Divisor Relation is Antisymmetric/Corollary/Proof 2

Corollary to Divisor Relation is Antisymmetric

Let $a, b \in \Z$.

If $a \divides b$ and $b \divides a$ then $a = b$ or $a = -b$.

Proof

Let $a \divides b$ and $b \divides a$.

Then by definition of divisor:

$\exists c, d \in \Z: a c = b, b d = a$

Thus:

\(\ds a c d\)

\(=\)

\(\ds a\)

\(\ds \leadsto \ \ \)

\(\ds c d\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds d\)

\(=\)

\(\ds \pm 1\)

Divisors of One

\(\ds \leadsto \ \ \)

\(\ds a\)

\(=\)

\(\ds \pm b\)

as $a = b d$

$\blacksquare$

Sources

• 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 6$: The division process in $I$
• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(ii)}$
• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$