Real number inequalities Suppose that ${\alpha,\beta,\gamma}$ are real numbers with ${\beta<\gamma}$.
(a) Prove that ${\alpha+\beta\leq\alpha+\gamma}$
(b) Prove that ${\alpha+\beta\neq\alpha+\gamma}$
I have gone about this various ways and I think it comes down to confusion as to what is meant by ${\alpha}$ and ${\beta}$.  Because the definition of their sum is ${\gamma=\{a+b:a\in \alpha,b\in \beta\}}$ 
We are using the dedekind cut definition of real numbers and in this sense, these real numbers beta and gamma CONTAIN all numbers less than them.  so why couldn't i pick a number in gamma that is less than the number i pick in beta.  in which case, this isn't true.  or is the definition of sum the sum of the supremums of both cuts.  In which case, why is a sum defined like this - as if it is ANY element of beta plus ANY element of gamma.  Clearly, this shouldn't be a hard proof but I am lacking some fundamental understanding here.  
 A: With the Dedekind cut definition is that $\beta$ or any other real number, is viewed as a subset of $\mathbb Q$; a subset with properties $\beta\ne \emptyset, \beta \ne \mathbb R$ and for any $q \in \beta$ then for an $r\in \mathbb Q$ so that $r < q$ then $r\in \beta$; and that $\beta$ has no largest element.
Now I assume you have proven already that if $\alpha$ and $\beta$ are such cuts that the set $\tau = \{a+b|a\in \alpha; b\in \beta\}$ is also a Dedekind cut.
So you need to prove that $\alpha, \beta, \gamma$ are cuts that if $\beta \subsetneq \gamma$ (that's what $\beta<\gamma$ means) that $\alpha + \gamma \subseteq \alpha + \beta$.

[In this post when I use a roman letter such as $a,b,$ or $c$ it will be assumed they are rational numbers.  If I compare $a < b$ this is to be assumed to be the usual rational number order and not the Dedekind Cut  order.  $\beta < \gamma$ will mean $\beta \subsetneq \gamma$ and $\beta \le \gamma$ will mean $\beta \subseteq \gamma$]

So if $m\in \alpha +\beta$ then there are an $a\in \alpha$ and $b\in \beta$ so that $m = a+b$.  Now $b \in \beta\subset \gamma$ so $b\in \gamma$ and so $a+b \in \alpha + \gamma$. 
That's all there is to it.  
And to prove $\alpha + \beta \ne \alpha + \gamma$ is actually harder.
The problem is you can't do: let $c\in \gamma; c\not \in \beta$ (that's legit) and so let $a\in\alpha$ so $a+c> a+b$ for any $b\in \beta$ (which is true); but there could be an $a' >a$ so that $a' + b = a + c$.
We need a lemma and the notion that a cut has no largest element.
Lemma:  For any positive rational $d$ there is an $a\in \alpha$ so that $a + d\not \in \alpha$.
Pf: Let $a'\in \alpha;$ $c \not  \in \alpha$ then $c > a'$. Let $e=c-a'$.  Archimedian principal says there is a positive integer $n$ so that $nd > e$.
Consider $a' + kd$ for integers $k=0... n$.  $a'+0*d\in \alpha$ and $a'+nd > a'+e=c \not \in \alpha$. So there is some first element $k$ so that $a'+kd\not \in \alpha$ and a last element $a'+(k-1)d \in \alpha$.  Let $a=a'+(k-1)d$ and that is an $a\in \alpha$ so that $a+d \not \in \alpha$.
Okay.  Let $c_1\in \gamma; c_1\not \in \beta$. There is a $c_2\in \gamma; c_2 > c_1$.  Let $d=c_2 - c_1$.  Let $a\in \alpha$ so that $a + d\not \in \alpha$.
$a + c_2 \in\alpha +  \gamma$ but we can show that $a+c_2\not \in \alpha + \beta$.

Let $m=a'+b\in \alpha + \beta$ with $a'\in \alpha $ and $b\in \beta$. we will show that $a'+b$ can not equal $a+c_2$.

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$b < c_1 < c_2$ so $b < c_2-d$.  So $a'+b < (a'-d) +c_2$.  $a'-d < a'$ so $a'-d\in \alpha$ and $a'=(a'-d)+d\in \alpha$. 

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But $a+d \not \in \alpha$ so $a+d >a'$ and $a > a'-d$.

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So $a'+b < (a'-d)+c_2 < a+c_2$. So $a'+b$ can not be equal to $a+c_2$ so $a+c_2 \not \in \alpha + \beta$.

