Answer to Question #257691 in Real Analysis for dave

Question

R is an ordered eld which properly contains Q as an ordered subfield.

(a) Does R have elements 0 and 1 such that x+0 = x for all x 2 R and x
1 = x for

all x 2 R.

(b) Prove that 0 = 0 and 1 = 1 where 0 and 1 are the integers that we know very

well.

Answer 1

Field is abelian group with respect to + and abelian group on R\{0} withrespext to * and $0 \ne1$ . Therefore elements $1=1_R,0=0_R$ exist.

Let $0_Q$ zero in subfield Q. We have identity $0_Q+0_Q=0_Q$

Let $(-0_Q)$ inverse in (R,+) element of $0_Q$

Tnen

$(-0_q)+0_Q+0_Q=(-0_Q)+0_Q;\\ 0+0_Q=0;\\ 0=0_Q$

By the same way $1_Q\cdot 1_Q=1_Q$

$1_Q\ne 0$ because $0=0_Q$

Let  $1_Q^{-1}$ be inverse of $1_Q$ in R

Then

$1_Q^{-1}\cdot 1_Q\cdot 1_Q=1_Q^{-1}\cdot 1_Q=1=1_R$

$1_R^ \cdot 1_Q=1_R\\ 1_Q=1_R$