Re: with Q.
- From: "Nobody" <Nobody@xxxxxxxxxxx>
- Date: Wed, 18 Jan 2006 07:41:38 -0500
"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
news:dqlb9m$g1m$1@xxxxxxxxxxxxxxxxxxxx
> hello......doctor~
>
> there does not exist "x" in rational number
> such that x^2 =12.
>
> ---------------------------------------
> if there exists "x" in Q,
> x = n/m , (n,m)=1. n,m in Z.
>
> so, (n/m)^2 = 12
>
> => n^2 = 12.m^2
>
> so, n^2 is even => n is even.
>
Change the line above to:
" so, 3 divides n^2 => 3 divides n "
and continue to show that 3 divides m as well.
> so, n=2k form.
>
> since n^2 = 12.m^2, (2k)^2 = 12.m^2.
>
> => 4k^2 = 12.m^2
>
> => k^2 = 3.m^2
>
> i want to deduce that m is even and (m,n) =/= 1.
> in the end, "x" does not exist by contradiction.
>
> but i can't induce the my idea.
>
> is this impossible work ?
>
> i want to know a correct solution in this case.
>
> thank you very much for your advice.
>
>
.
- References:
- with Q.
- From: mina_world
- with Q.
- Prev by Date: Re: zero to the power zero
- Next by Date: Re: rotation matrices- PitchYaw
- Previous by thread: with Q.
- Next by thread: Re: with Q.
- Index(es):
Relevant Pages
|
