Question

Prove that there exists a real number 'x' satisfying x^2 = 2

Answer 1

\[T = \left\{ t \in R : t^{2} < 2\right\}\]
set x = least upper bound of T.
We need to prove x^2 = 2

Answer 2

Let supT = "least upper bound of T"
To prove this, we shall consider this:
If a and b are real numbers, then either a < b or a = b or a > b.

To prove that x^2 = 2 is supT
We need to prove that x^2 < 2 is false, and x^2 > 2 is false.

Answer 3

Case 1: Prove that x^2 < 2 is false by proof by contradiction.

Assume x^2 < 2
Then
\[(x + \frac{1}{n})^{2} < 2\]
Is not an element of set T

Answer 4

\[(x + \frac{1}{n})^{2} < 2\]
then
\[=x^2 + \frac{2x}{n}+\frac{1}{n} < x^{2} + \frac{2x}{n}+\frac{1}{n} = x^{2}+\frac{2x+1}{n}\]

By the archimedean property, for any real number x, there exists an n such that:
\[\frac{1}{n} < x \]

Answer 5

But the archimedean property is useless here because I'm trying to find an 'n' that makes the previous equation greater than x^2 and less than 2.

Instead, choose n such that n is a natural number, and
\[\frac{1}{n} < \frac{2 - x^{2}}{2x+1}\]

Answer 6

Then
\[\frac{2x + 1}{n} < 2 - x^{2}\]
then
\[(x + \frac{1}{n})^{2} < x^{2} + (2 - x^{2}) = 2\]
then
(x+1)/n is an element of T. Contradicting the fact that x is an upper bound of T.
Thus x^2 < 2 is false.

Answer 7

Case 2: Prove the same thing about x^2 > 2.