Exercise 3.8 of Ross' Elementary Analysis Let $a,b \in  \mathbb{R}$. Show if $a \leq b_1$ for every $b_1 > b$, then $a \leq b$.
It took my a while to realise that there is no mistake in this formulation. And then someone helped me with the strategy.
Proof by contradiction:
Let $a>b$ and let $b_1 = \frac{a+b}{2}$. But how do I derive a clean contradiction from this?
I multiply with 2:
$2(b_1) = b_1 +b_1 = a+b$.
Now I can see some contradiction. Since $b_1 >b$, it has to be the case that $a > b_1$ for the equation to work out, which contradicts our assumption. But how do I write that down in formalese?
Secondly, I have a hard time seeing how the false assumption that $a>b$ is what leads to the contradiction. $b_1 +b_1 = a+b$ would have been false anyway for $a =b_1$ at best and $b_1 >b$.
I feel really stupid!
 A: As we have $a>b$, then by this assumption we have $\dfrac{a+b}{2} > b$, so the choice of $b_1$ is dependent on the assumption of $a>b$. Now if we put
$ b_1 \ge a $, we immediately  get $b \ge a$, which is a contradiction
A: Inspired by Aditya, I write down an easy to understand proof.
Assume $a>b$.
$a>b \iff a+b>b+b \iff a+b>2b \iff \frac{a+b}{2}>b$. Now we let $b_1=\frac{a+b}{2}$ and take the assumption that $a \leq b_1$. It follows that $\frac{a+b}{2} \geq a \iff a+b \geq2a \iff a+b \geq a+a \iff b\geq a$. And that last inequality is in contradiction to our assumption from which we wanted to derive a contradiction.
A: Alternative approach:
Assume 
$\displaystyle a > b \implies (a - b) > 0$ 
and establish a contradiction.
Let $t = (a - b)$ and  $b_1 = a - \dfrac{t}{2}.$
This yields a contradiction because:

*

*$t > 0 \implies \dfrac{t}{2} > 0.$


*$b = a - t \implies b_1 = b + \dfrac{t}{2} \implies b_1 > b. $


*$a - b_1 = \dfrac{t}{2} > 0 \implies a > b_1.$
A: Your proof will not hold, because you made two assumptions, so a contradiction will yield two possible false assumptions. The proof is not trivial:
By request, $\{a;b\}\in\mathbb{R}$, $a\leq b_1$ whenever $b_1>b$. This means that we are left only with two possible intervals $\{A\ni a;B\ni b\}\subset\mathbb{R}$:
$$A=(-\infty;b_1]$$
$$B=(-\infty;b_1)$$
and the propositions in the exercise are equivalent to:
$$a\in A\Leftrightarrow b\in B$$
This fact together with $b_1>b$ proves $a=b_1\Leftrightarrow b<a$  and that $a\text{ and }b$ are in $A\cap B$.
Going back to the order relations, we have that $a<b_1$ and $b<b_1$, so the only thing we can surely state about $a$ and $b$ is that they share a least upper bound $b_1$. If you make the claim: $a>b$, then there exists an $a$ wich is an upper bound for all $b$‘s, but $a<b_1$, so $a$ is a lower bound than $b_1$, wich is a contradiction, so it must be the case that $a\leq b$ (Trivially true, because it can’t be greater than), hence your proof is over.
However, if you make another assumption (like the $b_1=\dfrac{a+b}{2}$) you’ll have to prove first that the assumption you made is true before you can discard the first one, but that is impossible, since the only way you can prove $b_1=\dfrac{a+b}{2}$ is by showing that $a$ and $b$ both equal $b_1$, but $a+b<2b_1$, hence $a=b≠b_1$. This shows that the two assumptions would cause the proof to fail.
A: Too long for a comment
‘ Secondly, I have a hard time seeing how the false assumption that $a>b$ is what leads to the contradiction.’ Intuitively speaking, you can imagine the real line, with $x>y$ if $y$ is on the right side of $x$. Then, if $a>b$, there is definitely a point $c$ between $a$ and $b$. Namely $c$ is at the right side of $b$, but left side of $a$, which contradicts what is stated: points ‘righter’ than $b$ is ‘non-righter’ than $a$. But this intuition is quite informal, and you should construct a $c$ to make it rigorous. In your post you have constructed $c=\dfrac{a+b}{2}$. There are definitely other constructions, for instance $\lambda a+(1-\lambda)b$ for any $0<\lambda<1$.
