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Worked Examples · Example 9

Q.Let AA be the set of all natural numbers. Define R={(x,y); xy<0; x,y∈A}R = \{(x, y);\ \frac{x}{y} < 0;\ x, y \in A\}.

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✓ Free question

Since natural numbers are all positive, the ratio x/yx/y can never be negative — so RR is the empty (void) relation.

A=N={1,2,3,4,… } (all positive integers),R={(x,y):xy<0, x,y∈A}A=\mathbb{N}=\{1,2,3,4,\dots\}\ (\text{all positive integers}),\qquad R=\left\{(x,y): \frac{x}{y}<0,\ x,y\in A\right\}

A relation with no elements satisfying its defining condition is the void (empty) relation, R=∅R=\varnothing.

  1. Recall the domain. AA is the set of natural numbers, so every x∈Ax\in A and every y∈Ay\in A satisfies x>0x>0 and y>0y>0.

  2. Analyse the sign of xy\dfrac{x}{y}. The quotient of two positive numbers is always positive:

x>0, y>0 ⇒ xy>0for every choice of x,y∈Nx>0,\ y>0\ \Rightarrow\ \frac{x}{y}>0 \quad\text{for every choice of } x,y\in\mathbb{N}

  1. Compare with the condition defining RR. RR requires xy<0\dfrac{x}{y}<0. Since xy\dfrac{x}{y} is provably always >0>0 (never <0<0, never even =0=0, as x≠0x\neq0), no pair (x,y)∈A×A(x,y)\in A\times A can satisfy the defining condition.

  2. Conclude.

R={(x,y):x,y∈N, xy<0}=∅R=\{(x,y): x,y\in\mathbb{N},\ \frac{x}{y}<0\}=\varnothing

This is called the void relation (empty relation) on AA.

Self-check: Testing any sample pair, e.g. x=5,y=2x=5,y=2: 52=2.5>0\frac{5}{2}=2.5>0, not <0<0 — confirms no pair can ever qualify, for any positive x,yx,y.

✓Final answer

R=∅R=\varnothing — the void (empty) relation, since xy>0\dfrac{x}{y}>0 for all natural numbers x,yx,y, so the condition xy<0\dfrac{x}{y}<0 is never satisfied.

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←Previous section · 4.5Types of Relations→Next question · 4.5Example 10 — Let A = \1, 2, 3, 4, 5\. Define R = \(x, y) : x + y ∈ N;x, y ∈ A\.