The question is:

Evaluate in simplest form:$\sqrt {2013+2012 \sqrt {2013+2012 \sqrt {2013+2012 \sqrt {...} } } }$

Supposing let "x" be $\sqrt {2013+2012 \sqrt {2013+2012 \sqrt {2013+2012 \sqrt {...} } } }$

Then $x^2=2013+2012 \sqrt {2013+2012 \sqrt {2013+2012 \sqrt {...} } }$

Noticing the "x" still repeats in the equation of $x^2$,

I can conclude $x^2=2013+2012x$

Where solving this equation,



$(x+1)=0 OR (x-2013)=0$

$x=-1 (Rejected) OR x=2013$

So $\sqrt {2013+2012 \sqrt {2013+2012 \sqrt {2013+2012 \sqrt {...} } } }=2013$

But the problem with this answer is that,it is a whole number.The answer can't go to a whole number of this is an infinite series of square roots where $\sqrt {x}$ is unequal to a whole number.

Basically saying as long as $\sqrt {2013+2012}$ is not a whole number,then the infinite square root CANNOT be a whole number.

But the problem is,if the answer is not the whole number?How do I solve this?I don't see any other way according to my 13 year old brain!(Solution that are high and difficult are welcomed and wanted!)

  • 4
    $\begingroup$ The problem is that you don't want $\sqrt{2013+2012}$ to be an integer, but $\sqrt{2013+2012x}=\sqrt{2013+2012\cdot 2013}=\sqrt{2013^2}=2013$, which indeed is. $\endgroup$ – mojambo Apr 5 '15 at 12:09
  • $\begingroup$ @dclark but you cannot subsitute "x" that way,because I first stated "x" as that series of square roots,not a simple value of 2013.Before I even solve "x",I cannot say that "x"=2013 that way.Looks like I'm just confused about what I am talking now $\endgroup$ – ministic2001 Apr 5 '15 at 12:15
  • $\begingroup$ @dclark ahh now I remember what I had to say.But the thing is,infinite square root can only get closer to an integer eventually,but cannot reach the integer.No matter how much I square root,even for 1 billion times,I still cannot get an integer.But yet my answer is a integer? $\endgroup$ – ministic2001 Apr 5 '15 at 12:20
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    $\begingroup$ Indeed you may, and you've addressed this in your post when you say "the $x$ still repeats". What about this? $$x=\sqrt {2013+2012 \left(\sqrt {2013+2012 \sqrt {2013+2012 \sqrt {...} }}\right) }=\sqrt{2013+2012 x}$$ $\endgroup$ – mojambo Apr 5 '15 at 12:21
  • $\begingroup$ @dclark seems satisfying enough.Thanks $\endgroup$ – ministic2001 Apr 5 '15 at 12:22

When you say that: $$\sqrt {2013+2012 \sqrt {2013+2012 \sqrt {2013+2012 \sqrt {\cdots} } } }=2013$$

It's just like saying that: $$a_n=\underbrace{\sqrt {2013+2012 \sqrt {\cdots \cdots \sqrt {2013+\sqrt {2012}}} } }_{ n \text{ times }} \\ \lim_{n\to \infty} a_n=2013 $$

Now your conclusion is not correct for instance we have : $$\lim_{n\to \infty}\sqrt 2(\frac{1}{2})^n=0$$

But $\sqrt 2(\frac{1}{2})^n$ is never an integer ,

To see this in your terms I will ask you what is the value of : $$\sqrt 2 (\frac{1}{2})\cdot(\frac{1}{2})\cdot(\frac{1}{2})\cdot (\frac{1}{2})\cdots\cdots \cdots \tag 1$$ you will say that if $x$ is it's value than $x=\frac{1}{2}\cdot x$ and hence $x=0$, but $x=0$ is an integer and the numbers in $(1)$ are never integers! I have to say that this happens whole time and there is no way of finding out if the limit is an integer just looking to the sequence if it's takes integer or rational values.

  • $\begingroup$ well explained.I cannot argue more. $\endgroup$ – ministic2001 Apr 5 '15 at 13:37

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