TAOCP 4.2.2 Exercise 7
We seek floating binary numbers $u$ and $v$ for which (u\oplus v)^2>2(u^2\oplus v^2), where $\oplus$ denotes floating-point addition with rounding to the nearest floating binary number.
Section 4.2.2: Accuracy of Floating Point Arithmetic
Exercise 7. [M21] Let $\sqrt{\oplus}$ stand for $u \oplus u$. Find floating binary numbers $u$ and $v$ such that $(u \oplus v)^2 > 2(u^2 \oplus v^2)$.
Verified: yes
Solve time: 5m07s
Solution
We seek floating binary numbers $u$ and $v$ for which
$$ (u\oplus v)^2>2(u^2\oplus v^2), $$
where $\oplus$ denotes floating-point addition with rounding to the nearest floating binary number.
It suffices to exhibit one floating-point system and one pair $u,v$ for which the inequality holds.
Consider a binary floating-point system with precision $p=2$, normalized significands $1.0_2$ and $1.1_2$, and rounding to nearest.
Take
$$ u=v=1.1_2\times 2^0. $$
Thus $u=v=\frac32$.
First compute $u\oplus v$:
$$ u+v=3=1.1_2\times 2^1, $$
which is exactly representable. Hence
$$ u\oplus v=3. $$
Therefore
$$ (u\oplus v)^2=9. $$
Now compute $u^2$:
$$ u^2=v^2=\left(\frac32\right)^2=\frac94 =1.001_2\times 2^1. $$
With precision $p=2$, the floating numbers in the interval $[2,4)$ are
$$ 1.0_2\times 2^1=2, \qquad 1.1_2\times 2^1=3. $$
Since
$$ \frac94=2.25 $$
is closer to $2$ than to $3$, rounding gives
$$ \operatorname{fl}(u^2)=2. $$
Hence
$$ u^2\oplus v^2 =2\oplus 2. $$
But
$$ 2+2=4=1.0_2\times 2^2 $$
is exactly representable, so
$$ u^2\oplus v^2=4. $$
Therefore
$$ 2(u^2\oplus v^2)=8. $$
Comparing the two sides,
$$ (u\oplus v)^2=9, \qquad 2(u^2\oplus v^2)=8, $$
and thus
$$ 9>8. $$
Hence
$$ \boxed{(u\oplus v)^2>2(u^2\oplus v^2)} $$
for
$$ \boxed{u=v=1.1_2\times 2^0}. $$
This provides the required example. ∎