# 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.

• It helps to use the definition of Dedekind cut (and addition of them) to clarify what the hypotheses and desired conclusions are. We're assuming that $\beta\subset\gamma$ and that $\beta\ne\gamma$ (so that, for instance, there exists $x\in\gamma$ with $x\notin\beta$). We're trying to prove that $\{a+b\colon a\in\alpha,\,b\in\beta\}\subset\{a+c\colon a\in\alpha,\,c\in\gamma\}$ and that the two sets are not equal (so it would suffice to prove, for example, that there exists $y\in\{a+c\colon a\in\alpha,\,c\in\gamma\}$ such that $y\notin\{a+b\colon a\in\alpha,\,b\in\beta\}$. Can you go from there? – Greg Martin Feb 14 at 6:07
• " so why couldn't i pick a number in gamma that is less than the number i pick in beta." That would only tell you information about one element. Nothing about the sets as a whole. "as if it is ANY element of beta plus ANY element of gamma" we aren't finding one sum. It is a set of all possible sums. – fleablood Feb 14 at 7:47
• Example $\alpha =\{q\in\mathbb Q|q^3 < 5\}; \beta=\{q\in\mathbb Q|q\le$ or a circle will circumference of $q$ will have a diameter$< 1\};\gamma=\{q\in\mathbb Q|q^5<1023\}$. Then you must prove $\alpha + \beta=\{m+n| m\in \alpha;n\in \beta\}\subsetneq\alpha+\gamma=\{m+n|m\in\alpha;n\in \gamma\}$. Only we have no way to refer to identify the supremas. (In this example $\alpha\approx \sqrt[3]{5}$, $\beta\approx \pi$ and $\gamma\approx \sqrt[5]{1023}$. – fleablood Feb 14 at 8:04
• greg martin yes, thank you. – mayalarson Feb 14 at 14:28
• I guess one thing is that for cuts $\alpha \le \beta$ doesn't mean the values $a \in alpha;b\in \beta$ are such that $a \le b$ (which isn't true as you point out) but that $\alpha \subseteq \beta$. It's a different ordering. Some text us different notatation such $\alpha \underline{\prec}\beta$ means $\alpha \subseteq \beta$. This forms an ordering. – fleablood Feb 14 at 17:41

## 1 Answer

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 $$a+c_2 \not \in \alpha + \beta$$.