Question

plz help me
prove that he set S ={q belongs to Q : q > 0 and q2 < 2}This set has
no largest number.

Answer 1

Do you mean
\[S = \{q \in \mathbb{Q} : q \gt 0~ \mathrm{and }~q^2 \lt 2\}\]

?

Answer 2

wow you get a medal just for decoding.

Answer 3

:) this looks same as proving x^2=2 is not solveable in rational numbers ?

Answer 4

Can you use density of \(\mathbb{Q}\) in \(\mathbb{R}\)?

Answer 5

its an non empty open interval, so density would take care of it...

Answer 6

ok as i said before its the same as proving rational number are infinite, i got dull for a second cuz of ganesh comment but now ok :P

Answer 7

so basically , show that the set of
q^2 is an infinite countable subset of rational

Answer 8

nvm my comments im an amateur in topology/analysis lol

Answer 9

How does this prove it?

Answer 10

its prove cuz of this :-
infinite=no absolute supremum .

Answer 11

ok continue @zzr0ck3r

Answer 12

er \(x_0\in (x,\sqrt{2})\) and we are done

Answer 13

but just because something is infinite does not mean a proper subset is

Answer 14

humm , well i cant see ur way yet :O
what i mean is this :-
show that q^2 is closed on rational xD

Answer 15

this is first step

Answer 16

are we reading the same question? I am reading this.

Show that there is no largest element in \((0,\sqrt{2})\)

Answer 17

:)

Answer 18

best way to show that \(\mathbb{Q}\) is countable is \(f(\frac{m}{n}) =(m,n)\in \mathbb{N}\times \mathbb{N}-\{0\}\) where \(\frac{m}{n}\) is in reduced form.

I only say this because I just learned it today, and its quick and easy:)

Answer 19

p.s. that was bad notation when I put (0,sqrt2) I mean the rationals in there....

Answer 20

zzr your first argument looks a lot like \(\forall x \in (0, \sqrt{2}), \)
\[x\lt \dfrac{x+\sqrt{2}}{2} \lt \sqrt{2} \]

(i had to convert like this as im not familiarty with topology terms like dense etc.. )

Answer 21

let me make more clear

Answer 22

you cant use sqrt(2) in your proof, since sqrt(2) is not defined (yet)

Answer 23

thats a good point

Answer 24

and we are working only with q , such that 0<q^2 < 2 ,

Answer 25

consider the set \(S=\{x\in \mathbb{Q}| x\in (0,\sqrt{2})\}\). Suppose by way of contradiction there exists a greatest element \(x\) in \(S\). Then by the density of the rationals there is a \(y\in \mathbb{Q}\) s.t. \(x<y<\sup S\). This is a contradiction.

Answer 26

maybe lets assume for now that we have created Reals already to continue with these cool proofs :)

Answer 27

well i said my view , in case u learned it yet or not
simply show that q^2<2 is closed in Q , and then show its infinite for q>0
done .

Answer 28

This of course only works if they have proved that the rationals are dense in the reals. i.e. between any two numbers a<b there is a rational number c s.t. a<c<b

I wish the user would tell us what they can use:)

Answer 29

yeah I don't follow. Are we talking the topological definition of closed?

Answer 30

i liked this one clear and proper.

http://prntscr.com/5cqseu

also arkhamedian use same thing , about rational

Answer 31

in his bath?

Answer 32

still going thru previous replies.. just a quick q\[f(\frac{m}{n}) =(m,n)\in \mathbb{N}\times \mathbb{N}-\{0\}\]

this is like cantor's diagonalization proof where they build a matrix and count diagonally right ?

Answer 33

the product of two countable sets is countable or somehting..

Answer 34

yeah. Its similar, but he didn't talk about 1-1 in that proof

Answer 35

product and sum

Answer 36

you might just want to clarify that your sup S is a rational number

Answer 37

why?

Answer 38

it might be confused with the square root 2

Answer 39

Q is dense in R

Answer 40

the supremum need not be in the set, and thus might not be rational.

Answer 41

I assume we are talking about R being our universe here....

Answer 42

the set S is only rational numbers

but the proof hinges on density of rationals, which hasnt been proven, and sqrt(2) which is undefined . so you are taking a lot for granted in your proof

Answer 43

are you in the class?

Answer 44

should we start from set axioms? I have said about 5 times now if you simply read that this proof is valid if and only if density is allowed.

Answer 45

sorry i didn't read the entire thread . (take it easy)

Answer 46

I am easy :) lol Just pointing out what I already said

Answer 47

but sqrt(2) was not given in the original statement. im not sure if you should treat it like a number

Answer 48

I feel like this is really easy to prove by contradiction. Basically, this boils down to proving that there is no greatest element\[0 < q < \sqrt{2}, q \in Q\]
So suppose x is the greatest such value. Then let\[0 < y < \sqrt{2} - x, y \in Q\]
Then x + y is a value in Q that satisfies the conditions for q.

Answer 49

then remove sqrt of two and call is S. its the same set....

Answer 50

this is the same thing I did .... you are using density

Answer 51

yes its same as showing the average of x and sqrt(2) is greater than x

Answer 52

if you can show such a y there is no need to do the sqrt(2) -x

just pick y between x and sqrt(2)

Answer 53

or the sup of the set if we don't like irrationals...

Answer 54

but the sup of the set is irrational so I am unclear what the point was there...

Answer 55

hey i found this proof that uses only rational numbers, but i dont get it completely

Answer 56

i remember same page in my book :O lol

Answer 57

looks it is using complicated algebra instead of density and other sophisticated tools

Answer 58

prob cooler.... This is why I hate analysis... The solutions are boring and microscopic.

Answer 59

@zzr0ck3r Ah, you are correct. I was just skimming posts and missed the one where you actually wrote the proof.
That said, I wrote it the way I did because its easy, without a lot of work, to prove that a rational number will always exist in a neighborhood around 0. (Obviously this can be extrapolated for any neighborhood of Q in R, but I was going for something simple to understand for people who might not have as much set theory background).

Answer 60

Note that for any real number there is a sequence of rational numbers that converges to it. So there is a sequence in in our set that converges to the sup of our set. So for any positive number given there is an element of our sequence between our "largest element" and the supremum of our set, and we are done.