Induction
2/24/12
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The Idea of Induction
Color the integers ≥ 0
0, 1, 2, 3, 4, 5, …
I tell you, 0 is red, & any int
next to a red integer is red,
then you know that
all the ints are red!
2/24/12
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Induction Rule
R(0)
and (" n)(R(n)fi R(n+1))
R(0),("
R(1m
), R(2),…,R(n),…
)R(m)
2/24/12
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Like Dominos…
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Second level
Third level
Fourth level
Fifth level
Example Induction Proof
Let’s prove:
2
1+r +r +
(for r ≠ 1)
n
+r
=
(n+
1)
r
-1
r -1
Example Induction Proof
Statements in magenta form a
template for inductive proofs:
•
•
Proof: (by induction on n)
The induction hypothesis, P(n), is:
1+r +r2 +L +r n =
(for r ≠ 1)
(
n+
1
)
r
-1
r -1
Example Induction Proof
Base Case (n = 0):
? r 0+1 - 1
2
0
1+r +r +L +r =
r -1
1
OK!
r -1
=
=1
r -1
Example Induction Proof
• Inductive Step: Assume P(n) for some n ≥ 0
and prove P(n+1):
2
1+r +r +L +r
n+1
r
=
(n+1)+1
-1
r -1
Example Induction Proof
Now from induction
hypothesis P(n) we have
r
1+r +r +L +r =
2
so add r
n
n+1
n+1
-1
r -1
to both sides
Example Induction Proof
adding r
n+1
to both sides,
r - 1 n+1
( 1+r +r +L +r ) +r = r - 1 ÷+r
n+1
n+1
r - 1+r (r - 1)
This proves
=
r -1
P(n+1)
2
n
completing the
proof by induction.
n+1
r
=
n+1
(n+1)+1
-1
r -1
an aside: ellipsis
“” is an ellipsis.
Means you
should see a pattern:
2
1+r +r +L +r
n
• Can lead to confusion (n = 0?)
• Sum notation more precise
n
=∑ r
i=0
i