Fiber of Truth/Examples/x^2 = 2 in Reals
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Example of Solution Set
Let $x$ denote a variable whose domain is the set of real numbers $\R$.
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 $\set {\sqrt 2, -\sqrt 2}$.
Proof
From Difference of Two Squares, we have:
- $x^2 - 2 = \paren {x - \sqrt 2} \paren {x + \sqrt 2}$
from which it follows that:
- $x = \sqrt 2$
or:
- $x = -\sqrt 2$
As both $\sqrt 2 \in \R$ and $-\sqrt 2 \in \R$ the result follows.
$\blacksquare$
Sources
- 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): $\S 1$: Some mathematical language: Variables and quantifiers