Zero Divides Zero
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Theorem
Let $n \in \Z$ be an integer.
Then:
- $0 \divides n \implies n = 0$
That is, zero is the only integer divisible by zero.
Proof
\(\ds 0\) | \(\divides\) | \(\ds n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists q \in \Z: \, \) | \(\ds n\) | \(=\) | \(\ds q \times 0\) | Definition of Divisor of Integer | |||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(=\) | \(\ds 0\) | Integers have no zero divisors, as Integers form Integral Domain. |
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: The Integers: $\S 10$. Divisibility
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
- 1994: H.E. Rose: A Course in Number Theory (2nd ed.) ... (previous) ... (next): $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(ii)}$