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TRANSCRIPT
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Mathematical Induction I
Lecture 4: Sep 16
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Proving For-All Statements
Objective: +rove
It is ver common to prove statements of this form. Some ,*amples:
-or an odd number m' mi is odd for all nonne#ative inte#er i.
)n inte#er n / 1 is divisible b a prime number.
&0auchSchwar inequalit( -or an a1'2'an' and an b1'2bn
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Universal Generalization
( )
. ( )
A R c
A x R x
→
→∀valid rule providin# c is independent of )
3ne wa to prove a forall statement is to prove that &c( is true for an c'but this is often difficult to prove directl
&e.#. consider the statements in the previous slide(.
Mathematical induction provides another wa to prove a forall statement.
It allows us to prove the statement step-by-step.
Let us first see the idea in two e*amples.
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O Po!ers Are O
-act: If m is odd and n is odd' then nm is odd.
+roposition: for an odd number m' mi is odd for all nonne#ative inte#er i.
Let +&i( be the proposition that m i is odd.
$ +&1( is true b definition.$ +&5( is true b +&1( and the fact.
$ +&( is true b +&5( and the fact.
$ +&i71( is true b +&i( and the fact.
$ So +&i( is true for all i.
Idea of induction.
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"ivisibility by a Prime
Theorem# )n inte#er n / 1 is divisible b a prime number.
Idea of induction.
$Let n be an inte#er.
$If n is a prime number' then we are done.
$3therwise' n 8 ab' both are smaller than n.
$If a or b is a prime number' then we are done.
$3therwise' a 8 cd' both are smaller than a.
$If c or d is a prime number' then we are done.
$3therwise' repeat this ar#ument' since the numbers are
#ettin# smaller and smaller' this will eventuall stop and
we have found a prime factor of n.
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3b9ective: +rove
$ea o% $nuction
This is to prove
The idea of induction is to first prove +&( unconditionall'
then use +&( to prove +&1(
then use +&1( to prove +&5(
and repeat this to infinit2
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The $nuction &ule
and &from n to n 71('
proves ' 1' 5' '2.
+ &(' ∀n ∈; + &n (+ &n 71(
∀m ∈; + &m (
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This Lecture
$ The idea of mathematical induction
$ %asic induction proofs &e.#. equalit' inequalit' propert'etc(
$ Inductive constructions
$ ) parado*
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Proving an '(uality
Let +&n( be the induction hpothesis that the statement is true for n.
%ase case: +&1( is true
Induction step: assume +&n( is true' prove +&n71( is true.
because both L>S and >S equal to 1
That is' assumin#:
?ant to prove:
This is much easier to prove than provin# it directl'
because we alread =now the sum of the first n terms@
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Proving an '(uality
Let +&n( be the induction hpothesis that the statement is true for n.
%ase case: +&1( is true
Induction step: assume +&n( is true' prove +&n71( is true.
b induction
because both L>S and >S equal to 1
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Proving an '(uality
Let +&n( be the induction hpothesis that the statement is true for n.
%ase case: +&1( is true
Induction step: assume +&n( is true' prove +&n71( is true.
b induction
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Proving a Property
%ase 0ase &n 8 1(:
Induction Step: )ssume + &i ( for some i ≥ 1 and prove + &i 7 1(:
)ssume is divisible b ' prove is divisible b .
Aivisible b b inductionAivisible b
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Proving a Property
%ase 0ase &n 8 5(:
Induction Step: )ssume + &i ( for some i ≥ 5 and prove + &i 7 1(:
)ssume is divisible b 6
is divisible b 6.
Aivisible b 5b case analsis
Aivisible b 6b induction
+rove
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Proving an $ne(uality
%ase 0ase &n 8 5(: is true
Induction Step: )ssume + &i ( for some i ≥ 5 and prove + &i 7 1(:
b induction
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)auchy-Sch!arz
&0auchSchwar inequalit( -or an a1'2'an' and an b1'2bn
+roof b induction &on n(: ?hen n81' L>S B8 >S.
?hen n85' want to show
0onsider
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)auchy-Sch!arz
&0auchSchwar inequalit( -or an a1'2'an' and an b1'2bn
Induction step: assume true for B8n' prove n71.
induction
b +&5(
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This Lecture
$ The idea of mathematical induction
$ %asic induction proofs &e.#. equalit' inequalit' propert'etc(
$ Inductive constructions
$ ) parado*
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Gray )oe
0an ou find an orderin# of all the nbit strin#s in such a wa thattwo consecutive nbit strin#s differed b onl one bitC
This is called the Dra code and has some applications.
>ow to construct themC Thin= inductivel@
5 bit
1111
bit
11111111111
1
0an ou see the patternC
>ow to construct 4bit #ra codeC
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Gray )oe
bit
111
111111111
bit &reversed(
111111
111111
111111111111111
111111111
11
111111
4 bit
differed b 1 bitb induction
differed b 1 bitb induction
differed b 1 bitb construction
,ver 4bit strin# appears e*actl once.
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Gray )oe
n bit
222
222212
n bit &reversed(
1222
22222
222222212
11
111111
n71 bit
differed b 1 bitb induction
differed b 1 bitb induction
differed b 1 bitb construction12
2
222222
So' b induction'
Dra code e*ists for an n.
,ver &n71(bit strin# appears e*actl once.
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Doal: tile the squares' e*cept one in the middle for %ill.
n2
n
2
Puzzle
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There are onl trominos &Lshaped tiles( coverin# three squares:
-or e*ample' for E * E pule mi#ht tile for %ill this wa:
Puzzle
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Theorem: -or an 5n * 5n pule' there is a tilin# with %ill in the middle.
+roof: &b induction on n (
+ &n ( ::8 can tile 5n * 5n with %ill in middle.
%ase case: &n 8(
&no tiles needed(
Puzzle
&Aid ou remember that we proved is divisble b C(
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n
2
Induction step: assume can tile 5n
* 5n
' prove can handle 5n7 1 * 5n7 1.
15 7n
Puzzle
FowwhatCC
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Puzzle
$ea* It would be nice if we could control the locations of the empt square.
n2
n2
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"one+
Puzzle
$ea* It would be nice if we could control the locations of the empt square.
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The ne! iea*
+rove that we can alwas find a tilin# with %ill any!here.
Puzzle
Theorem %: -or an 5n * 5n plaa' there is a tilin# with %ill anwhere.
Theorem: -or an 5n * 5n plaa' there is a tilin# with %ill in the middle.
0learl Theorem % implies Theorem.
) stron#er propert
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Theorem %: -or an 5n * 5n plaa' there is a tilin# with %ill anwhere.
+roof: &b induction on n (
+ &n ( ::8 can tile 5n * 5n with %ill anwhere.
%ase case: &n 8(
&no tiles needed(
Puzzle
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Induction step:
)ssume we can #et %ill anwhere in 5n * 5n .
+rove we can #et %ill anwhere in 5n 71 * 5n 71.
Puzzle
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Puzzle
Induction step:
)ssume we can #et %ill anwhere in 5n * 5n .
+rove we can #et %ill anwhere in 5n 71 * 5n 71.
n2
n2
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Method: Fow #roup the squares to#ether'
and fill the center with a tile.
"one+
Puzzle
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Some &emar,s
ote .: It ma help to choose a stron#er hpothesis
than the desired result &e.#. G%ill in anwhereH(.
?e need to prove a stron#er statement' but in
return we can assume a stron#er propert in
the induction step.
ote /: The induction proof of G%ill anwhereH implicitl
defines a recursive al#orithm for findin# such a tilin#.
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0aamar 1atri2
0an ou construct an n*n matri* with all entries 71 and
all the rows are ortho#onal to each otherC
Two rows are ortho#onal if their inner product is ero.
That is' let a 8 &a1' 2' an( and b 8 &b1' 2' bn('
their inner product ab 8 a1b1 7 a5b5 7 2 7 anbn
This matri* is famous and has applications in codin# theor.
To thin= inductivel' first we come up with small e*amples.
1 1
1 1
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0aamar 1atri2
Then we use the small e*amples to build lar#er e*amples.
Suppose we have an n*n >adamard matri* >n.
?e can use it to construct an 5n*5n >adamard matri* as follows.
>n >n
>n >n
So b induction there is a 5= * 5= >ardmard matri* for an =.
It is an e*ercise to chec= that the rows are ortho#onal to each other.
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$nuctive )onstruction
This technique is ver useful.
?e can use it to construct:
codes
#raphs
matrices
circuits
al#orithms
desi#ns
proofs
buildin#s
2
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This Lecture
$ The idea of mathematical induction
$ %asic induction proofs &e.#. equalit' inequalit' propert'etc(
$ Inductive constructions
$ ) parado*
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Parao2
Theorem: )ll horses have the same color.
+roof: &b induction on n (
Induction hpothesis:
+ &n ( ::8 an set of n horses have the same color
%ase case &n 8(:
Fo horses so obviousl true@
…
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&Inductive case(
)ssume an n horses have the same color.
+rove that an n7 1 horses have the same color.
Parao2
…n+1
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2
-irst set of n horses have the same color
Second set of n horses have the same color
&Inductive case(
)ssume an n horses have the same color.
+rove that an n7 1 horses have the same color.
Parao2
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2
Therefore the set of n 71 have the same color@
&Inductive case(
)ssume an n horses have the same color.
+rove that an n7 1 horses have the same color.
Parao2
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?hat is wron#C
+roof that + &n ( → + &n 71(
is false if n 8 1' because the two
horse #roups do not overlap .
-irst set of n8 1 horses
n 81
Second set of n8 1 horses
Parao2
&%ut proof wor=s for all n 3 1(
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4uic, Summary
ou should understand the principle of mathematical induction well'
and do basic induction proofs li=e
$ provin# equalit
$ provin# inequalit
$ provin# propert
Mathematical induction has a wide ran#e of applications in computer science.
In the ne*t lecture we will see more applications and more techniques.