Ok, sorry. It's a proof for the Theorem:
"For each positive real number a and each integer n > 1, there is a unique real number b such that$\displaystyle b^n = a$
In the special case $\displaystyle b^2 = 2 $
The proof begins with
$\displaystyle 1^2 < 2 < 2^2$
$\displaystyle (1.4)^2 < 2 < (1.5)^2$
$\displaystyle (1.41)^2 < 2 < (1.42)^2$
etc.
to give decimal b^2 = (1.414...)^2 = 2
Then the method uses a table
b | b | b^2 |
1 | 1 | 1 |
1.4 | 1.4 | 1.96 |
1.41 | 1.41 | 1.9881 |
1/414 | 1/414 | 1.999396 |
etc.
to prove so.
Then to prove the least upper bound of the set of numbesr in the third column of the table is 2, we check that M=2 is an upper bound of E, which follows from the inequalities at the top. Then to show if M' < 2 there is a number in E that is greater then M'.
We put$\displaystyle x_0 = 1, x_1 = 1.4, x_2 = 1.41, ...$
Then we have
$\displaystyle 2 < (x_n + \frac{1}{10^n})^2 = x^2_n + \frac{2x_n}{10^n} + (\frac{1}{10^n})^2$
so
$\displaystyle x^2_n > 2 - \frac{1}{10^n}(2x_n + \frac{1}{10^n})$
$\displaystyle > 2 - \frac{5}{10^n}$