\(\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!\sqrt{3!....}}}}}}}\)

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)

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## Comments

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

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$\sqrt6\approx2.449$

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

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

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

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Thanks @Suchet Sadekar

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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):

$6 = N$

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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

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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.

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

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

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@Siddharth Chakravarty, I understand now!

Oh, ok ThanksLog in to reply

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

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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 ....

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

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

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

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$\sqrt{6\sqrt{x}} > 0$. If so, you understood what they are trying to say and why 0 is an absurd solution

Well do you agree or not thatLog in to reply

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)

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

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

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@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.

Ugh, there are 2 valuesLog in to reply

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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.

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extraneous solution wiki

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