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