Q45. - Algebra 1

Question

CHALLENGE Determine whether the following statement is always true. If not, provide a counterexample.

If $\mathbf{2}\left(\mathbf{b}+\mathbf{c}\right)=\mathbf{2}\mathbf{b}+\mathbf{2}\mathbf{c}$, then $\mathbf{2}+\left(\mathbf{b}\cdot \mathbf{c}\right)=\left(\mathbf{2}+\mathbf{b}\right)\left(\mathbf{2}+\mathbf{c}\right)$.

Step-by-Step Solution

Answer

The following statement is not always true. A counterexample can be when $b=3$ and $c=5$. Then, the hypothesis, $2\left(3+5\right)=2\left(3\right)+2\left(5\right)$ is true. However, the conclusion, $2+\left(\left(3\right)\left(5\right)\right)=\left(2+3\right)\left(2+5\right)$ is false.

Step 1. State the concept.

For a conditional statement to be true, if the hypothesis is true, then the conclusion must also be true.

Step 2. List the given data.

The given statement is “If $2\left(b+c\right)=2b+2c$, then $2+\left(b\cdot c\right)=\left(2+b\right)\left(2+c\right)$”.

Step 3. Identify the hypothesis and conclusion.

The hypothesis is the part of the statement following “if”, that is, “$2\left(b+c\right)=2b+2c$” and the conclusion is the part of the statement following “then”, that is, “$2+\left(b\cdot c\right)=\left(2+b\right)\left(2+c\right)$”.

Step 4. Determine the truth of the statement.

In general, for any two real numbers, $b$ and $c$, $2\left(b+c\right)=2b+2c$ holds due to the distributive property of multiplication over addition.

 

However, for any two real numbers, $b$ and $c$, $2+\left(b\cdot c\right)=\left(2+b\right)\left(2+c\right)$ does not hold as addition is not distributive over multiplication.

 

So, in general, the hypothesis of the given statement is true but the conclusion is false. This implies that the given statement is not always true.

Step 5. Give counterexample.

Put $b=3$ and $c=5$ in $2\left(b+c\right)=2b+2c$ to get,

$\begin{array}{l}2\left(3+5\right)=2\left(3\right)+2\left(5\right)\\ 2\left(8\right)=6+10\\ 16=16\end{array}$ 

So, the hypothesis is true.

 

Put $b=3$ and $c=5$ in $2+\left(b\cdot c\right)=\left(2+b\right)\left(2+c\right)$ to get,

$\begin{array}{l}2+\left(3\cdot 5\right)=\left(2+3\right)\left(2+5\right)\\ 2+15=\left(5\right)\left(7\right)\\ 17=35\end{array}$ 

This is a contradiction. Thus, the conclusion is false.

 

So, $b=3$ and $c=5$ is a counterexample of the given statement.