HW 1The solutionsjust sketch the ideasof the proof Whenyou writedownyoursolutions you might have to write more details if needed

1 Suppose rtx RX EQThen rtx r rx t are also rational

XE Q E Hereand in thefollowing andfuturesolutions this symbol means contradiction

2 Let XE E 0Then LEX because it's a lowerbound

XE f because it's a upperbound

By transition defDM

3 Because A is bounded below so by the least upper bound

property infA P exists

Similarly for Hye A y x forsome XEASo F X E infA

A is bounded above s supC Also exists

Finally we show infA supEA

For t XEA we getX if A Xe infAX ssuptA x x supEA

So by infA is an upperboundof A

suptAl is a lower boundof A

infarsuptasuptAleinfA

infA suptA

am

4 jute 3 shoe 158 i cD i o EIf 1 0 L t c O I 3 Oc E

I 170 H C 1 L O E Da

5 Proveby induction

for nel fromthe hypothesis OcY eyeSuppose we haveOry c Yak forsome Kea

Oc y yi g cyzk.gcyzk.gs y'tso we have yic Yi for t neat

DM

6 If atabt o then a b by 5 I

7 a A B 21 31 1

b A B Q

A B E Q because a b EQ for ta bea

Conversly given qEQ 9 9 0 whereget 0GBG E A B A

e Note that for ta EA WEBinfA sup Bea b supA infBA B is also bounded So supA B exists

Next show the equality

By we have supA B s supA infB

AEA s.t.az supA Elaconversely for t o P'ok

be sit be'mfBt

a be supA E infBsupLAB a Supa infB E for 4220SupCAB a Supa infB

Overall we have supA B supa infBam