hung Woei Neoh - 5 years ago

@Hung Woei Neoh – Yes? I will try to help what help do you want?

Ashish Menon - 5 years ago

@Hung Woei Neoh – Yes we can only take positive roots.

Ashish Menon - 5 years ago

@Ashish Menon – Could you please explain why?

Rakshit Pandey - 5 years ago

@Rakshit Pandey – Yes first of all, $\sqrt{\phantom{0}}$ indicates principal root, so we can take only positive value. Secondly, if we substitute the negative value in $x$ , the LHS amd RHS would not be equal.

Ashish Menon - 5 years ago

@Ashish Menon – $x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 +\ldots}}}}$
$\Rightarrow x = \sqrt{2 + x}$
Squaring both sides,
$\Rightarrow x^2 = 2 + x$
Putting $x = -1$ ,
$\Rightarrow 1 = 2-1$
$\Rightarrow 1 = 1$

L.H.S. = R.H.S.

Rakshit Pandey - 5 years ago

@Rakshit Pandey – Look the step $x = \sqrt{2 + x}$ .
$x$ can be anything but $\sqrt{2 + x}$ is non-negative since $\sqrt{\phantom{0}}$ is principal root. Now $x = -1$ is negative. How can a negative number be equal to a non-negative number? The reason why this satisfies is that $\sqrt{-1}$ os an imaginary numbee and so it does not follow the usual rules of arithmetic calculation.

Ashish Menon - 5 years ago

Kok Hao
May 26, 2016

$2=\sqrt { 4 } =\sqrt { 2+2 } =\sqrt { 2+\sqrt { 4 } } =\sqrt { 2+\sqrt { 2+2 } } =\sqrt { 2+\sqrt { 2+\sqrt { 4 } } } =\sqrt { 2+\sqrt { 2+\sqrt { 2+2 } } } =\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 4 } } } } =\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2+2 } } } }$

As you might see, this pattern continues forever, thus the answer is $\boxed{2}$

Different approach, nice!

Mahdi Raza - 1 year ago

Mahdi Raza
Jun 12, 2020

$x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{\ldots}}}}}}$

$x = \sqrt{2 + x}$

$x^2 = 2 + x$

$\boxed{x = 2}$

Tanmoy Sarkar
May 26, 2016

Or in other way, as adding sqrt 2 continues, the value gets smaller, so we can only count the first and the second 2's and neglect others.

Interesting!

The answer is 2.

4 solutions

0 pending reports