Show that floating point $\sqrt{x \cdot x} \geq x$ for all long $x$. 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.
 A: 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.
