Question

Prove: Inequalities Use the properties of inequalities to prove the following inequalities.

Rule 6 for inequalities: If a, b, c and dare real numbers such that $a<b$and $c<d$, then $a+c<b+d$.[Hint: Use rule 1 to show that $a+c<b+c$ and $b+c<b+d$.use Rule 7.]

Short answer

If$a<b\text{and}c<d,\text{then}a+c<b+d.$

Step-by-step solution

Step 1. Apply the concept of inequalities.

An inequality looks like an equation, except that in the place of the equal sign is one of the symbols, $<,>,\le ,\ge$.

Example of an inequality is width="77" height="20" role="math">4x+7≤19.

An inequality is linear if each term is constant or a multiple of the variable. To solve a linear inequality, isolate the variable on one side of the inequality sign.

Rule 1:$A\le B\iff A+C\le B+C$ (Adding the same quantity to each side of an inequality gives an equivalent inequality).

Rule 6: $A\le B\text{and}C\le D\text{then}A+C\le B+D$If (Inequalities can be added).

Rule 7: If$A\le B\text{and}B\le C$ then$A\le C$ (Inequality is transitive).

Step 2. Simplify the inequality.

Need to prove that if$a<b\text{and}c<d,\text{then}a+c<b+d.$

As $a<b$, add c to both sides of the inequality and hence the following is obtained:

$a+c<b+c----\left(1\right)$

Also $c<d$. Hence, adding b to both sides of the inequality, the following is obtained:

$b+c<b+d----\left(2\right)$

Step 3. Prove the inequality.

Combining the inequalities (1) and (2) by using the rule 7, the following is obtained:

$\begin{array}{c}a+c<b+c<b+d\\ \Rightarrow a+c<b+d\end{array}$

Hence proved.