today : Proof by Contradiction
Induction
Ex_ampk: Roof by Contradiction
There are infinitely many Prime numbers
Translation : the IN FPEIN ( Prime Cp) ^ p> n
. )
.Discussion we want to prone 5.
Instead we 'll assume - S and
get a contradiction.
A Proof By contradiction is a specialcase of the
cmtrapositiiliSay he want to prone S
This is equivalent to True → s.
To prove Tme→S,
the conhaposite form
.
is ns → False-
a back to proving the existence of
infinitely many primes bycondition
S : then ⇒ pen ( Prime Cp) ^ p > no )
7 s : ^ V. not N FPEIN ( Prime ( Pnp > no )
=,
Fnoe IN tpe IN 1. Prime ( p) v
pen. )
; znoe N V. pen ( Prime (p) ⇒
p<no )
For sake of contradiction ,
Assume 'S.
( Nod we are Not proving is )let no EHV be such that HPEN ( Prime g) ⇒p£no)Let P= { p, R . - Pe } be the ( finite ) set
of all primes.
- all primes pathatgn.
Let P* =p ,
. Pi . .. epe + 1
We want to show : Prime (pty andp*
> no
to get a contradiction
P = { D, ,
...
,Pe } = set of all Prime numbers
4 that are ← no
@ Show poets Prime ← Why ?
We 'll come back to this later.
�2� Show P* > no showp*=(pipi . .pedtl> no
Assume for sake of contradiction p*= no ,
andp* is Prime
Then p* E P.
But it cant be
since for eery pie 8,
Pi< Pi . . :petl = p*
contradiction
.
Q= Lets try p* =p,
tpzt . . a pet I instead .
why doesn't this work ?'
Let no = 4
As P s { 2,3 }
2+3+1 = 6,
so Not Prime
Comment To see p* =P,
' Pi . -
' Pet I is Prime,
it . is important that P={ p ,. . pig are ay primes e no
fxampkzy Roof by contradiction ( informal proof )
if is irrational [ cannot be written as F,
a. bez ]
Assume that @ is rational .( for sake of contraction )
Then Vz =
Pg ,
where we can assume p, q
trace
No common divisors
then 2 = Pga
Sop2=2q2
,
so 15 and therefore p must
be even
But then 4 divides p2 - - 15=4 . K for some K
But then q2= Pg = 4€ = 2k Ts even,
so q'
is een
But this contradicts our assumption that
P , q had no common divisors !
inductive hugely Importantworkhorse behind most proofsversatile
,fun
n A great part of 1 higher arithmetic ) derives an
additional charm from the peculiarity of
important propositions ,
with the impress of
simplicity on them ,are often easily discovered
by induction,
& yet are of so profound a
character that we cannot find the demonstrations
til after manyrain actempts ,
a even then,
when we do succeed,
it is often by some
tedious or artificial process ,while the simple
methods may long remain concealed.
"
- Carl Friedrich gauss
"Science in its ultimate ideal consists of a set
of propositions arranged In a hierarchy,
the lowest
level of the hierarchy being concerned with
particular facts,
a the highest with some
general Law, governing elrythiry in the universe
.
The various levels In the hierarchy had a
2 fold logical connection ; the upwardconnection proceeds by induction
,the downward
by deduction"
- Bertrand Russell
Induction is usually used to provea
" for all " statement : then Pcn )
It is used when a direct proof is Not
obvious or possible
Idea is Like dominoes
TF← base case
← induction *
y KEN Pad ⇒P(k+Dstep
then can conclude th PCD. Say n=ioE¥¥I⇒EIii¥¥⇒#¥t
£so 3 so 4 So 10
one true true the
Example 1 - lt2t3t . ... + n
- o
Prove then §,i =ncntd
2
Discussion :
-Here
.
is a very clever argumentthat doesn't use induction
.
Say n= 15 .We want to show 1+2+3 + . - +15
= 15-(16)
Add up the numbers from 1 to 15 twice like this ;
2
It 2 + 3 + 4 t . . . - + 14 t 15
+ 15+14 t 13 +12 + . .t 2 + 1
.
+ 16= (6+16 t . . .
= ( (6) 15
Then we had to divide by 2 since he added them twice.
So d§gi = 451¥ .
same argument works
for all n.
Example 1 - Now let's PNU it by induction !
-
It won't require
Prove then a§i = n£ntI any clever idea
Let Pln ) .If,i = newItopwuun.CN#nducfmnn .
i. Base case PG )
:tsi
=o (esrutm ).
2. Jnductiu step .
Poore HKHNPCK) ⇒ PCKTD
Let KEN .Assume PCK) - ie
.
That g.ly i = Ktla
Now we want to prod PCKTD.
.
qq.net#IEE.YmmaEf⇐ I =£i+ Hts )
e-
is ) is )
= K¥+1 t ( Kti ) byinductee
'
hypothesis
= K[tIk +2CkztD=kCKtD[2CktD_ = (14*42)
Here -
is a slightly different way to
prod the inductive step ,this time
starting with inductee hypothesis :
gkzgi = KEY Ynhtuctiehyp'
added kt '
i. fly it Ctai ) = KIKI +( kti ) [ to both sides ]
Yeti= KEH +
a " ) finals,ii§inaf-
£+1 i =
C k¥+4 [ rearranging ]i= , 2
Induction
Statement to prow : then Pcn )
What we actually prove :
Plo ) ^ ( V. Ken Pck ) ⇒ p(k+ ,) ) } thenby' Maude
induction
- - then Pln)BASE INDUCTION STEP
CASE
ProofBase case : Let n=o
.
Want to pnupco )
Induction step : Let KEN and assume PCK)want to prove Pact ) )
-
INDUCTION
HYPOTHESIS
VARIATIONSBASEcase doesn't hau to be n=o Pcn )
-
Examfk Prove the IN n > 3 ⇒ 2^+1<2"
this is a proof by induction where base case is njf
Base Case : Show PC3 )3
show 2.3+1 = 7 < 8et 2
Inducted step : Let KEN,
1<=3.
Assume PCK) : 21<+1 < 2kKtl
We want to prove PCKH ) : 2CktDt1 < 2-
*2CKtD+1 = (2Ktl ) t 2
< 2kt 2 by and . hypothesis ¥+ zk
I 2kt 2k since K 's 3
= 2kt l.
Except Prone tx, ye ZL
'
,
the IN
six.
g)⇒ 51
#. yn ) 5- eeeuuaknta
then the ,yeI
* can't do induction on ZL. 51x.gs/xn@
Proof Let KYEZ .We want to poorInfidelityc-
payWeill prou this by induction No fall variables
Let Pcn ) :
Skx.
g)⇒
skxnyn)
in here
Basecasie Prove Plo )
want to prove
5k¥⇒
511×0. y° )
-
This holds since
5*070=0,
and go [ 510 " neo
.s=o]
Let KEN . assume PCK ) :
SK.
g)⇒
s|(xk. YK ) Indactnestep
want to poor :
Sky) ⇒
skxkttykt)← PCKTD
Assume Slx .
y .Let ce R such that
SET)
By Jndhyp 51×1'. y
's Letc
'eKsuch that
5.ci#yk* can write x
" '-
y" '
add a subtract same
= x" "
.
x*y+ xicy. yk
" [ quantity )= x
"( × . g) + y( ×
" .yk ) [ rewriting ]
=p .sc *y.sc' [ West
.ee#darIde9d=xk.yif=5(xk.c+Yc
' ) [ rewriting] .
:"
= xkc + yc'
.C
' 'EZL
Then 5K¥' '. ykt
') since 5. c' '
= xkt '. ykt
'
.
.
Exa=pk Prone t x. ye ZL,
then
six. g)⇒sllxn . yn )
t.t.ve#analternathewayfopnuth_Pnu:VneNPCn
)
where Now pcn ) : tqye Z SIX .
y ⇒ Slxn .ynPG ) : txyek �5�H . y⇒ 51×0 . y°
Let x. yek . then x°-y°=o so 51×0 . y°Inductee step ! Pml HKEINPCK ) ⇒ PCKTD
Let Keim.
Assume PCK ) : the ,yez slxg ⇒ 51×1'
. y's
want to prod PCKTD : thgyez slx .
y ⇒ s/×'s "
.
y't 1
assumed PCK )tx.ge#s1x.y ⇒ slxk .
y's
#want to prove Pckti ) : tx
, ye ZL SH .
y ⇒ SKH '. yktl
andlet TyeKg-
Assume SIX -
y .
So FCEZC such that 5. c=x .
yKt I
yet 1. yttl = ×
" '- XKY + ¥ Y
-Ypyassumption a a sexy
= XKCX - g) + y(×K .
y' ) and by ind . hypothesis
-
=x's
.es+ g.as
<) k¥544347
So sc'
Sc '=*''-yk
} rest similar to before
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