Question

What is the 100th digit to the right of the decimal point

Answer 1

Answer:

you know about “algebraic integers”, you know f(n)=(1+2–√)n+(1−2–√)n is a (positive) integer for n≥0, because it’s an algebraic integer that is also a rational number. Without using that theory, though, you can also prove it by induction.

This f(n) is a sequence of integers that satisfies a recurrence relation (like the Fibonacci numbers but with a different recurrence relation). For n=0 it is 2. For n=1 it is again 2. If n≥2 then (1+2–√)n+(1−2–√)n=(1+2–√)2∗(1+2–√)n−2+(1−2–√)2∗(1−2–√)n−2. Because (1+2–√)2=3+2∗2–√=2∗(1+2–√)+1 (and (1−2–√)2=3–2∗2–√=2∗(1−2–√+1) we can rewrite f(n) as (2∗(1+2–√+1))(1+2–√)(n−2)+(2∗(1−2–√)+1)∗(1−2–√)(n−2)=2∗f(n−1)+f(n−2).

So (1+2–√)3000 is an integer with (1−2–√)3000 subtracted from it. Because the exponent is even, this is a positive number. But (1−2–√)3000=(2–√−1)3000<(1/2)3000=((1/2)10)300<(0.001)300=10(−900). If we subtract such a small positive value from a positive integer we get a number whose first several hundred digits after the decimal point are all 9.

I hope I haven’t made a mistake.