Learn first: Natural numbers and operations
Integers express negative quantities — debt, frost, depth below sea level, downward motion. Working with them you must know the sign rules perfectly: most arithmetic slips in the exam involve a minus sign. The absolute value (modulus) is the idea of “distance”: it keeps reappearing in equations and inequalities, functions, geometry and coordinates. This lesson covers the set of integers, operations, parity, division with remainder and equations and inequalities with absolute values.
Z = {…, −3, −2, −1, 0, 1, 2, 3, …} — the natural numbers, their opposites and zero.
The integers are closed under addition, subtraction and multiplication: the sum, difference and product of integers are integers. Division does not always give an integer — which is why rational numbers are needed later.
Multiplication and division: (+)(+) = +, (−)(−) = +, (+)(−) = −
Addition: same signs — add the absolute values; different signs — subtract the smaller absolute value from the larger and take the sign of the number with the larger absolute value
Why does minus times minus give plus? To keep the distributive law: (−1)·(1 + (−1)) = (−1)·0 = 0. Expanding gives (−1)·1 + (−1)·(−1) = 0, i.e. −1 + (−1)·(−1) = 0, so (−1)·(−1) = 1.
Order of operations: brackets first, then powers, then multiplication and division (left to right), and finally addition and subtraction.
Calculate −3·(−4) − 12 : (−6) + (−1)²⁰²⁵.
Even numbers have the form 2k, odd numbers 2k + 1 (k ∈ Z); zero is even, −3 is odd.
even ± even = even; odd ± odd = even; even ± odd = odd
if a product has an even factor, the product is even
Dividing an integer by a natural number with remainder: a = bq + r, where 0 ≤ r < b. The remainder is never negative, even when a is negative. For example, −17 = 5·(−4) + 3: the quotient is −4 and the remainder 3 (not −3!).
Consecutive integers: of n, n + 1 one is always even; of n, n + 1, n + 2 one is always divisible by 3. Hence n(n + 1) is always even.
Find the sum of all integers from −10 to 12.
|a| = a if a ≥ 0; |a| = −a if a < 0
Do not read −a in the definition as “a negative number”: when a < 0, −a is positive. For example, |−5| = −(−5) = 5.
Geometric meaning: |a| is the distance on the number line from a to zero; |a − b| is the distance between points a and b. For instance, the distance between −7 and 4 is |4 − (−7)| = 11.
Properties:
|a| ≥ 0, |−a| = |a|, |ab| = |a|·|b|, |a/b| = |a|/|b|, |a|² = a², √(a²) = |a|
Triangle inequality: |a + b| ≤ |a| + |b|
Equality in the triangle inequality holds when a and b have the same sign (or one of them is zero).
|x| = a (a > 0): x = ±a; if a = 0, x = 0; if a < 0, no solutions
With an expression inside it works the same way: |x − 3| = 5 → x − 3 = ±5 → x = 8 or x = −2. Geometrically: the points 5 units from 3.
If the right-hand side contains x (|f(x)| = g(x)), first write the condition g(x) ≥ 0 or check every root.
With several absolute values use the interval method: the points where the expressions inside become zero split the line into intervals; on each interval remove the bars with the correct signs and check that the root found belongs to that interval.
Solve |x − 1| + |x + 2| = 5.
Geometric check: the distances from x to 1 and to −2 add up to 5; the gap between them is 3, so x lies outside the segment, 1 unit beyond either end.
Solve |x − 4| = 2x − 1.
|x| < a: −a < x < a; |x| > a: x < −a or x > a
In terms of distance: |x| < a — points closer than a to zero (one interval); |x| > a — points further than a (two rays).
How many integer solutions does |x − 2| ≤ 3 have?