Does This Sequence Converge in the 5-adic Metric?

[rednalino]

Metric Space and Topology HW help!

Let X be a metric space and let (s_n
)_n be a sequence whose terms are in X. We say that (s_n
)_n converges to s \ni X if
\forall \epsilon > 0 \exists N \forall n ≥ N : d(s_n,s) < \epsilon

For n ≥ 1, let j_n = 2[(5^(n) - 5^(n-1))/4].

(Convince yourself that 5^(n) - 5^(n-1) is always divisible by
4, so the exponent in the definition is always a positive integer.) The first few terms of this
sequence are
2; 32; 33554432; 42535295865117307932921825928971026432
so you would reasonably expect this sequence to diverge with respect to the usual metric on
Q (the one given by the usual absolute value).

However, show that |j²_n - (-1)|₅ ≤ 5^(-n) where ||₅ is the 5-adic absolute value.


My Attempt:
I started by writing the claim in terms of v₅(j²_n + 1). Then i tried to find a recurrence that looks like this:
(j²_n+ 1)^5 = (²_n+1+1) + (some other stuff).

I was thinking I can show that the sequence (j_n)_n is also Cauchy with respect to ||₅, so in Q₅,the completion of Q with respect to ||₅, the sequence (j_n)_n converges to a number j \niQ5 such that j² = -1. It follows that Q5 is not an ordered field, unlike the completion of Q with respect to the usual ||₅, which is our old friend R.

 

[Fredrik]

More people will be able to attempt to solve the problem if you define "5-adic absolute value".

 

[micromass]

More people will be able to attempt to solve the problem if you define "5-adic absolute value".

http://en.wikipedia.org/wiki/P-adic_number

Basically, you take the rational numbers \mathbb{Q} and you define some very weird norm on it. For a rational number q=a/b, you factor 5 out of it, so you get

q=5^n \frac{a^\prime}{b^\prime}

Then you define

|q|_5 = 5^{-n}

For example

\frac{23}{10} = 5^{-1} \frac{23}{2}

So |23/10|_5=5.

Also, |5^n|_5 = 5^{-n}. So 5^n converges to 0 in that norm.

Anyway. The OP should try to look at the expression

j_n^2+1

and he should try to factor out 5 as many times as he can. Maybe try it first for small n.

 

[Fredrik]

That is one crazy norm. 5ⁿ converges to 0.

 

[micromass]

That is one crazy norm. 5ⁿ converges to 0.

If p=10 (that's not a prime, I know), then you can show crazy things like

...9999999999 = -1

In the sense that

\sum_{k=0}^{+\infty} 9\cdot 10^k = -1

where convergence of the series is n the 10-adic norm.

It actually isn't so crazy. A naive student (who doesn't know that natural numbers have to have finitely many digitis), might do something like

Let x=...99999. Then 10x+9=x. So x=-1.
​

In some sense, the 10-adic numbers are a formalization of that (wrong) argument. In the 10-adic numbers, the argument does work

 

[rednalino]

http://en.wikipedia.org/wiki/P-adic_number

Basically, you take the rational numbers \mathbb{Q} and you define some very weird norm on it. For a rational number q=a/b, you factor 5 out of it, so you get

q=5^n \frac{a^\prime}{b^\prime}

Then you define

|q|_5 = 5^{-n}

For example

\frac{23}{10} = 5^{-1} \frac{23}{2}

So |23/10|_5=5.

Also, |5^n|_5 = 5^{-n}. So 5^n converges to 0 in that norm.

Anyway. The OP should try to look at the expression

j_n^2+1

and he should try to factor out 5 as many times as he can. Maybe try it first for small n.

starting with this
j_n^2+1
where n=1 which is 2 from the first term from the given sequence
gives |2² +1| = |5|₅ factored out 5 once.

repeated for the next term n=2, which gives |1025|₅
where 5 could be factored out twice.

repeated for the next term |33554432²+1| which could be factored out 9 times.

the last term given could have been factored at least 4 times.

how do i show that
5^{-V₅[j2_n+1]} ≤ 5^-n

 

[micromass]

You'll need to show that you can factor 5 from j_n^2 +1 at least n times.