Use the previous exercise (1.11, #4 parts (d) and (e)) to deduce a formula for and prove this formula is correct.

** Claim: **

* Proof. * Let , then we have

Thus, there exists some with such that

Hence, . Thus,

Then, for each with we have

Then, for , we have

So,

Hmmm… the latex formatting seems to have gone badly awry on my last comment. You can get it as a PDF at https://hursts.org.uk/proof-%5Bnx%5D-2.pdf

Claim:

Proof:

(1)

This allows us to derive a closed form of the summation:

(2)

For the first term on the RHS of \eqref{e:3}

Applying \eqref{e:2} we get

(3)

Thus, for condition \eqref{e:2}

and since is not involved in the formula, it is true .

Thanks for the work your doing here. But I’ve not been able to figure out the introduction of j into into the proof, even after reading the explanation you gave in the comment section.Please can you like explain again how your going about your proof, like how you know where your headed….

Remember when you multiply a number with a fraction, that fraction also gets multiplied and it might account for new integer(s) being added to the original integer part. So summing a j >= 0 to the original mn absolute value accounts for the possibility that x has a fraction that whe multiplied n times, becomes larger than mn.

Sorry to be commenting so much, but how do you know that you can choose j, such that 0 <= j < n in equation 3?

Hi, no problem, comment away! (Though if I get busy, responses could start to lag.)

We have

But we know for any number that it is between and for exactly one integer . But since we know that integer is somewhere between and , which means it is for some integer . So, then we have that is between and . Does that make sense? Basically, we are just saying that we know for some integer , but then it is more convenient to write it as since that will help us out later.

Not sure if this actually clears up your question. Let me know if not.