Real Number between Zero and One is Greater than Square/Proof 1
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Theorem
Let $x \in \R$.
Let $0 < x < 1$.
Then:
- $0 < x^2 < x$
Proof
We are given that $0 < x < 1$.
By direct application of Real Number Ordering is Compatible with Multiplication, it follows that:
- $0 \times x < x \times x < 1 \times x$
and the result follows.
$\blacksquare$