# Is #sqrt3# rational or irrational?

##### 2 Answers

Irrational

#### Explanation:

Suppose

#x = 1+1/(1+1/(1+x))#

#color(white)(x) = 1+(1+x)/(2+x)#

#color(white)(x) = (3+2x)/(2+x)#

Multiplying both ends by

#x^2+2x = 3+2x#

Subtracting

#x^2 = 3#

and hence:

#x = sqrt(3)#

So we have found:

#sqrt(3) = 1+1/(1+1/(1+sqrt(3)))#

#color(white)(sqrt(3)) = 1+1/(1+1/(2+1/(1+1/(2+1/(1+1/(2+...)))))#

Since this continued fraction does not terminate,

**Bonus**

Here's a fun method to calculate rational approximations to

First note that:

#7^2 = 49 = 48+1 = 3(4^2)+1#

Hence an efficient approximation to

Consider the quadratic with zeros

#(x-7-4sqrt(3))(x-7+4sqrt(3)) = (x-7)^2-48#

#color(white)((x-7-4sqrt(3))(x-7+4sqrt(3))) = x^2-14x+49-48#

#color(white)((x-7-4sqrt(3))(x-7+4sqrt(3))) = x^2-14x+1#

Now if

Define a related sequence of integers recursively by:

#{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 14a_(n+1)-a_n) :}#

Then the ratio between successive terms will tend to

The first few terms are:

#0, 1, 14, 195, 2716, 37829, 526890#

Let's stop there to find:

#7+4sqrt(3) ~~ 526890/37829#

So:

#sqrt(3) ~~ 1/4(526890/37829-7) = 1/4(262087/37829) = 262087/151316 ~~ 1.7320508076#

Irrational

#### Explanation:

#6/1# #18/3# #30/5#

We cannot express **exactly** as the ratio of two numbers. If we evaluate

**NOTE:**