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Question

If (2+3)n=I+f where I and n are +ive integers and 0 < f < 1, Show that I is an odd integer and (1f)(I+f)=1

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Solution

(2+3)n = I + f.
or I + f = 2n+nC12n13+nC22n2(3)3+nC32n3(3)3+
Now, 0 < 2 - 3 < 1. 0(23)n<1.
Let (23)n = f' where 0 < f' < 1.
f=2nnC12n13+nC22n2(3)2nC32n3(33)3+
Adding (1) and (2),
I + f + f' = 2[2n+nC22n2.3+.....]
or I + f + f' = even integer.
Now 0 < f < 1 and 0 < f' < 1.
0 < f + f' < 2.
Hence from (3) we conclude that f + f' is an integer between 0 and 2.
f + f' = 1 f' = 1 - f.
From (3) and (4) we get I + 1 = even integer.
I is an odd integer.
Now I + f = (2+3)n, f' = 1 - f = (23)n
(I + f) (1 - f) = [(2 + 3) (2 - 3)]n=(43)n=1.
(I + f) (1 - f) = 1

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