Chapter 4, Problem 6RE

Explanation of Solution

 Table:

a	b	Math.pow(a,b)	Math.max(a,b)	a/b	a%b	Math.floorMod(a,b)
2	3	8	3	0	2	2
3	2	9	3	1	1	1
2	-3	0.125	2	0	2	-1
3	-2	0.111	3	-1	1	-1
-3	2	9	2	-1	-1	1
-3	-2	0.111	-2	1	-1	-1

 Explanation:

 Given that “a” and “b” are two integer type values.

 Math.pow(a,b):

• The “Math.pow” is a power function used in mathematics to return the value of “x” to the power of “y”.
• The first argument in the pow() function is the base whereas second argument is its power.
• This power function is included in the math library in Java.
• Its syntax is as follows:

      Math.pow(base, number)

 When a=2, b=3;

     $\begin{array}{l}Math.pow\left(2,3\right)=2^3\\ =2*2*2\\ =8\end{array}$    

 When a=3, b=2;

     $\begin{array}{l}Math.pow\left(3,2\right)=3^2\\ =3*3\\ =9\end{array}$

 When a=2, b=-3;

     $\begin{array}{l}Math.pow\left(2,-3\right)=2^{-3}\\ =\frac{1}{2^3}\\ =\frac{1}{2*2*2}\\ =\frac{1}{8}\\ =0.125\end{array}$    

 When a=3, b=-2;

     $\begin{array}{l}Math.pow\left(3,-2\right)=3^{-2}\\ =\frac{1}{3^2}\\ =\frac{1}{3*3}\\ =\frac{1}{9}\\ =0.111\end{array}$    

 When a=-3, b=2;

     $\begin{array}{l}Math.pow\left(-3,2\right)={(-3)}^2\\ =-3*-3\\ =9\end{array}$

 When a=-3, b=-2;

     $\begin{array}{l}Math.pow\left(-3,-2\right)={(-3)}^{-2}\\ =\frac{1}{{(-3)}^2}\\ =\frac{1}{-3*-3}\\ =\frac{1}{9}\\ =0.111\end{array}$   

 Math.max(a,b):

• The “Math.max()” function is used to display the largest of zero or more numbers.
• It means that it returns the number with the highest value.

 When a=2, b=3;

     $Math$