Inequalities
2.0 Theorems
2.0 Theorems
1. We may add the same number to both sides of an inequality and the sign of inequality will not change i.e., If $$a>b$$ then $$a+c>b+c$$
Proof: If $a>b$, we can write $$a-b>0$$
Add and subtract $c$ both sides, we get $$a-b+c-c>0$$$$a+c>b+c$$
Similarly, we may subtract the same number to both sides of an inequality and the sign of inequality will not change i.e., If $$a>b$$ then $$a-c>b-c$$
2. We may multiply both sides of an inequality by same positive number and the sign of inequality will not change i.e., If $$a > b\quad {\text{and }}c > 0$$ then $$ca>cb$$
Example: $4>-5$, if we multiply both sides by $3$, we get $$12>-15$$
Similarly, we may divide both sides of an inequality by same positive number and the sign of inequality will not change i.e., If $$a > b\quad {\text{and }}c > 0$$ then $$\frac{a}{c} > \frac{b}{c}$$
3. If we multiply both sides of an inequality by same negative number, then the sign of inequality will change i.e., If $$a > b\quad {\text{and }}c < 0$$ then $$ca<cb$$
Example: $-4<5$, if we multiply both sides by $-4$, we get $$16>-20$$
Similarly, If we divide both sides of an inequality by same negative number, then the sign of inequality will change i.e., If $$a > b\quad {\text{and }}c < 0$$ then $$\frac{a}{c} < \frac{b}{c}$$
4. If we change sign on both sides of an inequality, then the sign of inequality will change i.e., If $$-a<-b$$ then $$a>b$$
Example: $-1<2$, then on changing sign on both sides, we get $$1>-2$$
The same can also be interpreted using theorem $3$ in which we multiply by the same negative number both sides and sign of inequality changes.
5. If $a$ and $b$ are both positive or both negative, then on taking reciprocals, the sign of inequality changes i.e., If $$a>b$$ then $$\frac{1}{a} < \frac{1}{b}$$
Example: $2<5$ then on taking reciprocals, we get $$\frac{1}{2} > \frac{1}{5} \Rightarrow 0.5 > 0.2$$
Question 1. Solve the following inequalities:
(a) $ - 3x + 4 \leqslant 0$
(b) $ - 1 \leqslant 4x + 3 \leqslant 1$
Solution: (a) Our first step is to make the coefficient of highest power of $x$ positive by multiplying the given inequality by $-1$. According to theorem $3$, we get $$\begin{equation} \begin{aligned} - ( - 3x + 4) \geqslant 0 \\ 3x - 4 \geqslant 0 \\ 3x \geqslant 4 \\ x \geqslant \frac{4}{3} \\ \therefore x \in \left[ {\frac{4}{3},\infty } \right) \\\end{aligned} \end{equation} $$
(b) $$\begin{equation} \begin{aligned} - 1 \leqslant 4x + 3 \leqslant 1 \\ - 1 - 3 \leqslant 4x \leqslant 1 - 3 \\ - 4 \leqslant 4x \leqslant - 2 \\ \frac{{ - 4}}{4} \leqslant x \leqslant \frac{{ - 2}}{4} \\ - 1 \leqslant x \leqslant - \frac{1}{2} \\ \therefore x \in \left[ { - 1, - \frac{1}{2}} \right] \\\end{aligned} \end{equation} $$
