Question

Let a and b be real numbers with a 14 years ago

Answer 1

anyone?

Answer 2

:cry:

Answer 3

it's analysis, got an exam tomorrow :/

Answer 4

we can do this

Answer 5

start with this: if x and y are positive numbers then there is a natural number n with
nx > y

Answer 6

this is more or less obvious (and is called "archemidian propery" i think
then consider a < b so
\[b-a>0\] and using the above you know there is an integer n with
\[n(b-a)>1\]

Answer 7

which is another way of saying that
\[na\] and nb\] differ by more than one. therefore there must be an integer m between them

Answer 8

i.e. there is an m in Z with
\[na<m<nb\]

Answer 9

didn't get where the 1 comes from in the n(b-a)>1

Answer 10

then divide by m and you are done
\[a<\frac{n}{m}<b\]

Answer 11

oh the 1 comes in to make the proof work.

Answer 12

the idea is that you want to assure that
\[na\] and \[nb\] differ by more than one, insuring an integer between them

Answer 13

you could have used 2, or 3 or anything really

Answer 14

i see. you are a legend, thanks!

Answer 15

i might be clearer if you tried this with actual numbers and see how it worked

Answer 16

we can do it if you want

Answer 17

no time, i'm gonna memorise it for the exam. this only came up once in the past papers. there's a whole lot of other stuff i have to work through. analysis is a feather :/

Answer 18

of course that wouldn't be a proof, but it might explain the mechanics of the proof

Answer 19

ok good luck!

Answer 20

thanks!

Answer 21

i have to add that i know this because i remember it. it would not have come to me off the top of my head. i have seen it before.

Answer 22

yw

Answer 23

haha you've got a hell of a memory :D