1yes if n=1 then it is less than one.so?
if (9+4√5)n=p+β where n and p are postive integers and β is a positive proper fraction, prove that (1-β)(p+β)=1 ?
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62if (9+4√5)n=p+β where n and p are postive integers and β is a positive proper fraction, prove that (1-β)(p+β)=1 ?
(9-4√5)n<1 because (9-4√5)<1 and n is a +ve integer.
also note that
S=(9-4√5)n+(9+4√5)n (let)
S will be an integer.
(9+4√5)n = S - (9-4√5)n
(9+4√5)n = S-1+ {1-(9-4√5)n}
Thus, β={1-(9-4√5)n}
1-β=(9-4√5)n
also, p+β=(9+4√5)n (from above)
Multiply the two, we get
(1-β)(p+β)=1
:)