Thread: How can an irrational number be finite?

1. 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!

2. 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

3. Thank you veeery much!

4. Hello, stressedIBstudent1!

A slightly different approach . . .

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

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

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, $\phi.$

5. You really explain things in a simple way, thank you very much!

6. How can an irrational number be finite?
The wording might better have been 'how can an irrational number be algebraic?'

7. 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.