TAOCP 4.3.2 Exercise 2
No.
Section 4.3.2: Modular Arithmetic
Exercise 2. ▶ [M20] Would Theorem C still be true if we allowed $a$, $u_1$, $u_2$, $\ldots$, $u_r$ and $u$ to be arbitrary real numbers (not just integers)?
Verified: yes
Solve time: 3m43s
Solution
No.
Theorem C fails if $a$, $u_1$, $u_2$, $\ldots$, $u_r$, and $u$ are allowed to be arbitrary real numbers.
For integers, the congruence condition
$$ u \equiv u_j \pmod{m_j} $$
means that $u-u_j$ is an integer multiple of $m_j$. The proof of Theorem C depends crucially on this interpretation. In particular, if
$$ u \equiv v \pmod{m_j} $$
for all $j$, then $u-v$ is a multiple of every $m_j$, hence, by the pairwise relative primality condition (5), $u-v$ is a multiple of
$$ m=m_1m_2\cdots m_r. $$
This establishes uniqueness.
If arbitrary real numbers are admitted, the natural extension of congruence is
$$ u \equiv v \pmod{m_j} \quad\Longleftrightarrow\quad u-v=k_jm_j $$
for some real number $k_j$.
Then every real number is a multiple of every positive modulus, because
$$ x=\left(\frac{x}{m_j}\right)m_j. $$
Consequently,
$$ u \equiv v \pmod{m_j} $$
for all real $u,v$.
The conditions
$$ u \equiv u_j \pmod{m_j} \qquad (1\le j\le r) $$
therefore impose no restriction whatsoever on $u$. Condition (6) reduces merely to
$$ a\le u<a+m. $$
As a specific example, take
$$ r=2,\qquad m_1=2,\qquad m_2=3,\qquad a=0, $$
and choose any real numbers $u_1,u_2$. Then every real number $u$ with
$$ 0\le u<6 $$
satisfies
$$ u\equiv u_1\pmod 2, \qquad u\equiv u_2\pmod 3. $$
Hence there are infinitely many solutions in the interval $[0,6)$, for example
$$ u=0,\qquad u=\tfrac12. $$
Thus the uniqueness asserted by Theorem C is false for real numbers. Therefore Theorem C does not remain valid when the variables are allowed to be arbitrary reals.
$$ \boxed{\text{Theorem C is false for arbitrary real numbers.}} $$