Section 2.2 Induction
Proof by induction:
to prove p(n) true, n
Idea: like playing the domino game.
Suppose dominos are placed correctly, then hitting the
1st domino, when it falls, we know the rest of them will
also fall.
( the 2nd will fall, in general, when the kth one falls, it
implies the (k+1)th falls. Since k is any arbitrary number,
actually every domino falls.)
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Induction:
Step1: (inductive base) or IB is to show that
p(1) true
Step2: (inductive step) or IS is to show that
p(k)  p(k + 1),
where p(k) is called inductive hypothesis or IH.
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e.g. show 1+3+5+…(2n-1) = n2, n  1
IB:
when n = 1, LHS = 1 = 12 = RHS
IS:
Assume 1+3+5+….(2k-1)=k2 and want to
Show 1+3+5+…(2k-1) + [ 2(k+1)-1]
=(k+1)2
1+3+5+…(2k –1) + (2k + 1) = k2 + 2k+ 1
k2 (by IH)
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= (k +1)2
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e.g.
n(n+1)
Show 1 + 2 + 3 + … n =
2
IB:
when n = 1, LHS = 1 =
IS:
Assume 1 + 2 + 3 + …k = k(k+1) and
,n1
1(1+1) = RHS
2
2
want to show 1 + 2 + 3 + …+ k + (k+1) = (k+1)(k+2)
2
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1 + 2 + 3 + …+ k + (k+1) = k(k+1) + (k+1)
2
k(k+1) (by IH)
k(k+1) + 2(k+1)
2
=
2
=
by direct proof:
(k+1) (k+2)
2
Let x =
1+
2 +
+) x =
k + k-1 + k-2 + … + 1
2x = (k+1)+(k+1)+
3 + …+ k
…
+(k+1)
k of them
 x = k(k+1)
2
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eg: Show 1+ 21+ 22 + … + 2n = 2n+1 – 1,  n 
IB:
when n = 1, LHS = 1+ 2 = 22 - 1 = RHS
IS:
Assume 1 + 2 + 22 + … 2k = 2k+1 – 1, and
Show 1 + 2 + 22 + … 2k + 2k+1 = 2k+2 – 1

1 + 2 + 22 + … 2k + 2k+1 = 2k+1 – 1 + 2k+1
2k+1– 1
(by IH)
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= 2*2k+1 – 1
= 2k+2 – 1
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eg:
Show 22n – 1 is divisible by 3,  n  1
IB:
when n = 1, LHS = 22 – 1 = 3 divisible by 3
IS:
Assume 22k – 1 is divisible by 3
(22k – 1 = 3m for some integer m),
and want to show 22(k+1) – 1 is divisible by 3.
 22(k+1) – 1 = 22k+2 – 1 = 22 * 22k – 1
= 22(3m+1) – 1 ( 22k – 1 = 3m
22k = 3m + 1)
=12m + 3
= 3(4m + 1)
 is divisible by 3
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where 4m + 1 is an integer
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1
1
1
n
eg : Show

 ... 

, n  1
1* 2 2 * 3
n(n  1) n  1
1
1
IB : when n  1, LHS 

 RHS
1* 2 1  1
1
1
1
k
IS : Assume

 ... 

and want to
1* 2 2 * 3
k (k  1) k  1
1
1
1
1
k 1
show

 ... 


1* 2 2 * 3
k (k  1) (k  1)( k  2) k  2
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1
1
1
1
k
1


 ... 



1* 2 2 * 3
k (k  1) (k  1)( k  2) k  1 (k  1)( k  2)
k__
k+1 (by IH)
k (k  2)  1

(k  1)( k  2)
(k  1)( k  1)

(k  1)( k  2)
(k  1)

(k  2)
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