Question

http://math.stackexchange.com/questions/767117/let-a-b-in-r-where-a-b-prove-that-there-exist-a-rational-number-c-and/767138?noredirect=1#767138

Answer 1

@experimentX

Answer 2

I am not using decimal expansion...too many numbers involved.

Answer 3

this is rational density - irrational density theorem.

Answer 4

what? never heard of it.

Answer 5

this is not that difficult ... but you wrote a whole lot of things on there lol

Answer 6

do you need to do it by decimal expansion?

Answer 7

because I thought decimal expansion would work, but like one person said, it's going to be cumbersome... I don't have too... I'm doing the other way ... easy way. I got the irrational part when I replied to firegarden

Answer 8

here is how I found a way to do it. but it is not the best method.
http://santoshlinkha.wordpress.com/2012/12/25/96/

Answer 9

that irrational part is pretty awkward

Answer 10

the idea is same as what robjohn replied ... just take the difference of two numbers, multiply it by some number N such that the difference is large enough to overtake some interval ...
that is
N(a-b) > 1 => Na < K < NB ... apparently you will find that K/B is in between a and b ... and since both being integer, it's rational.

Answer 11

here you will find very nice proof.
http://en.wikibooks.org/wiki/Real_Analysis/Properties_of_Real_Numbers#Corollary_.28Density_of_rationals_and_irrationals.29

Answer 12

like this?

N(a-b) > 1
Na-Nb > 1 ?

Answer 13

yep ... if Na - Nb > 1 ... then there must be some number between Na and Nb ... let it be K.

Answer 14

Na < K < Nb

and then

do I divide the whole thing by b ? or just K /b?

Answer 15

no ... by N ... where N is natural number.

Answer 16

Let me edit my blog proof ... it's kind of awkward.

Answer 17

if I divide by N then
a < k/N < b

Answer 18

yes ... both K and N are natural numbers ... so what you get?

Answer 19

that axiom made everything a bit easier... the dumb book doesn't have it...and the decimal expansion will make it worse.

Answer 20

natural .... rational numbers!?!

Answer 21

hell yes.

Answer 22

okay ... you proved that a rational numbers exist between them.

now tyme to prove irrational number between them.

Answer 23

woo hoo ^^

Answer 24

which is d being irrational in that question for a <d<b

Answer 25

let c = K/N ... it b < c < a ... where 'c' is rational number.

Answer 26

ok
b <k/n < a

Answer 27

first insert a irrational number between two rational number.

Answer 28

hmmm ... \[Q \ \in a+ \sqrt 2 , b + \sqrt 2\]
like this?

Answer 29

let me tell you something very easy ...

let
b < c < d < a ... where, where 'c' and 'd' are rational numbers.

now ... multiply (d-c) by some natural number M such that
M(d-c) > 1 + sqrt(2)

Answer 30

sorry ,,,
M(d-c) > sqrt(2)

Answer 31

Md-Mc > sqrt(2)

Answer 32

hold on a sec ... let put, Md - Mc > 2 ... just to be secure.

Answer 33

Now, ... there is some natural number T between Md and Mc such that
Mc < T + sqrt(2) < Md

Answer 34

then d is irrational because we have T + sqrt(2) . Suppose T - sqrt(2) must be irriational because T-(T-sqrt(2)) = T-T+sqrt(2) = sqrt(2) which made it rational when clearly it's not.

Answer 35

okay ... let me rewrite the whole proof.

Answer 36

all right done
http://santoshlinkha.wordpress.com/2012/12/25/96/

Answer 37

hit refresh!!

Answer 38

Dang!! n(b-a) > 3

Answer 39

yay ^^

Answer 40

there's broken latex but I can navigate through it ^^

Answer 41

I would have answered on M.SE but ... there are too many nice answers.

Answer 42

now I just got this nasty blackboard proof stuff to do ...
For any natural number m,[a], [a] not equal to 0, has a multiplicative inverse in Zm if and only if gcd(a,m) = 1

Answer 43

Zm - congruence classes modulo m
z 3 a,b,a~b <->m(a-b) a (three horizontal lines) b mod m

Answer 44

what theeeeee...and the prof was like it's just a half a line proof . el o el

Answer 45

what is [a] ??

Answer 46

we can add and multiply equivalence classes by adding and multiplying representatives [a][b]=[ab]

Answer 47

fundamental theorem of algebra

Answer 48

that a in the box is modulo ... but it's rarely written these days

Answer 49

which book are you using?

Answer 50

passage to abstract mathematics

Answer 51

hint
we need an inverse
gcd(a,m ) = 1 -> xa+jm = 1

Answer 52

every element except 0 has a multiplicative inverse... hmmm

Answer 53

i thnk u need to prove :

\(xa \equiv 1 \mod m \iff gcd(a,m) = 1\)

Answer 54

eeyup @ganeshie8

Answer 55

:O

Answer 56

what information are given on x, a and m ... are all these things integers?

Answer 57

assuming that these things are integers, ... since 'a' does not divide the expression, it doesn't divide either of them.
on the other way, we need to prove that 'm' doesn't divide 'a' either.

Answer 58

let me grab the definitions..

Answer 59

By definition 6.2.6, we let \( m \in N\) and \(a,b \in Z\). Then a is congruent to b modulo m written \( a b\) (mod m) if \( m l a-b\). The relation on \(Z\) of congruence modulo m is the set \([(a,b) \in Z \times Z a b \) (mod m).\\

Moreover, by definition 2.5.3, we let \(a, b \in Z\) with \(a\) and \(b\) not both 0. Let \(D(a,b)\) be the set of common divisors \(a\) and \(b\). We denote this element by \(gcd(a,b)\). Thus

\((\forall c \in D (a,b) [c \leq gcd(a,b)]\)

Answer 60

except that Z ab should be