Chapter 6.5, Problem 81HP

To determine

To find: a counterexample for the given statement.

Answer to Problem 81HP

One term is zero and other term is any integer.

Explanation of Solution

Given:

The given statement is $a^2+b^2={\left(a+b\right)}^2$ .

Calculation:

The given statement is $a^2+b^2={\left(a+b\right)}^2$ .

Suppose a combination of positive and negative term for the given statement.

Let $x=-2,y=3$ . Therefore,

  $\begin{array}{l}{a}^2+b^2={\left(-2\right)}^2+{\left(3\right)}^2=4+9=13\\ {\left(a+b\right)}^2={\left(-2+3\right)}^2=1^2=1\end{array}$

Thus, $a^2+b^2\ne {\left(a+b\right)}^2$ .

Let another example in which one term is zero, then

Let $a=7,b=0$ . Therefore,

  $\begin{array}{l}{a}^2+b^2={\left(7\right)}^2+{\left(0\right)}^2=49+0=0\\ {\left(a+b\right)}^2={\left(7+0\right)}^2=7^2=49\end{array}$

Hence, one term is zero and other term is any integer is the counterexample for the given statement.