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this holds true for all integers such that: (a/a-1)+a=(a/a-1)*a, for all integer values of a. This was what I found while solving maths today morning.

Note by Siddharth Singh
1 year, 10 months ago

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Try using \(\LaTeX\) !

" \dfrac{a}{a-1} + a = \dfrac{a}{a-1} \cdot a " produces \[ \dfrac{a}{a-1} + a = \dfrac{a}{a-1} \cdot a\] Azhaghu Roopesh M · 1 year, 10 months ago

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