Re: with Q.

In article <dqlb9m$g1m$1@xxxxxxxxxxxxxxxxx>,
mina_world <mina_world@xxxxxxxxxxx> wrote:
>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.
>
>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.

You won't be able to. There is no (discernible) problem with the
powers of 2 (since the 2-part of 12, namely 4, ->is<- a square). The
problem is with the powers of 3.

Try it with 3 instead of 2.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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