Fiber of Truth/Examples/x^2 = 2 in Rationals

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Example of Solution Set

Let $x$ denote a variable whose domain is the set of real numbers $\Q$.

Let $\map P x$ be the propositional function defined as:

$\map P x := x^2 - 2$

Then the solution set of $\map P x$ is the empty set $\O$.


From Difference of Two Squares, we have:

$x^2 - 2 = \paren {x - \sqrt 2} \paren {x + \sqrt 2}$

from which it follows that:

$x = \sqrt 2$


$x = -\sqrt 2$

As both $\sqrt 2 \notin \Q$ and $-\sqrt 2 \notin \Q$ the result follows.