MATHEMATICAL
INDUCTION
–
Howtodoi
Steph
the statement
It
true
is
any
is
be
n
shown
that
the
=
initial
value
g
for
n
=
n
=
k
for
which
.
Then
.
k -11
.
statement is
.
statement
the
Assume
:
value
true
value
an
true
is
to
for
Step
initial
Consider
:
prove
is
true
the
for
statement
Example
math induction
Ising
:
”
3
1
is
to
prove
multiple og 2 for
a
–
n
–
that
1,43
–
Steph
For
:
31
–
multiple
a
Step
n
–
It
I
1
=
of
2
,
3
=
Assume
:
k
n
1
=
2
which
is
.
3
”
.
means
–
that
–
Ii
is
371
true
is
for
true
.
,
.
–
Or ,
3k
it
–
means
1
is
that :
a
( It is
–
We
have
3kt
to
‘
–
prove
1
multiple of
assumption )
2
an
that
is
also
a
-_
ktn
a
,
multiple g
?
3kt
‘
I
–
=
=
2
3
=
3k
.
2.3k
+
+
3k
1
?
I
–
( 3k
t
–
)
¯m
/
the
last step
gas.TT?.nuipeega
2.3k is
a
3k
.
multiple og2 ( easy to
3kt
‘
-1
is
a
see
) !
multiple og 2
.
Conclusion
–
mathematical
By
we
3
deduce
that
I
is
multiple
=
A
can
”
–
(
n
induction
,
a
2
,
.
.
.
)
,
og
2
.
questions
–
Think
)
1
+
3
t
5
t
for
2)
induction
mathematical
Using
t
these
about
( ab )
1-
.
.
.
n
7
=
,
( 2n
2
.
2
)
=
n
.
.
”
=
a
”
ab
”
natural
for every
about
talk
we
(
on
,
n
–
:
these
Thursday ) !
is
true
number
examples
n
.
INDUCTION
MATHEMATICAL
–
FIL
1+3+5-1
:
for
Drove
n
1
=
.
.
2
,
+
,
Gn
.
.
11
–
=
n
‘
.
:
–
Steph
For
:
? Step
I
.
–
=
n
a)
1
=
1
:
‘
?
is
It’s correct
satisfied
.
,
Steps
for
n
Then
is
Let us
:
l
t
(
have
3
t
3 tst
It
have
it
is true
.
:
we
statement
k
=
true
we
the
assume
to
.
is
+
.
.
(2k
t
)
=
kid
assumption )
an
that
prove
–
for
n
–
tats ,
:
5 t
t
.
.
.
=
Gk
–
t
( kt , )2
) t
(Hkt l )
–
)
t
it
3
t
5
(2K
+
t
.
.
.
It
hate
3
+
t)
(2 Ckti )
x
( kt 1)
=
We
–
2
)
:
5 +
t
.
.
‘
=
.
.
Gk
1) + (21kt D
–
k
t
@
Ck I
–
1
be
+
be
+2k
2k
+
f- 2
I
)
–
=
–
?
–
? from the assumption
=
t
?
–
=
–
1)
g
n
–
I
( lat 1)
2
.
1)
)
k
then
is
i
satisfied
By
t
3 t
have
l
t
–
(Zcktn )
t
r
–
7)
=
Chat
.
mathematical
we
.
induction
,
:
3 t 5
t
:
.
.
kn
–
t
)
=
n
?
ME
End
Cab )
Drove
:
”
”
=
a
b
”
is true
every
natural
for
Sd
member
n
.
:
Steph
for
:
(ab )
Hence
n
–
‘
=
step
I
:
at .b
1
”
is
=
ab
satisfied
.
Step
for
Iet
:
n=
k
.
It
(ab )
the statement
assume
us
means
k
=
that
ah bk
is true
:
is
.
true
(Itis
an
we have
to
( ab )
htt
Indeed
or
prove
=
🙁 abyk
Cobh Cab )
.
Cab )
htt
=
=
that :
akt ‘
=
=
bkt it
.
also
ok bk
( ah bk ) Gb )
.
.
.
.
by ab )
.
-11
bk
(
)
holds
(Multiplying
both sides
Ca ? a) (bk b )
( ah -11 )
assumption )
=
aktl bktl
.
Then
step
Then
by
statement
(ab )
2
proved
is
induction
math
”
”
=
every
a
–
b
.
,
the
”
natural
is
true
number
for
on
.
Strong
induction
:
another
is
?
inductor
‘s
,
Pcn )
To
prove
form
of
mathematical
:
all
positive integers
for
the
following steps
using
that
It
? Steph
( Base step )
proves
the initial
PG )
proposition
is
true
n
,
:
:
:
true
? Steph :
the
condition
(Inductive step)
:
It
statement
[ PG ) n p (2) A PG ) A
.
.
.
Pcb)
.
proves
that
paste) is
) for positive
integers
?
true
k
.

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