Using the axioms of ordered field

[NoName3]

I only have the axioms of ordered plus the basic laws of equality and inequalities that follow from it.

1. Prove that $1 < 2$.

2. Prove that $0 < \frac{1}{2} < 1$

3. Prove that if $a, b \in \mathbb{R}$ and $a < b$ then $a < \frac{a+b}{2} < b$

1. Assume the opposite - that's, suppose that $2 < 1$, then:

$\begin{aligned}
~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \begin{aligned} \\& ~~~~~2 <1 \\& \text {(Assumption)}
\\& \iff 2+(-1) < 1+(-1) ~~~~~~~~~~~~~~~~~~~~~~~ \\& \text {(Addition axiom)}
\\& \iff 2+(-1) < 0 ~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~\\& \text{(Inverse law for addition)}
\\& \iff 2+(-1)+(-1) < - 1 ~~~~~~~~~~~~~~~~~~~~~~~~~~ \\& \text{(Addition law for order)}

\\& \iff 2+(-2) < -1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\& \text{(Associative law for addition)}

\\& \iff 0 < -1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\& \text{(Inverse law for addition)}

\\& \iff 0 < (-1) (-1) \\& \text{(Multiplication law of order)}

\\& \iff 0 < 1 \\& \text{(Inverse law for multiplication)}

\\& \iff 0 < (1)+(1) < 2 \\& \text{(Addition law for order)}

\\& \iff 1 < 2 \\& \text{(Contradiction to assumption)}.\end{aligned} \end{aligned} $

1. I realized that I could simply said since we have the rule $-1 < 0 < 1.$ We have $0 < 1 \iff 1 < (1)+(1) \iff 1 < 2$.

2. From part (1) we have $1 < 2 \iff 2^{-1} < 1 \iff \frac{1}{2} < 1$. Also we have since $0 < 1$ and $1 < 2$ then $0 < 2 \iff 0 < \frac{1}{2}$.

Putting this together we get $0 < \frac{1}{2} < 1$. However, I still can't prove part (3).

 

[Greg]

For 3, can you start with $a=\dfrac{a+a}{2}<\dfrac{a+b}{2}$?

 

[NoName3]

For 3, can you start with $a=\dfrac{a+a}{2}<\dfrac{a+b}{2}$?

So $a=\dfrac{a+a}{2}<\dfrac{a+b}{2} < \dfrac{b+b}{2} = b.$ Thus $a < \dfrac{a+b}{2} < b$.

Brilliant! Thank you!