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MATHEMATICAL INDUCTION
Suppose we want to prove a mathematical statement
A
(
n
) which is formulated
for every natural number
n
∈
N
, i.e. n=1,2,3,4,.
...
One very often uses the following method (referred to as “mathematical induc-
tion”, or simply “induction”) to verify that the statement is true
for all
n
∈
N
.
(i) One proves the statement for
n
= 1
, i.e.
A
(1)
.
(ii) One shows that for every
n
the statement
A
(
n
)
implies the statement
A
(
n
+1)
(in other words, for every
n
it is true that if
A
(
n
)
holds then
A
(
n
+ 1)
holds too).
If we have checked both (i) and (ii) then we know that
A
(2) is true (namely
A
(1)
is true by (i), and therefore, by (ii)
A
(2) is also true). But then
A
(3) is also true
(we have just seen that
A
(2) is true and by (ii) it implies
A
(3)). Etc. etc.
Remark:
The induction does not have to begin with
n
= 1. Suppose we know
that
A
(4) is true and that for all
n
≥
4 the statement “
A
(
n
) implies
A
(
n
+ 1)”
holds then we can conclude that
A
(
n
) is true for all
n
≥
4.
1.
Examples
Example 1.1.
Let’s prove for
n
= 1
,
2
, . . .
the statement
A
(
n
) :
n
X
k
=1
k
=
n
(
n
+ 1)
2
,
In english: the sum of the ﬁrst
n
natural numbers (beginning with 1) is equal to
n
(
n
+1)
2
.
To verify
A
(1) just note that
∑
1
k
=1
= 1 =
1
·
(1+1)
2
.
Now we need to check that for all
n
the truth of
A
(
n
) implies the truth of
A
(
n
+ 1). Fix
n
. We write
n
+1
X
k
=1
k
= (
n
X
k
=1
k
) + (
n
+ 1)
Since we are assuming the truth of
A
(
n
) (for our ﬁxed
n
) we have
∑
n
k
=1
k
=
n
(
n
+1)
2
and we can use this in the last displayed formula. Thus
n
+1
X
k
=1
k
=
n
(
n
+ 1)
2
+ (
n
+ 1) = (
n
+ 1)
±
n
2
+ 1
²
=
(
n
+ 1)(
n
+ 2)
2
and this yields
A
(
n
+ 1).
To summarize we have shown
A
(1) and for all
n
we have shown that
A
(
n
) implies
A
(
n
+ 1). Thus, by mathematical induction it follows that the assertion holds for
all
n
.
±
1

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Example 1.2.
Let’s prove the following statement:
For all
n
∈
N
the inequality
n <
2
n
holds.
We label this statement
A
(
n
).
Clearly
A
(1) holds (namely we have 1
<
2 = 2
1
).
Now let’s ﬁx
n
and show that the validity of
A
(
n
) implies the validity of
A
(
n
+1).
This follows from
n
+ 1
<
2
n
+ 1
<
2
n
+ 2
n
= 2
·
2
n
= 2
n
+1
.
where in the ﬁrst inequality we have used the induction hypothesis (i.e. the assumed
truth of
A
(
n
)) and in the second inequality the obvious statement that 1
<
2
n
for
all
n
= 1
,
2
, . . .
).
By mathematical induction it follows that

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