Download Is sqrt 2 a rational number

Is
2 a rational number?
It appears from the number line test, that
2 is a rational number.
However, we know from Geometric proofs that
number.
2 is NOT a rational
We also know that only perfect square roots are rational numbers. In
other words, the square root of 4, 9, 16, 25, etc are rational numbers.
The square of a non-perfect square number is not a rational number.
(I don’t know where I read this, but I do remember this from my math
education.)
This is because square roots of non-perfect square numbers cannot be
expressed as a/b where a and b are rational numbers.
Heron’s Method says that we can use the formula
1
a 

x ( n1)   x n 
2
x n 
Therefore, if a=5, and x0=5, then x (n 1) 
1
5
 5   which brings us to
2
5
the step that gives us x ( n 1) = ½ (6) or x ( n 1) = 3.
1
a 
 which gives us
x ( n 2)   x n1 
2
x n1 
1
5
1  14 
7
  3   which equals x (n 2)    or x ( n  2)  or 2.333.
2
3
2 3 
3
Step 2 is the same:
x (n 2)
This brings is even closer to the square root of 5, which is 2.236
Step 3 is the same:
x (n 3)
x ( n3)
1
a
x ( n3)   x n 2 
2
x n 2

 which gives us





17 5 
1  7 15 


x

   or
which
equals
(n

3
)
23 7 
2
3 7 


3

 1  93  93
    
or 2.214.
 2  21  42
And to do Heron’s Method again all of this brings us closer to the
square root of 5. First we had an answer of 5, then 3, then 2.3333
and then 2.214.
Obviously this brings us to a number that is too small for the square
root of 5.
I have shown my work on the Heron’s Method worksheet, and do not
get the right answer their either.
According to Weisstein, Eric W. "Newton's Method." From MathWorld--A
Wolfram Web Resource. http://mathworld.wolfram.com/NewtonsMethod.html Wolf
Newton's method, is a root-finding algorithm.
Newton's method is sometimes also known as Newton's iteration,
although in this work the latter term is reserved to the application of
Newton's method for computing square roots.
For
a polynomial, Newton's method is essentially the same as
Horner's method.
However, Newton’s method uses this formula:
x n 1  x n 
bg
bg
f xn
f  xn
I tried using f(5) to find the square root, and it didn’t work.
I also tried the equation using f(x) = x2 – x + 6.
If you refer to the Newton’s Method worksheet, you will see my work
for both a random number and for f(x) = x2 – x + 6
Obviously I have done something wrong with the equation.