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I'm going through Apostol's Calculus. And I'm not sure how to tackle this. In this section he introduced the least upper bounds, and Archimedian properties of the real number system. Any hints would be appreciated.

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up vote 6 down vote accepted

Hint: Example: if $x + y$ was rational, then also $$y = \underbrace{\underbrace{(x + y)}_{\text{rational}} - \underbrace{x}_{\text{rational}}}_{\text{rational}}.$$ Contradiction.

Use this idea for the other expressions.

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I don't seem to follow your reasoning. I think I'm missing something key here. –  AlexHeuman Mar 13 '13 at 18:47
I've added some information, is it clearer now? –  azimut Mar 13 '13 at 18:49
The idea is that $\Bbb{Q}$ is closed under addition, subtraction, multiplication, and division, as it is a field. –  Stahl Mar 13 '13 at 18:49
Okay, I get what you're saying now. Thanks for your help. –  AlexHeuman Mar 13 '13 at 18:54
You need to show that $x +y$ is irrational if $x \neq 0$ is rational and $y$ is irrational; we assume $y$ is irrational. If it were the case that $x + y$ is a rational number, then $(x+y) - x$ would also be a rational number. But that's just $y$, which contradicts our assumption that $y$ is irrational. –  anonymous Mar 13 '13 at 18:54
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Another example, multiplicatively:

Assume $x\in \mathbb Q,\; x\neq 0,\;$ and $\,y\in \mathbb R\setminus \mathbb Q$.

Note that $$x\neq 0, x\in \mathbb Q \iff \dfrac 1x \in \mathbb Q$$

Suppose for the sake of contradiction that $\;xy \in \mathbb Q.$
Then $$\dfrac 1x\cdot(xy) = \left(\dfrac 1x \cdot x\right)y = y \in \mathbb{Q}$$ since the set of rational numbers is closed under multiplication, and we have that $\dfrac 1x \in \mathbb Q$ and $xy \in \mathbb Q$. This contradicts the assumption that $y$ is not rational.

This can easily be modified and generalized for the cases $\dfrac xy$ and $\dfrac yx$.

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This is a nice example +1 –  Amzoti May 11 '13 at 0:38
Thanks, Amzoti! –  amWhy May 11 '13 at 0:38
Nice Observations. –  B. S. Jun 11 '13 at 0:47
Thanks, dear friend, @Babak ! –  amWhy Jun 11 '13 at 0:51
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It's true precisely because the rationals are a subset of reals closed under subtraction and division (by elements $\ne 0),$ i.e. they form additive and multiplicative subgroups of $\rm\,\Bbb R.\:$ If you know about groups then you can appreciate this from a more conceptual viewpoint. $ $ Let $\rm\:G\subset H\:$ be abelian groups, and $\rm\:\overline G = H\backslash G\:$ be the complement of $\rm\:G\:$ in $\rm\:H.\:$ Then

Lemma $\rm\ \ \ g\in G,\ \ \bar g\in \overline G\ \Rightarrow\ g - \bar g\in \overline G\quad (Proof\!:\ else\ \ g - \bar g\, =\, g_2\!\!\in G\:\Rightarrow\: \bar g\, =\, g\!\!-g_2\in G)$

So $\rm\,\ x \in \Bbb Q,\ \ y \in \overline{\Bbb Q}\ \Rightarrow\: x\!-\!y \in \overline{\Bbb Q},\ $ so $\rm\ 2x \in \Bbb Q\: \Rightarrow\: 2x\!-\!(x\!-\!y) = x+y \in \overline{\Bbb Q}.\ $ In the same way:

So $\rm\ x\in \Bbb Q^*\!,\, \ y \in \overline{\Bbb Q^*}\Rightarrow\, x/y \in \overline{\Bbb Q^*},\:$ so $\rm\: x^2\in \Bbb Q^*\Rightarrow\: x^2/\,(x\,/\,y)\, =\, x\, *\, y \in \overline{\Bbb Q^*}$.

Finally $\rm\ 1 \in \Bbb Q^*\Rightarrow\: 1/(x/y)\, =\, y/x \in \overline{\Bbb Q^*}.\ $ Note every inference is an instance of the Lemma.

Remark $\ $ This is a special case of the following complementary view of a subgroup.

Theorem $\ $ Let $\rm\,G\,$ be a nonempty subset of abelian group $\rm\,H,\,$ with complement set $\rm\,\overline G = H\backslash G.\,$ Then $\rm\,G\,$ is a subgroup of $\rm\,H\!\iff\! G - \overline G\, =\, \overline G. $

Proof $\ $ Recall that $\rm\,G\,$ is a subgroup of $\rm\,H\!\iff\! G\,$ is closed under subtraction (subgroup test).

$\begin{eqnarray}\rm Complementing, & &\ \ \rm G\text{ is not a subgroup of }\, H\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ is\ false\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \overline G\\ &\iff&\ \rm\ g_1\, -\ \bar g\ \ =\,\ g_2\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \overline G\\ &\iff&\ \rm\ G\ -\ \overline G\ \subseteq\ \overline G\ \ is\ false\quad\ {\bf QED} \end{eqnarray}$

Instances of this are ubiquitous in concrete number systems, e.g. below. For many further examples see various prior posts here.

enter image description here

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While this answer is highly edifying, the first sentence is basically saying, "what you are trying to prove is true because what you are trying to prove is true." The OP is asking how to prove that the "rationals are closed under subtraction and division". He could go with the group theoretical proof, which is the edifying part, but it seems like a long way around to go for something that can be proved from within the axioms for $\mathbb{Q}$, $\mathbb{R}$, and fields that he certainly has right in the book. –  Todd Wilcox Mar 13 '13 at 19:47
@Todd The OP is not asked to prove that rationals are closed under subtraction and division. Rather he is asked to prove statements about mixed operations on rationals and irrationals. As I show, these statements are equivalent to the rationals being closed under subtraction and division (by $\rm\:r\ne 0).\:$ Most students do not realize this equivalence, nor do they realize its group-theoretical essence. For a reader who knows the basics of groups, this viewpoint yields a shorter more conceptual proof, one that generalizes widely (cf. above examples and linked posts). –  Math Gems Mar 13 '13 at 19:59
Ok, good point. It does seem that to be rigorous he would have to prove both that $\mathbb{Q}$ is closed under subtraction and division and then prove the closure is equivalent to the desired proof. Again, edifying and also the long way around. –  Todd Wilcox Mar 13 '13 at 21:56
@Todd I added the details of the proofs from this viewpoint. Notice that each proof follows from a single subtraction or division. So doing it this way is not only more general, but also more efficient. –  Math Gems Mar 13 '13 at 22:05
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