Square root of the square root of the square root of a number =1?

taking the square root of a number:

(number)^(1/2)

if you do again it is

(number)^(1/2)^(1/2) = (number)^(1/4)

again

(number)^(1/2)^(1/2)^(1/2) = (number)^(1/8)

the exponential will go 1/2,1/4,1/8,1/16,etc. which approaches 0

any number raised to the 0 power = 1

Say you take the sqrt (100) = 10.

and then the sqrt (10) = 3 something

and then the sqrt (3 something) = 1.7 something

and then the sqrt (1.7 something) = 1.3 something

By taking the square root, you are finding what times what equals that number...eventually the calculator will be rounding to 1, because the numbers will be 1.000000000000003 or similar. The calculator can't handle that, so it rounds to 1.

And the sqrt (1) = 1, so that becomes the end of the story.

PS: have you tried taking the square root of 0.1 over and over again? Try it!

It shouldn't be. The reason you get one is because of the imprecision of your calculator.

Try it with your computer calculator. Square root is power 0.5. If you do this 30 times it is like X^0.5^30. Now 0.5^30 is equal to 0.000000000931322574615478515625. Type in a large number, say 99999999 and put to the power of 0.000000000931322574615478515625. The result is 1.0000000171555959544714939867674. On most calculator, this is rounded to 1 but you can see that with a more precise one (e.g. Windows calculator) it is not the case.

It works for decimals as well.

It's because, as you keep taking the square root, the number keeps getting closer to 1. Eventually, the number gets so close to 1 that the calculator cannot distinguish it from 1.

Yes, because the calculator doesn't have an infinite number of decimal places to work with. If it did, it wouldn't come up with just the number 1.

This is an introduction limits. Cool, no?

You are asking why the

lim (x)^1/n goes to 1 when ever n -> oo for all x > 0

This is a question often investigated as a part of learning calculus.

HTH

Charles

Let a be any finite real positive number.

√a = (a)^(0.5)

√(a)^(1/2) = a^(0.5^2)

.

.

.

= a^ (0.5^n).

When n approaches to infinity, we can use L'Hopital's rule to get the answer.

lim a^ (0.5^n), n->infinity

= lim e^[0.5^n lna], n->infinity

= e^0

= 1