Question

Previously,
http://openstudy.com/study#/updates/4f83fe86e4b0505bf084b9ce

This thread is now closed so I am posting it as a new thread.

Pinging the participants:

@AravindG @Mani_Jha @experimentX @apoorvk @TuringTest @dumbcow @asnaseer

Answer 1

Lemma: If \( (a + b\sqrt{c})^n = m+ f \), where \(m\) and \(n \) are positive integers and \( f\) is a positive fraction less than one, and \( 0\lt (a - b\sqrt{c})^n \lt 1 \) with \( (a + b\sqrt{c})^n \times (a - b\sqrt{c})^n =1 \) then \( (m + f) (1 – f) = 1\).

Proof: To be precise we have to show \((a - b\sqrt{c})^n= (1 – f) \)
Let \( X = (a - b\sqrt{c})^n \)

Now, \( m+f+g= (a + b\sqrt{c})^n + (a - b\sqrt{c})^n = \text{ Even integer }(T) (say) \tag{1} \).
Since, \( (x+a)^n+(x-a)^n = 2 \left[ \binom n 0 x^n + \binom n 2 x^{n-2} a^2 + \binom n 4 x^{n-4}a^4 +\cdots \right] \)

Thus, from \( (1) \):


\( \implies f + X = T –m \implies f + X \) is an integer
\( \implies f+X = 1 \)

Since, \( 0 < f < 1, 0 < X < 1 ∴ 0 < f + X < 2 \implies f + X \)
is an integer between 0 and 2 \( \implies f + X = 1 \)

That is \( X= (a - b\sqrt{c})^n= (1 – f) \)

Quod erat demonstrandum!

Answer 2

anyway this stuff is too advanced for me ...

Answer 3

This is high school standard.

Answer 4

I wish I did things like this in high school.

Answer 5

i wish i had too ...

Answer 6

Ok with that line of M+F+G...where did that G come from and what does it equal?

Answer 7

G=g=X (typo)

Answer 8

wow!

Answer 9

thx fool u r the best!!

Answer 10

No problem, I proved this lemma 3 years ago (I suppose). So thank you for posting the question and tagging me along.

Answer 11

wat do u mean by lemma?

Answer 12

lemma means theorem?

Answer 13

from wiki: lemma = a proven statement used as a stepping-stone toward the proof of another statement

Answer 14

thx evryone

Answer 15

Sorry, AravindG I was responding to someone else and yes Callisto is right. I think I should call this Proposition.

Main results are theorems, smaller results are called propositions. A Lemma is a technical intermediate step which has no standing as an independent result. Lemmas are only used to chop big proofs into handy pieces.

From here: http://mathoverflow.net/questions/18352/theorem-versus-proposition/18356#18356

Answer 16

k thx again

Answer 17

Beautiful work

Answer 18

Thanks dumbcow.

Answer 19

Very elegant proof @FoolForMath

Answer 20

Thanks @asnaseer