Mathematics · Glossary

What is Divisibility?

Definition 29.1 High School Mathematics · Chapter 29 — Arithmetic

Let a,bZa, b \in \Z. We say bb divides aa, written bab \mid a, if there exists kZk \in \Z with a=kba = kb. We also say aa is a multiple of bb.

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Definition 6.1 University Mathematics — Year 1 · Chapter 6 — Integer Arithmetic

For a,bZa, b \in \Z, bb divides aa (written bab \mid a) when a=bqa = bq for some qZq \in \Z. Basic consequences: if bab \mid a and bab \mid a' then b(ua+va)b \mid (ua + va') for all u,vZu, v \in \Z; if bab \mid a and a0a \neq 0 then ba\abs b \leq \abs a; and aba \mid b together with bab \mid a force b=±ab = \pm a.

Examples

Example 6.3 (Positional numeration by repeated division)

Write 20262026 in base 77. Divide repeatedly by 77, keeping the remainders:

2026=7×289+3,289=7×41+2,41=7×5+6,5=7×0+5.2026 = 7 \times 289 + 3, \quad 289 = 7 \times 41 + 2, \quad 41 = 7 \times 5 + 6, \quad 5 = 7 \times 0 + 5 .

Reading the remainders from last to first: 2026=(5623)72026 = (5\,6\,2\,3)_7. Check: 5×343+6×49+2×7+3=1715+294+14+3=20265 \times 343 + 6 \times 49 + 2 \times 7 + 3 = 1715 + 294 + 14 + 3 = 2026. The uniqueness of Euclidean division is exactly what makes each digit forced: at every step the remainder is the only integer in [ ⁣[0,6] ⁣]\intint06 congruent to the current value mod 77, so base-77 writing is unique — the fact silently used whenever the weekend problem manipulates “the digits of nn in base pp”.

Example 6.9 (Solving a linear Diophantine equation)

Find all (x,y)Z2(x, y) \in \Z^2 with 6x+10y=46x + 10y = 4. First, the existence test: gcd(6,10)=2\gcd(6, 10) = 2 divides 44, so solutions exist (if the gcd did not divide the right-hand side, the left side would always be a multiple of it and there would be none). Divide through: 3x+5y=23x + 5y = 2. A particular solution is visible: (x0,y0)=(1,1)(x_0, y_0) = (-1, 1). For the general one, subtract: 3(x+1)=5(y1)3(x + 1) = -5(y - 1), so 35(y1)3 \mid 5(y-1), and Gauss’s lemma (gcd(3,5)=1\gcd(3,5) = 1) gives 3y13 \mid y - 1: y=13ky = 1 - 3k, then x=1+5kx = -1 + 5k. Conversely every such pair works:

(x,y)=(1+5k, 13k),kZ.(x, y) = (-1 + 5k,\ 1 - 3k), \qquad k \in \Z .

The pattern is general: one particular solution plus the integer multiples of (bgcd,agcd)\bigl(\frac b{\gcd}, -\frac a{\gcd}\bigr) — the same “particular plus homogeneous” structure as in Chapter 5, with Gauss’s lemma playing the uniqueness role.

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