Math Doubts

Division Rule of inequalities

Dividing the quantities on both sides of an inequality by a quantity is called the division rule of inequalities.

Introduction

In inequalities, two unequal quantities are compared by displaying an unequal sign between them. Dividing a quantity on one side by a quantity but not dividing the quantity on other side by the same quantity imbalances the inequality. Therefore, dividing the quantities on both sides of an inequality balances the principle of inequality and it is called the division property of inequality.

$12 > 6$

$12$ and $6$ are the quantities on left and right-hand side of this exampled inequality respectively. It expresses that the number $12$ is greater than the number $6$.

Now, divide the number $12$ by $3$ but do not divide the number $6$ by the number $3$ to understand the inequality rule of division.

$\implies$ $\dfrac{12}{3} > 6$

$\implies$ $\dfrac{\cancel{12}}{\cancel{3}} > 6$

$\implies$ $4 > 6$

This mathematical statement in the above inequality is absolutely wrong because the number $4$ is not greater than the number $6$. Therefore, it should be written in the following form.

$\,\,\,\therefore\,\,\,\,\,\,$ $4 < 6$

$12 > 6$ is the exampled inequality but it becomes $4 < 6$ after dividing the quantity on left-hand side of the inequality by $3$.

According to the comparison, the quantities on the left-hand side of the inequalities are changed but nothing changed on the right-hand side of the inequalities. Similarly, the sign between them is also changed due to the improper division.

It cleared that dividing the quantity by a particular quantity on any side of the inequality affects the equilibrium between the quantities. So, divide the quantities on both sides of the inequality $12 > 6$ by the $3$ to observe the result.

$\implies$ $\dfrac{12}{3} > \dfrac{6}{3}$

$\implies$ $\dfrac{\cancel{12}}{\cancel{3}} > \dfrac{\cancel{6}}{\cancel{3}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $4 > 2$

Now, let’s study the both inequalities $12 > 6$ and $4 > 2$.

Unchangeability

The sign between the numbers $12$ and $6$ is greater-than $>$ in the inequality. After dividing the quantities $12$ and $6$ by $3$, they become $4$ and $2$ respectively. However, the sign between them is unchanged.

Equilibrium

$12 > 6$ is the example inequality and it can be written in factor form as follows.

$\implies$ $3 \times 4 > 3 \times 2$

Now, divide the quantities on both sides by a number $3$.

$\implies$ $\dfrac{3 \times 4}{3} > \dfrac{3 \times 2}{3}$

$\implies$ $\dfrac{3}{3} \times 4 > \dfrac{3}{3} \times 2$

$\implies$ $\dfrac{\cancel{3}}{\cancel{3}} \times 4 > \dfrac{\cancel{3}}{\cancel{3}} \times 2$

$\implies$ $1 \times 4 > 1 \times 2$

$\,\,\,\therefore\,\,\,\,\,\,$ $4 > 2$

The inequality $4 > 2$ is there in the inequality $12 > 6$ and it is revealed from the actual inequality by dividing the numbers $12$ and $6$ by the same quantity $3$. This fundamental operation is called the division law of inequality.

Examples

The following inequalities in algebraic form expresses how to divide the quantities on both sides of an inequality by a quantity $c$.

$x > y$ is an inequality, then $\dfrac{x}{c} > \dfrac{y}{c}$

$x < y$ is an inequality, then $\dfrac{x}{c} < \dfrac{y}{c}$

$x \ne y$ is an inequality, then $\dfrac{x}{c} \ne \dfrac{y}{c}$

$x \le y$ is an inequality, then $\dfrac{x}{c} \le \dfrac{y}{c}$

$x \ge y$ is an inequality, then $\dfrac{x}{c} \ge \dfrac{y}{c}$

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