Direct and Inverse Proportion (Algebraic)

Proportion is one of the most powerful tools in mathematics, describing how two quantities relate to each other as one or both change. When we say two variables are in proportion, we mean there is a consistent, predictable relationship between them that can be captured in a single algebraic equation. Direct proportion and inverse proportion are the two fundamental types, and understanding them algebraically — rather than just numerically — allows us to solve problems involving any values, not just the ones we have already seen. In direct proportion, as one quantity increases, the other increases at the same rate. In inverse proportion, as one quantity increases, the other decreases proportionally. But real-world mathematics goes further: quantities can be directly proportional to the square, cube, or square root of another variable. For example, the area of a circle is directly proportional to the square of its radius, and the period of a pendulum is directly proportional to the square root of its length. Recognising and writing these relationships algebraically is an essential skill in algebra, physics, engineering, and economics. This article develops the full algebraic language of proportion — introducing the proportionality constant k, showing how to find k from given data, and extending the concept to squared, cubed, and root relationships. By the end, you will be able to write, manipulate, and apply proportion equations confidently across a wide range of mathematical and real-world contexts.

Key takeaways

Direct Proportion: y ∝ x
Inverse Proportion: y ∝ 1/x
Square and Cube Proportion: y ∝ x² and y ∝ x³
Square Root Proportion: y ∝ √x
Inverse Square Proportion: y ∝ 1/x²
Finding k and Solving Problems

Direct Proportion: y ∝ x

We write y ∝ x (read 'y is directly proportional to x') when doubling x doubles y, tripling x triples y, and so on. Algebraically this becomes y = kx, where k is the constant of proportionality. For example, if y = 12 when x = 4, then k = y/x = 12/4 = 3, giving y = 3x. We can then find y for any x: when x = 7, y = 21.

Inverse Proportion: y ∝ 1/x

We write y ∝ 1/x when doubling x halves y. Algebraically: y = k/x, equivalently xy = k (constant product). If y = 8 when x = 5, then k = 8 × 5 = 40, so y = 40/x. When x = 10, y = 4. Notice that in inverse proportion the product xy is always the same, which is a useful check.

Square and Cube Proportion: y ∝ x² and y ∝ x³

Many real relationships involve powers. 'y is directly proportional to the square of x' is written y ∝ x², giving y = kx². To find k, substitute a known pair: if y = 45 when x = 3, then 45 = k(9), so k = 5 and y = 5x². Similarly y ∝ x³ gives y = kx³. The volume of a sphere V = (4/3)πr³ is a classic example, where k = 4π/3.

Square Root Proportion: y ∝ √x

Here y = k√x. This appears in physics: the period T of a simple pendulum satisfies T ∝ √L, where L is its length. If T = 2 s when L = 1 m, then k = 2. For L = 4 m, T = 2√4 = 4 s — doubling the period requires quadrupling the length, not doubling it.

Inverse Square Proportion: y ∝ 1/x²

Gravity and light intensity obey y ∝ 1/x², written y = k/x². If intensity I = 100 W/m² at distance d = 2 m, then k = 100 × 4 = 400. At d = 5 m, I = 400/25 = 16 W/m². This 'inverse square law' means intensity falls off rapidly with distance.

Finding k and Solving Problems

The universal method: (1) Write the proportion statement (e.g., y ∝ x²). (2) Replace ∝ with = k to get the equation (y = kx²). (3) Substitute the known values of x and y to calculate k. (4) Rewrite the equation with the numerical value of k. (5) Use this equation to find unknown values. Always state the final equation before calculating answers.

The Key Equation

y = kx^n \quad \text{or} \quad y = \frac{k}{x^n}, \quad k = \frac{y}{x^n}

y — dependent variable (varies): The quantity whose value depends on x
k — constant of proportionality (varies): A fixed number that scales the relationship; found by substituting a known (x, y) pair
x — independent variable (varies): The quantity that is changed or controlled
n — power index (dimensionless): The exponent: 1 for linear, 2 for square, 3 for cube, 1/2 for square root proportion

Worked Example

Problem. The electrical resistance R of a wire is inversely proportional to the square of its radius r. When the radius is 0.5 mm, the resistance is 64 Ω. (a) Find the equation connecting R and r. (b) Find the resistance when r = 2 mm. (c) Find the radius when R = 4 Ω.

R = \frac{16}{r^2}, \quad R(2) = 4\,\Omega, \quad r = 2\,\text{mm}

Answer. k = 16, so R = 16/r². When r = 2 mm, R = 4 Ω. When R = 4 Ω, r = 2 mm.

Figure

Diagram

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Swiss

Euler systematised much of the algebraic notation we use today, including the use of k and other letters as constants in equations. His work on functions formalised the idea that a single equation like y = kx² could describe an entire relationship between variables, laying the groundwork for modern algebra and calculus. His 1748 work Introductio in Analysin Infinitorum established functions as central objects of mathematics. Euler lost sight in his right eye at age 28 and became completely blind by age 59, yet he continued to produce mathematics at an extraordinary rate — roughly half of his total output came after he lost his sight, dictated from memory.

Ratio and Proportion Basics — Direct and inverse proportion extend the concept of ratio; understanding equivalent ratios is essential before working with the algebraic proportionality constant k.
Linear and Quadratic Functions — Direct proportion y = kx is a linear function through the origin, and square proportion y = kx² is a quadratic; recognising their graphs connects proportion to the wider study of functions.
Laws of Indices — Proportion involving squares, cubes, and roots uses index notation (x², x³, x^½); fluency with index laws is needed to manipulate and solve these equations.
Percentage Change and Multipliers — A common exam question asks how a percentage change in x affects y in a proportion relationship; this requires combining proportion algebra with percentage multipliers.