HWI solutiont.si) p Q ' (Pna)

T T F

T F T

F T T

F F T

Cii )p a TPV >Q

T T F

T F T

F T T

F F T

1.3 : a b aab a- b asbl l F T T

Z l F F F

I 2 T F T

-2 l T F T

2-3 :

p Q p ⇒ Q Q⇒ p p⇐ Q

T T T T T

T F F T F

F T T F F

F F T T T

2.4 : Ci )

P Q P⇒ ( PVQ)

T T T

T F T

F T T

F F T

3.3 : Prove if h EZ s n is even , then n' is even .

Pf : We can write h=2M for some MEZ .

So n' =

@m5t-Czm7C2m7-2CzmDCHere2mE2D.Sori is even by defn . Eh

3.6 : For a , b EIR-

, if acts , then of > b'.

ff : a ab

=> of > ba =L- b)f- a) > C- b) . L- b) = b'

by T T( tones said: " ) ( by - a > - b)

na

4. l : There does not exist integers min set . 14h -12M -400.

ff : Prove by contradiction : say I m , n EZ sit .

14Mt 21h = 100 .

⇒ 7 (2Mt 3h) = 100⇒ It too . G

.

AM

4.4 : Prove if aEIR St.

a- 7,7A , then Aso or an .

Pf : Prove by contradiction : suppose 0<9-7.

So a > O{ a -no .

⇒ a .la -Dao .

⇒ at >a.

an

Problem I :

4 : Ci ) (alls e a Ic) ⇒ a l Cbtc ) .

Pf : By defu , 7- him C-2 set .

{{ Iaahm ⇒ btc ant am

a ( htm)

where htm Ek .

⇒ By defa al Cbtc) .

A

Cii ) all, or a Ic ⇒ al b - o .

Pt : If al b , then Inez at . b- ah

so b - c = Ca - n) C = a. (nc) .

⇒ al b - c .

If al c , then 7- NEZ se . C - ah

so b - c = b. Can) = a Cbn ) .

⇒ al b - c .

Ee

(n EIN'T5 : Part one : 6 In ⇒ ( necessary) .

Necessary conditions ate :

i ) 31 n ( BK n = Gm =3 - Gm)for some MEZ) .

V) 6th ( Bk n'= @ m)-

= 616mi ).

for some m EZ ) .

Vi ) 21h 8 3 In .

( similar reason as i ) )

vii) - In or 3 In .

( vi ) ⇒ vii ) ) .

Part two : ( sufficient ) ⇒ (notNts 6 In) .

-

Iii ) 121 n . ( similar reason as above) .

iv) A- 12 ( 12=6.2).

V ) 6) n' ( Bk IMEI sat . n'=6m .

" {312: - 5th : his primes)⇒ 21ns 3 In ⇒ bln )

.

Vi ) 21h 8 31 n ( Bk n=2m , 8 3h .

So 31 Mi,

is prime) .SO MI =3 Mr

so h= 241=2 . 3Mz

= 6ms

so 6th.)

.

Ah

8 ii) : x'- X-2 > o ⇐ Xa - I or X> 2 .

Pf :(=D Prove by contradiction : suppose -1*2-2

So { zoo ⇒ CXTDCX-2) so .

⇒ x'- x -zeo.

⇐) Suppose Xa - I so X - 2e- I -2<0

So { X -11<0 ⇒ 1) 'Cx -2) > 0 .

X-2<0⇒ xz -X - 270 .

Suppose X > 2 , X -1172+170.

So { X-270 . ⇒ (XH ) -CX-2370 .

X -1170⇒ x'-X -270

a