# How can an irrational number be finite?

• Nov 23rd 2010, 12:38 PM
stressedIBstudent1
How can an irrational number be finite?
Hi, I'm an IB student from Mexico and I'm dying to solve something.
As part of a homework i must fin the exact value of an irrational number (our teacher told us it would be a trinomial) but as far as I'm concerned an irrational number can't be exact.
The number is:
[IMG]file:///C:/Users/ANAMAR%7E1/AppData/Local/Temp/moz-screenshot-2.png[/IMG]The root of: the root of 1 plus the the root of 1 and so on...
I wasn't able to type it here bit it looks as if the first root is "housing" the next one and so on.

Thanks for you help and I really hope someone knows how to solve this! :)
• Nov 23rd 2010, 12:53 PM
skeeter
Quote:

Originally Posted by stressedIBstudent1
Hi, I'm an IB student from Mexico and I'm dying to solve something.
As part of a homework i must fin the exact value of an irrational number (our teacher told us it would be a trinomial) but as far as I'm concerned an irrational number can't be exact.
The number is:
[IMG]file:///C:/Users/ANAMAR%7E1/AppData/Local/Temp/moz-screenshot-2.png[/IMG]The root of: the root of 1 plus the the root of 1 and so on...
I wasn't able to type it here bit it looks as if the first root is "housing" the next one and so on.

Thanks for you help and I really hope someone knows how to solve this! :)

first, note that image files will not show as a link to your hard drive. you have to upload it as an attachment.

let $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}$

$x^2 = 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}$

$x^2 = 1 + x$

$x^2 - x - 1 = 0$

you'll get two irrational roots ... one is the value of your "nested" radical.

also, see the following article ...

Golden ratio - Wikipedia, the free encyclopedia
• Nov 23rd 2010, 01:10 PM
stressedIBstudent1
Thank you veeery much!
• Nov 23rd 2010, 02:49 PM
Soroban
Hello, stressedIBstudent1!

A slightly different approach . . .

Quote:

$\text{Evaluate: }\sqrt{1 + \sqrt{1 + \sqrt{1 + \hdots}}}$

$\text{We have: }\;x \;=\;\sqrt{1 + \underbrace{\sqrt{1 + \sqrt{1 + \hdots }}}_{\text{This is }x}}$

So we have: . $x \;=\;\sqrt{1 + x}$

Square both sides: . $x^2 \;=\;1 + x \quad\Rightarrow\quad x^2 - x - 1 \:=\:0$

And we get the same quadratic equation that skeeter found.

Follow his advice and get the answer: . $x \;=\;\frac{1 + \sqrt{5}}{2}$
. . which happens to be the Golden Mean, $\phi.$

• Nov 23rd 2010, 03:43 PM
stressedIBstudent1
You really explain things in a simple way, thank you very much!
• Nov 23rd 2010, 04:21 PM
TheCoffeeMachine
Quote:

How can an irrational number be finite?
The wording might better have been 'how can an irrational number be algebraic?'
• Nov 23rd 2010, 04:35 PM
skeeter
Quote:

Originally Posted by TheCoffeeMachine
The wording might better have been 'how can an irrational number be algebraic?'

I think the OP was confusing "exact" as a number with only a finite decimal representation.