TITLE: Show that floating point $\sqrt{x \cdot x} \geq x$ for all long $x$.
QUESTION [56 upvotes]: I verified experimentally that in Java the equality
Math.sqrt(x*x) = x

holds for all long x such that x*x doesn't overflow. Here, Java long is a $64$ bit signed type and double is a IEEE binary floating point type with at least $53$ bits mantissa and sufficiently long exponent.
Mathematically, there are two imprecise functions involved:

Conversion from long to double which loses precision due to the mantissa being only $53$ bits where $63$ bits would be needed. This operation is guaranteed to return the closest representable result.
Computing square root, which is also guaranteed to return the closest representable result.

Mathematically, this can be expressed like
$$
\mathop\forall_{x \in {\mathbb N} \atop x \le 3037000499}
\mathrm{round}\left(\sqrt{\mathrm{round}(x^2)}\right)
= x
$$
where $\mathrm{round}$ is the rounding function from $\mathbb R$ into the set of all numbers representable as double.
I'm looking for a proof since no experiment can assure it works across all machines.

REPLY [45 votes]: The idea is simple: Find upper and lower bounds for
$$X := \sqrt{\mathrm{round}(x^2)}$$
and show that $\mathrm{round}(X) = x$.

Let $\mathrm{ulp}(x)$ denote the unit of least precision at $x$
and let $E(x)$ and $M(x)$ denote the exponent and mantissa of $x$, i.e.,
$$x = M(x) \cdot 2^{E(x)}$$
with $1 \le M(x) < 2$ and $E(x) \in \mathbb Z$. Define
$$\Delta(x) = \frac{\mathrm{ulp}(x)}x = \frac{\mu \cdot 2^{E(x)}}x = \frac\mu{M(x)}$$
where $\mu=2^{-52}$ is the machine epsilon.
Expressing the rounding function by its relative error leads to
$$X = \sqrt{(1+\epsilon) \cdot x^2} = \sqrt{(1+\epsilon)} \cdot x
< \big( 1+\frac\epsilon2 \big) \cdot x$$
We know that $|\epsilon| \le \frac12\Delta(x^2)$ and get (ignoring the trivial case $x=0$)
$$\frac Xx < 1 + \frac{\Delta(x^2)}4 = 1 + \frac\mu{4 M(x^2)}$$

By observing $M(x)$ and $M(x^2)$ e.g. over the interval $[1, 4]$,
it can be easily be shown that $\frac{M(x)}{M(x^2)} \le \sqrt2$ which gives us
$$\frac Xx < 1 + \frac{\mu\sqrt2}{4 M(x)}$$
and therefore
$$X < x + \frac{\sqrt2}4 \frac{\mu}{M(x)} \cdot x < x + \frac12 \mathrm{ulp}(x)$$

Analogously we get the corresponding lower bound. Just instead of
$$\sqrt{(1+\epsilon)} < \big( 1+\frac\epsilon2 \big)$$
we use something like
$$\sqrt{(1-\epsilon)} > \big( 1 - (1+\epsilon) \cdot \frac\epsilon2 \big)$$
which suffices, since we used a very generous estimate ($\sqrt2/4<\frac12$) in the last step.

Because of $|X-x|$ being smaller than $\frac12 \mathrm{ulp}(x)$, $x$ is the double closest to $X$, therefore $\mathrm{round}(X)$ must equal to $x$, q.e.d.