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Prove that any real number $r$ can be expressed as the sum of two irrational numbers $x$ and $y$.

Progress: I have a specific example for any rational number $r$: $x = r-\pi$ and $y = \pi$ (or replace $\pi$ with any irrational number.) However, I can't seem to find a way to prove this in general for irrational $r$.

Any help is greatly appreciated.

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9 Answers 9

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$\forall x\in\Bbb R$, the law of excluded middle tells us that $x$ must either be irrational or rational. This should be obvious since, after all, $x\in\Bbb Q$ or $x\in\Bbb R \setminus\Bbb Q$ by definition of rational and irrational.

If $x$ is irrational then $x = \frac{x}{2} + \frac{x}{2}$, where $\frac{x}{2}$ is irrational. This is clear since a rational, $\frac12$, times an irrational, $x$, is an irrational.

If $x$ is rational then $x = (\sqrt{2}) + (x-\sqrt{2})$, where both $\sqrt{2}$ and $x-\sqrt{2}$ are irrational. This is clear since a rational, $x$, added to an irrational $-\sqrt{2}$, is still irrational.

So all $x\in\Bbb R$ can be expressed as a sum of irrationals. $\;\;\;\blacksquare$


Try the general equality $x = (\frac12 x + z) + (\frac12 x -z)$, $\forall x \in\Bbb R$, where we choose an arbitrary $z\in\Bbb R$ such that $z\in\Bbb Q$ when $x\in\Bbb R\setminus \Bbb Q$ and $z\in\Bbb R\setminus\Bbb Q$ when $x\in\Bbb Q$. In this way both $(\frac12 x + z)$ and $(\frac12 x - z)$ are guaranteed irrational, and the sum is always $x$, regardless of which subset it comes from. If we further ensure that $z\ne 0$ then the two number forms are guaranteed distinct.

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If $r$ is irrational, then $r=\frac{r}2+\frac{r}2$.

To tell the truth, for $r$ rational we also have $r=\frac{r}2+\frac{r}2$, but it is not useful in this case. :-)

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    $\begingroup$ Or $r={1 \over 3}r+{2 \over 3}r$, if you want the numbers to be distinct :) $\endgroup$
    – psmears
    Apr 19, 2018 at 10:55
  • $\begingroup$ Yes I was going to ask if this is equivalent to showing that, if $r$ is rational and $s$ irrational, then $r-s$ is also irrational. $\endgroup$ Apr 20, 2018 at 0:58
  • $\begingroup$ @ZeroTheHero Yes, because if the sum $r+s=t$ would be rational, then $s=t-r$. $\endgroup$ Apr 20, 2018 at 1:03
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A cardinality-based proof:

Let $r - \mathbb{Q} = \{r-q | q \in \mathbb{Q}\}$. This clearly has the same cardinality as $\mathbb{Q}$ hence is countable. Let $A = \mathbb{Q}' \, \backslash \,(r-\mathbb{Q})$ be the set of irrational numbers with $r - \mathbb{Q}$ removed. $A$ is clearly uncountable. For any $x \in A$, both $x$ and $r-x$ are irrational and sum to $r$. Thus, not only can $r$ be expressed as the sum of two irrational numbers, it can be so expressed in uncountably many ways. Since there are only countably many pairs $(x,y)$ which contain at lease 1 rational number and sum to $r$, it follows that most pairs of numbers which sum to $r$ have both terms irrational.

The same proof works if the set of irrationals is replaced by any cocountable set of reals (a set whose complement is countable). For example, given any real number $r$, most pairs of numbers which sum to $r$ consists of two non-computable numbers.

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  • $\begingroup$ For this argument to work, $\mathbb{R} \setminus \mathbb{Q}$ can be replaced by any appropriately large set, e.g. of full Lebesgue measure, comeager, etc. $\endgroup$
    – Adayah
    Apr 19, 2018 at 19:53
  • $\begingroup$ @Hamsterrific Different notations are used in different contexts. My impression is that the prime notation is perhaps slightly more common in probability theory than in other branches of mathematics. I have used the c and the overbar notation as well. $\endgroup$ Apr 20, 2018 at 2:24
  • $\begingroup$ @JohnColeman I see. I just looked at Wikipedia and indeed the prime is listed as one of the notations. Cool, learning something new every day :) $\endgroup$
    – Pedro A
    Apr 20, 2018 at 2:25
  • $\begingroup$ What guarantees $r-x$ is irrational? Clearly $x$ is but presumably $r$ can be any value, including perhaps values that make $r-x$ rational. $\endgroup$ Jun 8, 2020 at 20:04
  • $\begingroup$ @CogitoErgoCogitoSum if $r - x = q \in \mathbb{Q}$ then $x = r - q \in r - \mathbb{Q}$ which contradicts the definition of $A$. $\endgroup$ Jun 8, 2020 at 21:31
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Hint

If $x$ is irrational, then $x/2$ is irrational

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You have $x = \sqrt2 + (x-\sqrt2)$ and $x = -\sqrt2 + (x+\sqrt2)$. The numbers $x+\sqrt2$ and $x-\sqrt2$ cannot both be rational, because their difference $2\sqrt2$ is irrational.

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Another cardinality based proof:

We prove that, for any real $r$, there exists $x$ making both $r/2-x$ and $r/2+x$ irrational. From this, since their sum is $r$, we can conclude.

By contradiction, suppose that, for any $x$ at least one of these is rational. We can then define, for any $x>0$,

$$ f(x) = \begin{cases} r/2 - x & \mbox{if rational} \\ r/2 + x & \mbox{otherwise} \end{cases} $$

By hypothesis, this is a function $f : \mathbb{R}^+ \to \mathbb{Q}$. It is also injective, since its inverse can be computed as $|f(x)-r/2|=x$. This implies the cardinality inequality $card(\mathbb{R})=card(\mathbb{R}^+)\leq card(\mathbb{Q}) = card(\mathbb{N})$. We know this does not hold, reaching a contradiction.

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  • $\begingroup$ Well, you can really just take $x=0$ if $r$ is irrational, and $x=\sqrt{2}$ otherwise. $\endgroup$
    – Vincent
    Apr 19, 2018 at 20:26
  • $\begingroup$ @Vincent Sure! That is indeed the proof I like the most, and it was posted above. I wanted to share another proof, to show a different approach, even if less elegant. $\endgroup$
    – chi
    Apr 19, 2018 at 20:36
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Here is a measure theoretical proof. It is a theorem that if $A, B \subset \mathbb{R}$ have positive Lebesgue measure then $A+B := \{a + b | a \in A, b \in B\}$ contains an interval. Let $A = B =$ the set of irrational numbers. Then $A+B$ contains an interval. But $A$ is invariant under translations by rationals so $A+B$ is as well. Therefore $A+B = \mathbb{R}$.

And another cardinality proof. Let $x \in \mathbb{R}$ and let $\mathbb{I}$ be the irrationals. Let $x - \mathbb{I} = \{x - i | i \in \mathbb{I}\}$. $i \mapsto x - i$ is injective and $|\mathbb{I}| > |\mathbb{Q}|$ so $x - \mathbb{I} \not \subset \mathbb{Q}$. Therefore $(x - \mathbb{I}) \cap \mathbb{I} \ne \emptyset$, which means $x$ is the sum of two irrationals.

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Let $r \in \mathbb{R}$, $A := \mathbb{R}\setminus \mathbb{Q}$ and $B := \{r - x \ \vert \ x \in \mathbb{R}\setminus \mathbb{Q}\}$. Then both $A$ and $B$ are countable intersections of dense open subsets of $\mathbb{R}$, so $A \cap B$ is a countable intersection of dense open subsets. By Baire's theorem, it is dense, and hence it is not empty. Let $x \in A\cap B$. Then $r = x + r- x$ and both $x$ and $r-x$ are irrational.

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A somewhat more general proof:

If $r \in \mathbb{Q}$ select $a \in \mathbb{R}\setminus\mathbb{Q} \neq \emptyset$. Since $\mathbb{R}$ is a field, $b := q - a \in \mathbb{R}$. Suppose $b\in\mathbb{Q}$, then $a = q - b \in \mathbb{Q}$ in contradiction to $a \in \mathbb{R}\setminus\mathbb{Q}$. Thus $a, b \in \mathbb{R} \setminus \mathbb{Q}$ and $a+b=q$.

If on the other hand $r \in \mathbb{R}\setminus\mathbb{Q}$, then use $a := b := \frac{r}{2} \in \mathbb{R}\setminus\mathbb{Q}$ (because, if $a,b\in\mathbb{Q}$ was true, $r = 2a \in \mathbb{Q}$ yielding a contradiction) with $a + b = r$.

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