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The program initializes f to 1 and then multiplies it by the input number n and all the positive integers that are less than n. So 5! = (5)(4)(3)(2)(l) = 120, and S! = (8)(7)(6)(5)(4)(3)(2)( 1) = 40,320.

Reasons 1. Given 2. A geometric figure may be moved without changing its size or shape. Equal lines may be made to coincide. 3. Given 4. The whole equals the sum of its parts. 5. The sides of a straight angle lie in a straight line. 6. Given 7. If two sides of a triangle are congruent, the angles opposite these sides are congruent. 8. SAA.

A loop is controlled by three separate parts: an initialization, a continuation condition, and an update. For example, in the program in Example 3.3, the loop control variable is n; its initialization is c in >> n, its continuation condition is n > 1, and its update is n- -. When these three parts are simple, the loop can be set up as a for loop, which is usually simpler than its equivalent while loop and do. . . while loop. The syntax for the for statement is

Given: Circle O, diameter AB ' CD To Prove: CE > ED, BC > BD, AC > AD Plan: Congruent triangles are formed when radii are drawn to C and D, proving CE > ED. Equal central angles are used to prove BC > BD; then the Subtraction Postulate is used to prove AC > AD.

EXAMPLE 3.4 Sum of Squares Again This program has the same effect as the one in Example 3.2:

PROOF:

int n, sum = 0; tout << "Enter a positive integer: "; tin >> n; for (int i = 1; i <= n; i++) sum += i*i; tout << 'The sum of the first ' << n << ' squares is ' << sum << endl;

Here the initialization is int i = 1, the continuation condition is i < = n, and the update is i + +.

3.33 (a) (b)

Statements 1. Draw OC and OD. 2. 3. 4. 5. 6. OC > OD AB ' CD /OEC and /OED are right angles. OE > OE ^OEC > ^OED

It is customary to localize the declaration of the control variable in the initialization part of a

Reasons 1. A straight line may be drawn between two points. 2. Radii of a circle are congruent. 3. Given 4. Perpendiculars form right angles. 5. Reflexive property 6. hy-leg 7. Corresponding parts of congruent triangles are congruent. 8. In a circle, congruent central angles have congruent arcs. 9. A diameter bisects a circle. 10. In a circle, congruent arcs are equal arcs; Subtraction Postulate

for loop. For example, the control variable i in the program above is declared to be an int within the initialization part in t i = 1. This is a nice feature of C++. However, once the control variable is declared this way, it should not be redeclared in a later for loop. For example, for (int i = 0; i < 100; i++) sum += i*i; for (int i = 0; i c 100; i++) // ERROR: i has already been declared tout CC i*i*i;

6. An angle inscribed in a circle is measured by one-half its intercepted arc. Case I: The center of the circle is on one side of the angle.

The same control variable can be used again; it just cannot be redeclared:

Given: /A is inscribed in circle O. O is on side AC. To Prove: A/5 1BC 2 Plan: When radius OB is drawn, isosceles ^AOB is formed. /A is proved to be equal in measure to one-half central /1, which is measured by BC.

for (i = 0; i < 100; i++) // OK tout CC i*i*i; If you have the choice between a for loop and a while or do. . while loop, you should probably use the for loop. As the next example illustrates, a for loop is usually easier to

Compare this program with the one in Example 3.3:

PROOF:

main0 int n, f = 1; tin >> w tout << "Enter a positive integer: for (int i = 2; i <= n; i++) f *= i; tout << n CC fl factorial is ' CC f CC endl;