Chapter T, Problem 10ADT

State whether each equation is true or false.

1. (a) (p + q)² = p² + q²
2. (b) $\sqrt{ab}=\sqrt{a}\sqrt{b}$
3. (c) $\sqrt{a^2+b^2}=a+b$
4. (d) $\frac{1+TC}{C}=1+T$
5. (e) $\frac{1}{x-y}=\frac{1}{x}-\frac{1}{y}$
6. (f) $\frac{1/x}{a/x-b/x}=\frac{1}{a-b}$

To determine

To state: Whether the equation ${\left(p+q\right)}^2=p^2+q^2$ is true or false.

Answer to Problem 10ADT

The equation ${\left(p+q\right)}^2=p^2+q^2$ is false.

Explanation of Solution

Suppose the values of $p=2\text{and}q=1$.

Substitute the value of $p=2\text{and}q=1$ in the L. H. S. of the equation ${\left(p+q\right)}^2=p^2+q^2$,

$\begin{array}{c}L.H.S.={\left(2+1\right)}^2\\ =3^2\\ =9\end{array}$

Substitute the value of $p=2\text{and}q=1$ in the R. H. S. of the equation ${\left(p+q\right)}^2=p^2+q^2$,

$\begin{array}{c}R.H.S.=2^2+1^2\\ =4+1\\ =5\end{array}$

Therefore, $L.H.S.\ne R.H.S.$

In general, ${\left(p+q\right)}^2=p^2+2pq+q^2$.

Thus, the equation ${\left(p+q\right)}^2=p^2+q^2$ is false.

To determine

To state: Whether the equation $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true or false.

Answer to Problem 10ADT

The equation $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true.

Explanation of Solution

As long a and b are nonnegative real numbers.

Simplify the above expression $\sqrt{ab}=\sqrt{a}\sqrt{b}$ as follows,

$\begin{array}{c}L.H.S.=\sqrt{ab}\\ ={\left(ab\right)}^{\frac{1}{2}}\\ =a^{\frac{1}{2}}{b}^{\frac{1}{2}}\\ =\sqrt{a}\sqrt{b}\\ =R.H.S.\end{array}$

Suppose the values of $a=9\text{and}b=4$.

Substitute the value of $a=9\text{and}b=4$ in the L. H. S. of the equation $\sqrt{ab}=\sqrt{a}\sqrt{b}$,

$\begin{array}{c}L.H.S.=\sqrt{9\times 4}\\ =\sqrt{36}\\ =6\end{array}$

Substitute the value of $a=9\text{and}b=4$ in the R. H. S. of the equation $\sqrt{ab}=\sqrt{a}\sqrt{b}$,

$\begin{array}{c}R.H.S.=\sqrt{9}\times \sqrt{4}\\ =3\times 2\\ =6\end{array}$

Therefore, $L.H.S.=R.H.S.$

As long a and b are nonnegative real numbers.

Thus, the equation $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true.

To determine

To state: Whether the equation $\sqrt{a^2+b^2}=a+b$ is true or false.

Answer to Problem 10ADT

The equation $\sqrt{a^2+b^2}=a+b$ is false.

Explanation of Solution

Suppose the values of $a=2\text{and}b=1$.

Substitute the value of $a=2\text{and}b=1$ in the L. H. S. of the equation $\sqrt{a^2+b^2}=a+b$,

$\begin{array}{c}L.H.S.=\sqrt{2^2+1^2}\\ =\sqrt{4+1}\\ =\sqrt{5}\end{array}$

Substitute the value of $a=2\text{and}b=1$ in the R. H. S. of the equation $\sqrt{a^2+b^2}=a+b$,

$\begin{array}{c}R.H.S.=2+1\\ =3\end{array}$

Therefore, $L.H.S.\ne R.H.S.$

Thus, the equation $\sqrt{a^2+b^2}=a+b$ is false.

To determine

To state: Whether the equation $\frac{1+TC}{C}=1+T$ is true or false.

Answer to Problem 10ADT

The equation $\frac{1+TC}{C}=1+T$ is false.

Explanation of Solution

Suppose the values of $T=3\text{and}C=2$.

Substitute the value of $T=3\text{and}C=2$ in the L. H. S. of the equation $\frac{1+TC}{C}=1+T$,

$\begin{array}{c}L.H.S.=\frac{1+3\times 2}{2}\\ =\frac{1+6}{2}\\ =\frac{7}{2}\\ =3.5\end{array}$

Substitute the value of $T=3\text{and}C=2$ in the R. H. S. of the equation $\frac{1+TC}{C}=1+T$,

$\begin{array}{c}R.H.S.=1+3\\ =4\end{array}$

Therefore, $L.H.S.\ne R.H.S.$

Thus, the equation $\frac{1+TC}{C}=1+T$ is false.

To determine

To state: Whether the equation $\frac{1}{x-y}=\frac{1}{x}-\frac{1}{y}$ is true or false.

Answer to Problem 10ADT

The equation $\frac{1}{x-y}=\frac{1}{x}-\frac{1}{y}$ is false.

Explanation of Solution

Suppose the values of $x=3\text{and}\text{y}=2$.

Substitute the value of $x=3\text{and}\text{y}=2$ in the L. H. S. of the equation $\frac{1}{x-y}=\frac{1}{x}-\frac{1}{y}$,

$\begin{array}{c}L.H.S.=\frac{1}{3-2}\\ =\frac{1}{1}\\ =1\end{array}$

Substitute the value of $x=3\text{and}\text{y}=2$ in the R. H. S. of the equation $\frac{1}{x-y}=\frac{1}{x}-\frac{1}{y}$,

$\begin{array}{c}R.H.S.=\frac{1}{3}-\frac{1}{2}\\ =\frac{2-3}{6}\\ =-\frac{1}{6}\end{array}$

Therefore, $L.H.S.\ne R.H.S.$

Thus, the equation $\frac{1}{x-y}=\frac{1}{x}-\frac{1}{y}$ is false.

To determine

To state: Whether the equation $\frac{\frac{1}{x}}{\frac{a}{x}-\frac{b}{x}}=\frac{1}{a-b}$ is true or false.

Answer to Problem 10ADT

The equation $\frac{\frac{1}{x}}{\frac{a}{x}-\frac{b}{x}}=\frac{1}{a-b}$ is true.

Explanation of Solution

As long a and b are nonnegative real numbers.

Simplify the above expression $\frac{\frac{1}{x}}{\frac{a}{x}-\frac{b}{x}}=\frac{1}{a-b}$ as follows,

$\begin{array}{c}L.H.S.=\frac{\frac{1}{x}}{\frac{a}{x}-\frac{b}{x}}\\ =\frac{\frac{1}{x}}{\frac{a-b}{x}}\\ =\frac{1}{x}\times \frac{x}{a-b}\\ =\frac{1}{a-b}\\ =R.H.S.\end{array}$

Thus, the equation $\frac{\frac{1}{x}}{\frac{a}{x}-\frac{b}{x}}=\frac{1}{a-b}$ is true.