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

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

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

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