A History of Mathematics From Mesopotamia to Modernity

Modernity and itsAnxieties 233

z∈V′. The other way round is harder. Supposez^2 >6, we must decomposezas a product. For

somen,z^2 > 6 +( 1 /n). Choose a rational numberawitha^2 >2 anda^2 < 2 +( 1 / 3 n). Then,

z^2

a^2

>

6 +( 1 /n)

2 +( 1 / 3 n)

= 3

soz=aa′as required.

(c) Here it is much easier to use the lower classes. LetLibe the lower class belonging toxi,

and simply defineLto be theunionof theLis: all rational numbersawhich are in at least one

of theLis. ThenLisnotall of the rationals; in fact, ifa>M,ais in none of theLis, so not inL.

And it is easy to check thatLsatisfies the requirements for a lower class. Hence,Ldefines a real

numberM 0.

Since eachLiis contained inL,M 0 is greater than eachxi. Now supposeyis a real number

<M 0. Then there is a rational numberasuch thaty<a<M 0. From the definition ofM 0 ,a

is inLifor somei. Hence,a<xiand soy<xi.

(d) I put this in because, naively, one might think that defining real numbers as infinite

decimal fractions was good enough, and one needs to realize that adding them is com-

plicated by the possibility of having to ‘carry’ indefinitely far back. Consider the difference

between:

0.12345+0.87654=0.99999

and

0.12345+0.87655=1.00000

A change in the fifth decimal place changes all the results of the addition.

2. (a) Socks being identical, one needs a function which to each pair of socksSiassigns just one
   socksi. The axiom of choice will ensure this (but you cannot do it by hand). On the other hand
   with shoes, there is a natural rule: simply choose the left shoe of each pair. [In set-theoretic
   terms, it is a difference between ordered and unordered pairs.]
   (b) The number

√

2

√

(^2) is either rational or irrational. If it is rational, we have the example
we are looking for (since

√

2 is irrational anyway). If it is irrational, then takex=

√

2

√

2

, and

y=

√

2.xy=

√

2

2

=2 is rational and we are done in this case too.

Intuitionist critique: The procedure is not constructive; since, if we do not know which of

the two alternatives is true, we do not know which procedure to follow.

[Further complication; it can be proved that

√

2

√ 2

is irrational. If we use this, then of course

we have a proof without the law of the excluded middle.]

(c) Leta 0 =0 andb 0 =1. We construct inductively an interval[an,bn]of length 1/ 2 n.

Suppose[an,bn]constructed. Then:

1. if[an,(an+bn)/ 2 ]contains infinitely many elements of the set, letan+ 1 =anandbn+ 1 =
   (an+bn)/2;


2. if not (so the other half contains infinitely many elements), letan+ 1 =(an+bn)/2 and
   bn+ 1 =bn.

The sequencesan,bn, clearly both converge, to the same limitx; and at least one of them must

contain infinitely many distinct elements. This is the sequence we are looking for.