Even and Odd Integers; Divisibility

Chapter 1: Number Systems and Their Properties

Learning objectives

  • Determine if an integer is even or odd
  • Apply divisibility rules
  • Prove basic properties of even and odd numbers

Parity is the smallest fact in mathematics that pays off the most often. Whether a number is even or odd is a single bit of information, but that one bit decides whether equations are solvable, whether games can be won, whether 2\sqrt{2} is rational (it is not, and the proof is one parity argument). Divisibility is the same idea sharpened: instead of asking "is this a multiple of 2?" we ask "is this a multiple of nn?" for any nn we like.

Definitions you can prove things with

An integer is even if it can be written as 2k2k for some integer kk. It is odd if it can be written as 2k+12k + 1. These are not visual descriptions — they are algebraic forms, and that is what makes them useful for proofs. To show that the sum of two odd numbers is even, write the odds as 2m+12m+1 and 2n+12n+1, add them: 2m+2n+2=2(m+n+1)2m + 2n + 2 = 2(m + n + 1). The result has the form 2k2k. Done.

The parity table

even+even=evenodd+odd=even\text{even} + \text{even} = \text{even} \quad\quad \text{odd} + \text{odd} = \text{even}
even+odd=oddeven×anything=even\text{even} + \text{odd} = \text{odd} \quad\quad \text{even} \times \text{anything} = \text{even}
odd×odd=odd\text{odd} \times \text{odd} = \text{odd}

Memorise the rules, then forget them — the proofs above (and below for multiplication) recover any rule in a few seconds from the 2k2k and 2k+12k+1 definitions.

Mod ClockInteractive figure — enable JavaScript to interact.

(Set the modulus to 2 in the widget above and type an integer — the clock face shows even (0) and odd (1) as two halves, and the hand sweeps to the right one. Change the modulus to 3, 5, or 12 to see divisibility by other numbers the same way.)

Divisibility: a generalised parity

We say aa divides bb, written aba \mid b, if there exists an integer qq with b=aqb = aq. Examples: 3123 \mid 12 because 12=3412 = 3 \cdot 4; 505 \mid 0 because 0=500 = 5 \cdot 0; but 373 \nmid 7 because no integer qq gives 7=3q7 = 3q.

Divisibility behaves the way you would hope: if aba \mid b and aca \mid c, then aa also divides every "linear combination" bm+cnbm + cn. And divisibility is transitive: aba \mid b and bcb \mid c together force aca \mid c. (When aba \mid b we call aa a divisor of bb.)

The division algorithm — the engine

For any integers aa and bb with b>0b > 0, there exist unique integers qq (the quotient) and rr (the remainder) such that a=bq+ra = bq + r with 0r<b0 \leq r < b. Every divisibility statement, every congruence-class argument, every modular-arithmetic identity that you will ever meet rests on this one theorem.

Try it

  • Write the form 2k+12k+1 for the odd numbers 7,3,997, -3, 99. What is kk in each case?
  • Prove from the 2k2k form that the square of any even integer is divisible by 44.
  • Apply the division algorithm to a=23a = -23, b=5b = 5: find the unique (q,r)(q, r) with 0r<50 \leq r < 5.
  • Is 696 \mid 9? Is 969 \mid 6? Why is divisibility not symmetric?

Try it in code

A trap to watch for

It is tempting to reverse the linear-combination rule: if a(b+c)a \mid (b + c), surely aba \mid b and aca \mid c? The converse is false. Try a=2a = 2, b=1b = 1, c=1c = 1: then b+c=2b + c = 2 and 222 \mid 2, but 212 \nmid 1. The brain makes the mistake because addition feels symmetric — if a factor "survives" the sum, it must have been present in each piece. It need not. The correct one-way statement: aba \mid b and aca \mid c together force a(bm+cn)a \mid (bm + cn) for any integers m,nm, n, but a divisor of the sum tells you nothing about the parts. The fix: before splitting a divisibility statement across an addition, write out the actual integer quotients explicitly.

What you now know

You can classify integers by parity and divisibility using algebraic forms (not visual hunches), and you can use the division algorithm to convert any divisibility question into a clean arithmetic test. The next section moves beyond Z\mathbb{Z} to the rational numbers Q\mathbb{Q}, where division stops failing and the four laws keep working.

Quick check

References

  • Lang, S. (1971). Basic Mathematics. Springer. Chapter 1, §4 — even / odd / divisibility framed as the first step into number theory.
  • Hardy, G. H. and Wright, E. M. (2008). An Introduction to the Theory of Numbers, 6th ed. Oxford. Chapter 2 — the division algorithm proved cleanly with the well-ordering principle.
  • Niven, I., Zuckerman, H., and Montgomery, H. (1991). An Introduction to the Theory of Numbers, 5th ed. Wiley. Chapter 1 — divisibility, GCD, and the Euclidean algorithm.

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