CK-12-Pre-Calculus Concepts

http://www.ck12.org Chapter 12. Discrete Math

Base Case(s):Two base cases are shown however only one is actually necessary.

13 = 12

13 + 23 = 1 + 8 = 9 = 32 = ( 1 + 2 )^2

Inductive Hypothesis:Assume the statement is true for some numberk. In other words, assume the following is
true:
13 + 23 + 33 +···k^3 = ( 1 + 2 + 3 +···k)^2
Proof:You want to show the statement is true fork+1. It is a good idea to restate what your goal is at this point.
Your goal is to show that:
13 + 23 + 33 +···k^3 +(k+ 1 )^3 = ( 1 + 2 + 3 +···k+(k+ 1 ))^2
You need to start with the assumed case and do algebraic manipulations until you have created what you are trying
to show (the equation above):
13 + 23 + 33 +···k^3 = ( 1 + 2 + 3 +···k)^2
From the work you have done with arithmetic series you should notice:
1 + 2 + 3 + 4 +···+k= 2 k( 2 +(k− 1 )) =k(k+ 21 )
Substitute into the right side of the equation and add(k+ 1 )^3 to both sides:

13 + 23 + 33 +···k^3 +(k+ 1 )^3 =

(k(k+ 1 )

2

) 2

+(k+ 1 )^3

When you combine the right hand side algebraically you get the result of another arithmetic series.

13 + 23 + 33 +···k^3 +(k+ 1 )^3 =

((k+ 1 )(k+ 2 )

2

) 2

= ( 1 + 2 + 3 +···k+(k+ 1 ))^2
∴
The symbol∴is one of many indicators like QED that follow a proof to tell the reader that the proof is complete.
Concept Problem Revisited
If you forget the base case in an induction proof, then you haven’t really proved anything. You can get silly results
like this “proof” of the statement: “1=3”
Base Case:Missing
Inductive Hypothesis:k=k+1 wherekis a counting number.
Proof:Start with the assumption step and add one to both sides.

k=k+ 1

k+ 1 =k+ 2

Thus by transitivity of equality:

k=k+ 1 =k+ 2

k=k+ 2

Sincekis a counting number,kcould equal 1. Therefore:
1 = 3