How and why you want to "choose r" in the answer key? Then why use naturals, it looks redundant.

[robust granite]

I understand the part of the proof solution quite good until the "choose r" part.
Question and the book solution is provided
It is from Calculus 9 (Purnell) if it helps

[robust granite]

my question doesnt get any love 😦

[static ibex]

well, they choose r between a and b. If it happens that r=(k0-1)/n = b then they take new r=b-1/n, so that it is placed inside (a,b). There is a kind of vague misprint in the line with 'Note...' as they have chosen new r, which is equal to b-1/n.

[robust granite]

ah, i understand now... only this part is left

[static ibex]

this is a second question in the problem

they construct infinitely many such rational numbers

[robust granite]

yes

[robust granite]

the proof looks missing though... how do you choose r2 such that a < r2 < r1?

[static ibex]

you have just proven it is possible. Just take b=r1

[robust granite]

that contradicts the fact that r1 < b

[static ibex]

how?

r1<b

just call r1 as b. And apply what you proved right now.

there is a rational number r2 between a and r1. You have proved it just above

At each step you insert another rational number between the two according to the fact that there is a rational number between any a and b.

[robust granite]

.... oh, i thought you want to change the value of r1 to b, now i understand

[static ibex]

in fact we do exactly this. We kind of rename r1 into b

[robust granite]

so its like a value of r2 that satisfies a < r2 < r1 can be found by applying the same argument

[static ibex]

yes

[robust granite]

and we need to go back to the first step so the argument can be applied all over again

[static ibex]

we have proven it for any a and b. So yes you can take ANY two numbers and the statement that there is a rational between them is true

[robust granite]

well, i will definitelt need to write "following the same argument" to my proof for now, since the answer key seems to assume im intuitive enough to skip explanation like that

[robust granite]

hooray, thanks a lot for guiding me through this wonderful process

i cannot close the thread though

[static ibex]

try to type .close

[robust granite]

why "let n be a natural number" instead of real number? is it just to bound the set into positive values only, since positive real numbers seems to work with the argument as well

[robust granite]

i am still curious of one more thing 😅

[static ibex]

nope

you cannot

You have to get rational r

and it is (k0-1)/n

so n should be integer

[robust granite]

oh.. this is a really impressive proof than, very well thought proof

[robust granite]

.close