Appendix 6B

fig.1a Jack's proof for the ArchPr 

Suppose false then n≤a

fiN is bounded above

fi(Compl.Ax) $ supNŒ¬

n≤a "nŒN

e>0 $neŒN a-e<ne

let e=1 then a-1<n1

a<n1+1

Return to Extract 6.1.

 

fig.1b The tutor's proof for the ArchPr

N has a supremum

($aŒ¬) (n≤a, "nŒN)

"e>0 $nŒN a-e<n

In particular there exists n1ŒN

such that a-1<n1. But then a<n1+1.

Equivalent to the ArchPr:

If x,yŒ¬ and x>0 then ($nŒN) (nx>y)

("eŒ¬) e>0 fi ($nŒN) (1/n<e)

Return to Extract 6.1.

 

fig.2a Ben's proof for CD2.1ii

$x which is irrational

we can find a rational s

st a/x < s < b/x,

a,b rational

Return to Extract 6.2

Return to Section 6(ii).

 

fig.2bThe tutor's proof for CD2.1ii

Suppose that a, bŒ¬ and a<b.

Suppose also that all real numbers x such that a<x<b are rational.

We prove that a< 2a+b /3 < a+2b /3 <b.

By our assumption 2a+b /3, a+2b /3 ŒQ

2a+b /3 < 2a+b /3 + b-a /3√2 < a+2b /3 is irrational by part (i)

and lies between a and b.

Return to Extract 6.2

 

fig.3 The tutor's proof for CD2.2

Seeking to prove |x+y|≤ |x| + |y|

|x+y| -(x+y) x+y

|x|=x x -x

|y|=y y

...

Inductive hypothesis: |x1+...+xk| ≤ |x1|+...+|xk|

|x1+...+xk+xk+1| ≤ |x1+...+xk|+ |xk+1| ≤ |x1|+...+|xk|+|xk+1|

Return to Extract 6.3

 

fig.5 Andrew's proof for CD2.4ii

given h define S = {hn: nŒN}

m≥hn

suppose that

$nŒN st n≥hn "nŒN

nŒN, n+1ŒN and (n+1)nŒS

Return to Extract 6.5

 

fig.6a The tutor's proof for CD2.5i

[reconstruction from recording]

S»T must be non-empty and bounded above

in order to have a supremum. It is non-empty

because S and T are. Also

xŒS x≤supS

If xŒS»T then or . So or

xŒT x≤supT

that is in any case x≤ max {supS, supT}. So

max{supS, supT} is an upper bound for S»T.

[he continues and proves it is the lowest upper bound]

 Return to Extract 6.6

 

 fig.6b Alan's proof for CD2.5i

[reconstruction from recording]

let a= supS»T= supT

then a is an upper bound of S»T therefore a>x for all x in S»T

if aœS»T then a< supS

if xŒS»T then xŒS or xŒT and if xŒS then contradiction

 Return to Extract 6.6

 

 fig.7 The tutor's proof for CD2.5i

[reconstruction from recording]

[Cornelia has proved that ksupS is a lower bound]

let b>ks, where s=supS

then b/k < s. Therefore $aŒS: a>b/k. Then ka<b and kaŒkS.

 Return to Extract 6.7

 

fig.8a Jack's proof for CD3.3i

(xŒ¬) (x<≠0)

By the Archimedean Property

$nŒN such x-1 < n ≤ x

Therefore

0 ≤ x-n < 1

0 ≤ e < 1 where e = x-n.

Thus x=n+1 where e = x-n.

Return to Extract 6.8

 

fig.8b The tutor's rephrasing of CD3.3ii

Let r be a natural number >1. Show that for any tŒN

there exists nŒN and there exist a0, ..., an ŒN such that

0 ≤ as < r (for all s) and t = a0 + a1r + ... + anrn.

Return to Extract 6.8