# [Haskell-cafe] Re: 0/0 > 1 == False

Pierre-Evariste Dagand pedagand at gmail.com
Sat Jan 12 10:05:28 EST 2008

2008/1/12, Cristian Baboi <cristi at ot.onrc.ro>:
> Suppose lim a_n = a , lim b_n = b, c_2n = a_n, c_2n+1 = b_n.
>
> What is lim c_n ?

This reminds me of these good old days in "Classe prépa" (something
typically french) doing these silly things with sequences and
Epsilons...

So let's try to do it :

First, if we assume that a = b, that's ok : lim c_n = a = b

Second, we consider that a \neq b. We will show that c_n do not have a limit.
By contraction, we assume that  c_n has a limit c.

Then, by definition of limit, we have :

\exists N, \forall n \geq N, | c_n - c | \leq \epsilon

Therefore, this result holds for the suits (c_2n) and (c_{2n+1}), ie. :

\exists N, \forall n \geq N, | c_2n - c | \leq \epsilon     and
\exists N, \forall n \geq N, | c_{2n+1} - c | \leq \epsilon

By definition of c_2n and c_{2n+1}, we can replace them by a_n and b_n :

\exists N, \forall n \geq N, | a_n - c | \leq \epsilon      and
\exists N, \forall n \geq N, | b_n -c | \leq \epsilon

By definition of limit, this means that :

lim a_n = c      and    lim b_n = c

By unicity of the limit, this means that :
a = c               and     b = c

Leading to a contradiction as we have assumed that a \neq b.

Q.E.D.

But I'm still wondering about the relation between this and Haskell...

--
Pierre-Evariste DAGAND


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