An Important Theorem
Category : JEE Main & Advanced
If \[{{(\sqrt{A}+B)}^{n}}=I+f\] where \[l\] and \[n\] are positive integers, \[n\] being odd and \[A-{{B}^{2}}=K>0\] then \[(I+f)\,.\,f={{K}^{n}}\]
where \[A-{{B}^{2}}=K>0\] and \[\sqrt{A}-B<1\].
Hence L.H.S. and I are integers. \[\therefore \] \[f+{f}'\] is also integer
\[\Rightarrow \]\[f+{f}'=1\]; \[\therefore \]\[{f}'=(1-f)\]
Hence \[\left| \,\frac{y}{x}\, \right|<1\] \[={{(\sqrt{A}+B)}^{n}}{{(\sqrt{A}-B)}^{n}}\]
\[={{(A-{{B}^{2}})}^{n}}={{K}^{n}}\].
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