beginnersmind
  • beginnersmind
For fun: If n is a positive integer prove that the fraction \(\frac{n^2}{2n+1}\) cannot be simplified.
Mathematics
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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

beginnersmind
  • beginnersmind
For fun: If n is a positive integer prove that the fraction \(\frac{n^2}{2n+1}\) cannot be simplified.
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

myininaya
  • myininaya
\[\text{ let } k=\frac{n^2}{2n+1} \\ k(2n+1)=n^2 \\ n^2-2kn-k=0 \\ n=\frac{2k \pm \sqrt{4k^2+4k}}{2} \\ \text{ so } n \text{ is an integer } \\ \] just showing my thinking and this thinking though
myininaya
  • myininaya
\[n=k \pm \sqrt{k^2+k} \\ n=k \pm \sqrt{k(k+1)} \\ k>0 \] so I guess the equivalent question here is to show for integer k>0 we have k(k+1) is not a perfect square
myininaya
  • myininaya
I think it makes sense to say since to say there is no two consecutive positive integers such that when multiplied give us a perfect square

Looking for something else?

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

More answers

myininaya
  • myininaya
example: 1*2 2*3 3*4 4*5 5*6 6*7 7*8 both factors would have to be the same
myininaya
  • myininaya
of course k(k+1) is a perfect square when k=0 or k=-1
myininaya
  • myininaya
but as you said k>0
beginnersmind
  • beginnersmind
Seems to be correct, though not the proof I had in mind.
beginnersmind
  • beginnersmind
I'll leave this up with the hint that there's a 3 line proof if someone else wants to try it.
thomas5267
  • thomas5267
\[ \frac{n^2}{2n+1}=\frac{(n+1)^2-2n-1}{2n+1}=\frac{(n+1)^2}{2n+1}-1 \] \(n\) and \(n+1\) shares no common factors?
beginnersmind
  • beginnersmind
I'm not sure I see how that helps.
beginnersmind
  • beginnersmind
Ok, I sort of see it now. If n+1 and n share no common factors that neither do (n+1)^2 and n^2. So n^2 can't have a common factor with 2n+1 because that common factor would be a common factor between n^2 and n^2+2n+1, which is (n+1)^2. It's actually similar to the solution I had.
beginnersmind
  • beginnersmind
Which is: gcd(2n+1,n) = 1 so n and (2n+1) share no prime factor. therefore n^2 and (2n+1) share no prime factor
thomas5267
  • thomas5267
Pretty sure that \(n\) and \(n+1\) shares no common factor. \[ n=p_1^{k_1}p_2^{k_2}\cdots\\ n+1=p_1^{k_1}p_2^{k_2}\cdots+1\\ n+1=1\pmod {p_k} \]
beginnersmind
  • beginnersmind
Yeah, n and n+1 definitely share no common factor. :) I was trying to reconstruct how you used that fact to prove that n^2 and 2n+1 share no common factor.
thomas5267
  • thomas5267
\(2n+1\) cannot share factors with \(n\) and \(n+1\) at the same time so there is at most one of them is reducible. WLOG, let \(\dfrac{n^2}{2n+1}\) be reducible and \(\dfrac{(n+1)^2}{2n+1}\) be irreducible. So \(n^2\equiv 0\pmod{2n+1}\) and \((n+1)^2\not\equiv0\pmod{2n+1}\). But \((n+1)^2-(2n+1)=(n+1)^2-0\not\equiv 0\pmod{2n+1}\) and \((n+1)^2-(2n+1)=n^2\equiv 0\pmod{2n+1}\). So it must be the case that both \(\dfrac{n^2}{2n+1}\) and \(\dfrac{(n+1)^2}{2n+1}\) are irreducible.
beginnersmind
  • beginnersmind
Right, got it. Another way to show that n and (2n+1) share no common factors, is by realizing that any common factor d divides the expression (2n+1) - 2(n) = 1. So d|1 => d = 1.

Looking for something else?

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