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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.

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    $\begingroup$ 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? $\endgroup$ – Greg Martin Feb 14 at 6:07
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    $\begingroup$ " 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. $\endgroup$ – fleablood Feb 14 at 7:47
  • $\begingroup$ 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 $\gamma\approx \sqrt[5]{1023}$. $\endgroup$ – fleablood Feb 14 at 8:04
  • $\begingroup$ greg martin yes, thank you. $\endgroup$ – mayalarson Feb 14 at 14:28
  • $\begingroup$ 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. $\endgroup$ – fleablood Feb 14 at 17:41
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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 $a+c_2 \not \in \alpha + \beta$.

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