#Infinite rationals between two reals
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So essentially for any x and any n you can find m/n at distance at most 1/n from x
yes
so b-a > 1/n is enough?
Essentially, m/n is in (x - 1/n, x + 1/n)
Can you pick x and n smartly so that this interval is included in (a, b)?
Try drawing a picture
wait let me try
Sure
x=a, 1/n = (b-a)/2
No because m/n can still be outside of (a, b)
It could be less than x
You just know it is at distance less than 1/n from it
Sure
x=(a+b)/2, 1/n=(b-a)/2?
You can't say 1/n = (b-a)/2 but it's a good start
Take x as the midpoint
Can you find n such that the conclusion holds?
2 < (b-a)*n ?
such a natural n exists right cause of archimedian property?
Ok fair
This is fine
what were you expecting though?
Just that
Or give me an explicit expression with the floor or something
Like just prove to me that you can find one such n
Such that the conclusion holds
Now you just take this x and n candidates and verify everything fits
taking x=a+b/2 and 1/n<b-a/2 implies, there exists a rational number between x-1/n, x+1/n which is subset of (a,b) so there's a rational in (a,b)
and archimedian property gives such a n
@naive salmon also if you don't mind me asking can i ask about studying math related query?
What's up
so im studying this type of formal math for the first time so should i go ahead and refer a textbook first and then solve these assignments or just try solving these after lectures?
like my professor just taught lub property, archimedian property and gave this assignment
I probably dont need to know anything else to solve but its hard and time consuming to find whats the right approach
like you mentioned about interpreting it as rationals lying at max at 1/n dist, while I was playing around with some inequalities
A picture would have immediately solved the problem here
I mean I think formalizing is necessary in any case
regardless of your source material
But I think you should also find the way to learn that works best for you
there are people with basically zero geometric intuition
And it works for them anyway so whatever
there are people who will also get the problem with just a quick drawing
hmm probably i also need to draw a mental picture
also one last thing
here in (a), isn't it obvious cause S is bounded above or am i missing something?
cause F = Q or R includes R right?
and non empty too
Well what are the conditions for sup S to exist in R?
It is true that S is bounded from above and therefore you can use the least bound property
set must bounded above and non empty
Then yes, it suffices to prove that the set is not empty and that there is an upper-bound
The first two sentences argue that S is nonempty
The remaining argues that there is exists an upper-bound m
so the hint is just to help showing S is non empty and bounded?
Yes

So you're literally quite expected to go into details
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