TuringTest
  • TuringTest
Let f be a function from the set of integers to the set of positive integers. Suppose that, for any two integers m and n, the difference\[f(m)−f(n)\]is divisible by\[f(m−n)\]Prove that, for all integers m and n with\[f(m)≤f(n)\]the number\[f(n)\] is divisible by \[f(m)\]
Meta-math
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{f(m) - f(n)}{f(m-n)}=k\]Setting n = 0, \[f(m) | f(0)\]Setting m = 0, \[f(-n)| f(n)\]Setting m = 0, n = -n, \[f(n)| f(-n)\]\[f(n) = f(-n)\]Setting m = 2n, \[f(n) |f(2n)\]\[f(2n) \ge f(n)\]Setting n as -n \[f(m+n)|(f(m) - f(n))\]\[f(m) - f(n) \ge f(m+n)\] Assume \[f(m+n) > f(m)\], then\[f(m+n) > f(m) - f(n)\]However, this contradicts to \[f(m) - f(n) \ge f(m+n)\], which means \[f(m+n) \] is either smaller or equal to f(m). Assume \[f(m+n) < f(m)\]then setting m = n, \[f(2n) < f(n)\], which contradicts to \[f(2n) \ge f(n)\], Thus \[f(m+n) = f(m)\] Substitute this back to the original equation\[\frac{f(m+n)-f(n)}{f(m)}\]Thus, we proved \[f(m) | f(n)\]
TuringTest
  • TuringTest
Wow moneybird this looks really great! I'll have to give it special attention on a non-holiday though, I don't think I'm smart enough to really evaluate it. Certainly not right now at least. I'll be sure to get others to look at this, this was an IMO problem, yeah?
anonymous
  • anonymous
thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
Having had a bit more time to digest this I can say I follow the logic and see no mistakes. That said I don't think I'm qualified to give you the all clear myself. I'll be curious to hear other opinions. Again, excellent job!
Zarkon
  • Zarkon
I don't see a problem
Zarkon
  • Zarkon
though I would add more detail.
TuringTest
  • TuringTest
like? what cases are not covered that should be?
Zarkon
  • Zarkon
all the cases are covered...just more words to explain what is going on
TuringTest
  • TuringTest
well he still deserves a medal then :)
Zarkon
  • Zarkon
I don't like this sub forum...I'm only level 2 here :)
TuringTest
  • TuringTest
there ya go! on the way to guru again.
JamesJ
  • JamesJ
Interesting proof, because it seems you've proven something stronger. When you get to the statement \[ f(m+n) = f(m) \] m and n are arbitrary, and hence f is constant.
asnaseer
  • asnaseer
I'm not clear on how setting m=0 leads to:\[f(-n)| f(n)\]on the 5th line of the proof. same goes for the 7th line of the proof where setting m=0 and n=-n seems to lead to:\[f(n)| f(-n)\]how do we reach these conclusions?
anonymous
  • anonymous
Actually I missed one step, which is setting m = -n and n equal to zero.\[\frac{f(-n)-f(0)}{f(-n)} \implies f(-n)| f(0)\]
asnaseer
  • asnaseer
is that a legal step? you are (in the same step) setting m=-n AND n=0. isn't that equivalent to setting m=0?
anonymous
  • anonymous
This means that f(0) is divisible by f(-n). w When m = 0, \[\frac{f(0)-f(n))}{f(-n)} \implies f(-n) | f(n)\]
Zarkon
  • Zarkon
you already had f(m)|f(0)...m was arbitrary...thus...
asnaseer
  • asnaseer
I understand the 1st step where you showed \(f(m)|f(0)\), but I am afraid I don't follow the rest of the arguments. this is probably beyond my expertise as I canot see how the next steps are deduced?

Looking for something else?

Not the answer you are looking for? Search for more explanations.