Solution for Section 8.4.2

Solution for Section 8.4 Question 2

2. You might like to refer to Example 8.4.5 on pages 469 and 470 of your textbook.

Recall that to show a statement P(n) is true by mathematical induction, you must show it is true for n = 1 (or some other basis value), then you assume that the statement is true for n = k, and finally you must show that the statement is true for n = k+1.

Let P(n) be the statement that for any positive integer n, if a₁, a₂, ..., a_n and b₁, b₂, ..., b_n are real numbers then:

n	(a_i · b_i)	=	n	a_i	·	n	b_i.
P			P			P
i=1			i=1			i=1

Suppose that a₁ and b₁ are real numbers. Then by the definition of product, P :

1	(a_i · b_i)	=	a₁ · b₁	=	1	a_i	·	1	b_i
P					P			P
i=1					i=1			i=1

So the statement P(n) is true for n = 1.

Now assume that P(n) is true for n = k. That is, we assume that the following is true:

k	(a_i · b_i)	=	k	a_i	·	k	b_i.
P			P			P
i=1			i=1			i=1

Now you must show that the statement is true for n = k+1.

k+1

(a_i · b_i)

k	(a_i · b_i)	·	(a_k+1 · b_k+1 )
P
i=1

by definition of P

i=1

k	a_i	·	k	b_i	·	(a_k+1 · b_k+1 )
P			P
i=1			i=1

by inductive hypothesis

k	a_i	·	a_k+1	·	k	b_i	·	b_k+1
P					P
i=1					i=1

by associative and commutative laws

k+1	a_i	·	k+1	b_i
P			P
i=1			i=1

by defintion of P

So the statement P(n) is true for n = k+1.

Therefore, for any positive integer n, if a₁, a₂, ..., a_n and b₁, b₂, ..., b_n are real numbers then:

n	(a_i · b_i)	=	n	a_i	·	n	b_i.
P			P			P
i=1			i=1			i=1

Back to Section 8.4