# Infinite Root of 3!

$$\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!....}}}}}}}$$

How the heck does this business equal 6, I understood the algebraic approach -

$\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!....}}}}}}}$ = y

$\sqrt{3!y}$ = y

$6y$ = $y^{2}$

$y$ = $6$

But how is $\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{6....}}}}}}}$ = 6, in a practical sense, please help! (3! = 1x2x3 = 6)

# (I got the answer from few people, so now you can post whatever you want in this note.) Note by Percy Jackson
10 months ago

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You can view the original expression as $6^{\frac12}\times 6^{\frac14}\times 6^{\frac18}\times...=6^{\frac12+\frac14+\frac18+...}=6^1$

- 10 months ago

$\sqrt6\approx2.449$

$\sqrt{6\sqrt6}\approx3.83$

$\sqrt{\sqrt{6\sqrt6}}\approx4.76$

## Plug it in a calculator, the answer will tend toward 6 as you go on.

- 10 months ago

- 10 months ago

Mahdi's right ... but the equation might be better written as:

$\sqrt{6\sqrt{6\sqrt{6\sqrt{...}}}} = N \neq 0$

Square both sides:

$6\sqrt{6\sqrt{6\sqrt{6\sqrt{...}}}} = N^2$

Divide by the original equation (which isn't equal to 0):

$6 = N$

- 10 months ago

What the heck! 100 notifications in 1 night and 60% of them from this note! I'm unsubscribing from the comments. I appreciate everybody's work in proving and disproving different formulas. I've learnt a lot. Thanks everyone! @Siddharth Chakravarty, @Mahdi Raza, @David Hairston, @X X, @Suchet Sadekar, @Pi Han Goh

- 10 months ago

It's the same thing

\begin{aligned} x &= \sqrt{6\sqrt{6\sqrt{\ldots}}} \ne 0 \\ x &= \sqrt{6x} \ne 0 \\ x^2 &= 6x \ne 0 \\ x &= \boxed{6} \end{aligned}

Credit to @David Hairston and @Pi Han Goh for pointing out that $x=0$ is not a valid solution. Because otherwise for any number instead of 6, let's say 5, the two expressions equal to zero, and hence we would reach a point where we could say $6=5$. But that's ridiculous. Plus, the value of 0 comes from an extraneous root as mentioned by David again.

- 10 months ago

x = 0 doesn't work, but x = 6 does. i understood it now, i was getting the problem wrong, thanks @Mahdi Raza !

- 10 months ago

x=0 works, see below:

0=0........(1)

0=$\sqrt { 6\times 0 }$.......(2)

Substituting (2) in (1)

0=$\sqrt { 6\times 0 }$.

Again substituting (2) for the zero inside the root.

0=$\sqrt { 6\sqrt { 6\times 0 } }$

Continuing to substitute (2) in every zero inside the root we get,

0=$\sqrt { 6\sqrt { 6\sqrt { 6\sqrt { 6\sqrt { ... } } } } }$

- 10 months ago

Oh, ok Thanks @Siddharth Chakravarty, I understand now!

- 10 months ago

0 is not a valid solution. Read up cobweb plot.

- 10 months ago

But it does work. I don't understand if it is still necessary to check something else If I have shown and expanded it above. Would still want to understand what you exactly mean!

- 10 months ago

Like I said, read up cobweb plot.

- 10 months ago

But what is the flaw in the expansion then?

- 10 months ago

0 is not a solution to the original infinite product.

0 is an extraneous solution introduced when the original equation was squared.

You have to be mindful of extraneous solutions when you raise the power of a relationship, otherwise strange things happen.

Consider a simple example that proves that 1 = –1:

x = 1

x•x = 1•1

x = –1

Ugh ....

- 10 months ago

Can I know the flaw than in the expansion I did above? And in the example, you have given x=1 and later when you find the root of 1, you have 2 choices either to choose 1 or -1 it does not mean x necessarily has to be -1, and it is already given x=1 so x cannot be -1.

- 10 months ago

different expansions ... $\sqrt{6\sqrt{6\sqrt{...}}} \neq \sqrt{0\sqrt{0\sqrt{...}}}$

otherwise you could easily show that 6 = 5 because they both are equal to the 0 expansion.

- 10 months ago

Well, well there are 2 values as it is quadratic, so you took the wrong value. So you misinterpreted this, your approach is same as saying x^2 has two roots -1 and 1 so -1=1 which is not right.

- 10 months ago

The original equation was not quadratic. You introduced a quadratic relationship to solve the problem. However, the quadratic relationship had an extraneous solution, unintended by the original equation. Please know that the root of a non-zero number cannot be zero.

Yes $0 = \sqrt{0}$ but that was not the original expansion.

- 10 months ago

No, and @Mahdi Raza you also followed the herd without understanding. The ridiculous result 6=5 comes because the continued fraction is never-ending, so it could either be expanded by 6 or 0 and 0 is thus not an extraneous solution. So you assumed that the equation which is expanded by 0 is the same as the equation expanded by 6 and thus u got 6=0, dividing by 6 and then adding 5 gives you ur wrong result 6=5. I would say you the equation which is expanded by 6 is an extraneous solution in this equating of both the equations i.e the step of 6=0 and thus led to 6=5. You cannot just do rubbish in maths, its logic. The thing you can see that there is no flaw in the expansion by 0 so it is correct.

- 10 months ago

Well do you agree or not that $\sqrt{6\sqrt{x}} > 0$. If so, you understood what they are trying to say and why 0 is an absurd solution

- 10 months ago

Check this below expression:

0=$\sqrt{6\times 0}$

0=$\sqrt{6\sqrt{6\times0}}$

0=$\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{...}}}}}$ ( And this can be continued similarly, and let this be {1})

Now see this below expansion:

6=$\sqrt{6\times6}$

6=$\sqrt{6\sqrt{6\times6}}$

0=$\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{...}}}}}$ ( This could also be continued similarly, let this be {2})

Now why did 6=5 come, because you took {1} and {2} equal to each other, but they are not because they are both continued radicals of 6 but are expanded by 2 different numbers, which is either 0 or 6. So thus both the solutions turn out to be valid, because that continued radical could either be expanded by 0 or 6 as I showed, and hence 2 solutions.

But wait, even -6 can also be expanded similarly because 1 root of √36 is -6 but we didn't get it as an answer, why?

Look at the expansion of -6

-6=$\sqrt{6\times6}$

-6=$\sqrt{6\sqrt{6\times6}}$

( This can be done as 1 root is 36 also)

-6=$\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{...}}}}}$

But now let's see how you found out the values for 6 and 0,

x=$\sqrt{6\sqrt{6\sqrt{6\sqrt{6\sqrt{...}}}}}$

x=$\sqrt{6x}$

In this above step, you assumed that the inside radical has the same value as the whole outside radical, but in the case of -6 it is not, but they are the negative of each other, so we can write x as -x inside, which is:

x=$\sqrt{-6x}$

x^2=-6x

Which gives the 2 values, -6 and 0 again which we wanted and is right. Do you all get it? But yes, I know the √ is generally meant for positive roots (I'm talking for -6)

- 10 months ago

Firstly $\sqrt{x} > 0$. For example $\sqrt{4}$ is always 2 and not -2. Secondly consider this:

$0 = \sqrt{6\sqrt{6\sqrt{6\sqrt{6\ldots}}}}$ $0 = \sqrt{5\sqrt{5\sqrt{5\sqrt{5 \ldots}}}}$ $\implies 5 = 6 ?$

- 10 months ago

Ugh, there are 2 values @Mahdi Raza you are doing the same mistake, how many times should I tell you all, both are continued fractions BUT ONE IS EXPANDED BY 0 AND OTHER BY 6 OR 5 so they are not equal. I understand what you mean by $\sqrt{x}$ > 0 because it normally said as the positive root, but 0 is a valid solution if and only the fraction is continued in the Question. Don't approach like $\sqrt{6}$ and then $\sqrt{6\sqrt{6}}$ goes on approaching 6 because this is wrong as this is a finite fraction.

- 10 months ago

I still don't understand why don't you edit your solution. Or you delete it. What is wrong is wrong, you guys didn't understand what it means, even after me explaining the same thing lot. You get 6=5 because you substituted wrong values because true things cannot lead to false things.

- 10 months ago

But you are making the radical finite and in the question you are finding the value of a infinite continued radical.

- 10 months ago

From my example, x = 1, which has the solution ... x = 1.

This relationship is also a solution to the quadratic equation x•x = 1.

The quadratic equation has an extraneous solution x = –1, not intended by the original equation.

Solutions to the quadratic equation are not necessarily solutions to the original equation, x = 1.

You can use the quadratic equation to solve the original equation but you have to be mindful of the extraneous solutions introduced by raising the power of the relationship.

- 10 months ago