# $A\ very\ simple,yet\ elegant\ NT\ proof(corrected)$

$It\ is\ a\ well\ known\ fact\ that\ if\ in\ a\ fraction,the\ denominator\ is\ of\ the\ \\ form\ 2^{m}*5^{n}\ where\ m\ and\ n\ are\ non-negative\ integers\ the\ fraction\ \\ has\ a\ \\ terminating\ decimal.$ $Here\ is\ the\ proof:$ $Let\ us\ say\ that\ there\ is\ a\ fraction\ in\ which\ the\ numerator\ is\ N\ and\ \\ the\ denominator\ be\ 2^{m}*5^{n}\ where\ m+n=x$ $The\ fraction\ can\ be\ written\ as:\\ \dfrac{N}{2^{x-n}*5^{x-m}}$ $Multiplying\ by\ \dfrac{2^{n}}{2^{n}}\ and\ then\ by\ \dfrac{5^{m}}{5^{m}}\\ we\ have\ a\ fraction\ which\ has\ a\ new\ numerator,let\ us\ call\ it\ A.$\ $The\ new\ denominator\ is\ 10^{x}\\ Thus,the\ new\ fraction\ is\ \dfrac{A}{10^{x}}.$\ $Which,obviously,has\ a\ terminating\ decimal\ expansion.$

6 years, 10 months ago

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- 6 years, 10 months ago

WOW @Krishna Ar, this is called popularity!

- 6 years, 10 months ago

hehe

- 6 years, 10 months ago

-_-

- 6 years, 10 months ago

(y)

- 6 years, 10 months ago

Nice note, Adarsh. You may also like to add the number of decimal digits after a number of such a form terminates. It's closely related to your proof. Also, you don't need to latex the whole thing. If you are using a latex editor, then there's no need. If you want to learn it, you can easily do so here. Thanks.

- 6 years, 10 months ago

thanx.But I recently learned LATEX and now I can't get enough of it.

- 6 years, 10 months ago

BTW which class r u in?

- 6 years, 10 months ago

I'm in class 10.

- 6 years, 10 months ago

oh,ok.

- 6 years, 10 months ago

I don't understand "I can't get enough of it", you can always learn it. It's really easy to learn.

- 6 years, 10 months ago

"I can't get enough of it" means now I can't resist writing it.It is an idiom,I think.

- 6 years, 10 months ago

Do you have the proof of the property of recurring decimal expansions that they repeat for n-1 digits in a fraction 1/n?

- 6 years, 10 months ago

That's not necessary. It is only true in case of cyclic numbers. I was referring to the number of after-decimal digits after $\frac{x}{2^m5^n}$ terminates. It's the larger of $m$ and $n$.

- 6 years, 10 months ago

hmmm

- 6 years, 10 months ago

It is a very simple proof which was in my RD SHARMA but I thought it too elegant not to share it with you guys.

- 6 years, 10 months ago

expect the unexpected !

- 5 years, 8 months ago