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If $x$ and $y$ are irrational, is $x + y$ irrational? Is $x - y$ irrational?

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What is $\sqrt{2} - \sqrt{2}$? – BlueRaja - Danny Pflughoeft Jun 12 '12 at 17:29
$ (\pi) $ and $ (1 - \pi) $ are irrational. But $ (\pi + (1 - \pi)) = 1$ is not. – hkBattousai Jun 13 '12 at 4:11
If $x$ and $y$ are irrational, then at least one of $x+y$ , $x-y$ is irrational. Maybe that's what you want. – fermesomme Dec 9 '13 at 6:33
up vote 25 down vote accepted

$(2+\sqrt{2}) +(2-\sqrt{2}) = 4$ and $(2+\sqrt{2}) -(-2+\sqrt{2}) = 4.$

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The short answer to your question is that is not necessarily true. For instance, $\sqrt{2}, \sqrt{2}-1,1-\sqrt{2}$ are all irrational but $$\sqrt{2} + (1-\sqrt{2}) = 1 \in \mathbb{Q}$$ and $$\sqrt{2} - (\sqrt{2}-1) = 1 \in \mathbb{Q}$$

However, it is worth noting that if $x$ and $y$ are irrational, then either $x+y$ or $x-y$ is irrational i.e. $x+y$ and $x-y$ cannot be both rationals. The proof for this is given below.


If both $x+y$ and $x-y$ are rational, then we have that $x+y = \dfrac{p_1}{q_1}$ and $x-y = \dfrac{p_2}{q_2}$, where $p_1,p_2 \in \mathbb{Z}$ and $q_1,q_2 \in \mathbb{Z} \backslash \{0\}$.

Hence, $x = \dfrac{\dfrac{p_1}{q_1} + \dfrac{p_2}{q_2}}{2} = \dfrac{p_1q_2 + p_2q_1}{2q_1q_2}$ and $y = \dfrac{\dfrac{p_1}{q_1} - \dfrac{p_2}{q_2}}{2} = \dfrac{p_1q_2 - p_2q_1}{2q_1q_2}$.

Now $p_1q_2 + p_2q_1, p_1q_2 - p_2q_1 \in \mathbb{Z}$, whereas $2q_1q_2 \in \mathbb{Z} \backslash \{0\}$.

This contradicts the fact that $x$ and $y$ are irrationals. Hence, if $x$ and $y$ are irrational then either $x+y$ is irrational or $x-y$ is irrational.

Below are some statements worth knowing.

1 Sum of two rationals is always a rational.

Proof: Let the two rationals be $\dfrac{p_1}{q_1}$ and $\dfrac{p_2}{q_2}$, where $p_1,p_2 \in \mathbb{Z}$ and $q_1,q_2 \in \mathbb{Z} \backslash{0}$. Then $$\dfrac{p_1}{q_1} + \dfrac{p_2}{q_2} = \dfrac{p_1q_2 + p_2q_1}{q_1q_2}$$ where $p_1q_2 + p_2q_1 \in \mathbb{Z}$ and $q_1 q_2 \in \mathbb{Z} \backslash \{0\}$. Hence, the sum is again a rational.

2 Sum of a rational and an irrational is always irrational.

Proof: Let the rational number be of the form $\dfrac{p}{q}$, where $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$ while the irrational number be $r$. If $r + \dfrac{p}{q}$ is a rational, then we have that $r + \dfrac{p}{q} = \dfrac{a}{b}$ for some $a \in \mathbb{Z}$ and $b \in \mathbb{Z} \backslash \{0\}$. This means that $r = \dfrac{a}{b} - \dfrac{p}{q} = \dfrac{aq-bp}{bq}$ where $aq-bp \in \mathbb{Z}$ and $bq \in \mathbb{Z} \backslash \{0\}$. This contradicts the fact that $r$ is irrational. Hence, our assumption that $r + \dfrac{p}{q}$ is a rational is false. Hence, $r + \dfrac{p}{q}$ is a irrational.

3 Sum of two irrationals can be rational or irrational.

Example for sum of two irrationals being irrational

$\sqrt{2}$ is irrational. $\sqrt{2} + \sqrt{2} = 2 \sqrt{2}$ which is again irrational.

Example for sum of two irrationals being rational

$\sqrt{2}$ and $1-\sqrt{2}$ are irrational. (Note that $1-\sqrt{2}$ is irrational from the second statement.) But, $\sqrt{2} + (1-\sqrt{2}) = 1$ which is rational.

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$\sqrt 2 - 1$ is irrational being the sum of a rational and an irrational. – Abhishek Parab Jun 12 '12 at 5:14
@Abhishek: Yes, that’s the point: Marvis has given examples of irrational numbers with rational sum or difference. – Brian M. Scott Jun 12 '12 at 5:17
For the first proof, wouldn't this be simpler: if $x + y, x - y \in \mathbb{Q}$, then $(x + y) + (x - y) = 2x \in \mathbb{Q}$, so $x \in \mathbb{Q}$, and similarly for $y$? – ecatmur Jun 12 '12 at 8:06
good answer... + $\sqrt 2 +(1-\sqrt 2)$ – Phillip Schmidt Jun 12 '12 at 14:31
@Dancrumb I'd assume we've already established that $\mathbb{Q}$ is a field. Alternatively we can use 1 above as a lemma. – ecatmur Jun 12 '12 at 14:47

The two are the same question, since $x-y=x+(-y)$, and $-y$ is irrational if and only if $y$ is irrational. The answer is not necessarily: $\sqrt2+(-\sqrt2)=\sqrt2-\sqrt2=0$, which is rational.

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Using the language of group theory, a couple of the observations in user17762's excellent answer can be generalized considerably.

Proposition 0. Let $H$ denote a subgroup. Then: $$HH^c \subseteq H^c \,\mbox{ and }\,H^c H \subseteq H^c$$

That is: "An element of a subgroup times a non-element will always be a non-element."

Proof. We're trying to show:

$$\mathop{\forall}_{x,y \in G}(x \in H \wedge y \in H^c \rightarrow xy \in H^c)$$


$$\mathop{\forall}_{x,y \in G}(x \in H \wedge xy \in H \rightarrow y \in H)$$

So assume $x \in H$ and $xy \in H$. Then $x^{-1} \in H$, so $x^{-1}(xy) \in H$, so $y \in H$. QED

Remark. The above proposition doesn't hold if "subgroup" is replace by "submonoid." For example, let $H$ denote the set $\{0\} \cup \{2,3,4,\cdots\},$ viewed as an submonoid of the group $\mathbb{Z}.$ Then $2 \in H$ and $2+1 \in H$, but it would not be correct to say that $1 \in H$.

Anyway, as a special case, we deduce:

Corollary. $$\mathbb{Q} +\mathbb{Q}^c \subseteq \mathbb{Q}^c$$ That is: "A rational number plus an irrational number will always itself be irrational."

The opening observation of user17762's answer is trickier. It would be nice if the following were true:

Wishful thinking (Version A). Let $H$ denote a subgroup of $G$. Then: $$\mathop{\forall}_{x,y \in G} \left(x \in H^c \wedge y \in H^c \rightarrow xy \in H^c \vee xy^{-1} \in H^c\right)$$

But this is completely false, even in the Abelian case. For instance, let $H$ denote the subgroup $2\mathbb{Z}$ of $\mathbb{Z}$. Then $3$ and $1$ are non-elements of $H$, but despite this, both $3+1$ and $3-1$ are elements of $H$. In short: something has gonna horribly, awfully wrong!

Before going on, lets spend a few paragraphs to "improve" our wishful thinking in a very sensible way, despite that it will still be false. Then, later, we will add a premise to make it true.

Lets start off by strengthening Proposition 0 to the following:

Proposition 1. Let $H$ denote a subgroup of $G$. Then: $$\mathop{\forall}_{x,y \in G} \left(xy \in H \rightarrow (x \in H \leftrightarrow y \in H)\right)$$

Once again, the (extremely straightforward) proof is left as an exercise to the reader.

Remark. As before, the above proposition doesn't hold if "subgroup" is replace by "submonoid." For example, once again let $H$ denote the set $\{0\} \cup \{2,3,4,\cdots\},$ we observe that $2+1 \in H$, but it would not be correct to say that $2 \in H \leftrightarrow 1 \in H$.

Anyway, we can use the above theorem to remove improve our wishful thinking. Observe that under contraposition, the (false) claim under question is logically equivalent to

$$\mathop{\forall}_{x,y \in G} \left(xy \in H \wedge xy^{-1} \in H \rightarrow x \in H \vee y \in H\right)$$

But since we're assuming $xy \in H$, the statements $x \in H$ and $y \in H$ are equivalent, so this becomes:

$$\mathop{\forall}_{x,y \in G} \left(xy \in H \wedge xy^{-1} \in H \rightarrow x \in H \wedge y \in H\right)$$

Now contraposing again, we obtain:

Wishful thinking (Version B). Let $H$ denote a subgroup of $G$. Then: $$\mathop{\forall}_{x,y \in G} \left(x \in H^c \vee y \in H^c \rightarrow xy \in H^c \vee xy^{-1} \in H^c\right)$$

The case of the missing premise.

In order to see the problem with this (still bogus) "theorem," let us make a tentative attempt at proving the contrapositive. So assume: $$(0)\: xy \in H \qquad (1)\: xy^{-1} \in H.$$

The goal is to show that $x \in H$ and that $y \in H.$ Well from $(1)$, we know that $yx^{-1} \in H$. But $(0)$ says that $xy \in H.$ Hence, we can deduce $yx^{-1}xy \in H,$ or in other words we can deduce $y^2 \in H$. Unfortunately, we're now stuck. Fortunately, we've just found our missing premise! This becomes a definition:

Definition 0. Whenever $G$ is a group, call subgroup $H$ "square-root closed" iff it satisfies $$\mathop{\forall}_{x \in G}(x^2 \in H \rightarrow x \in H).$$

In symbols: $H^{1/2} \subseteq H$.

Remark. I don't think this is equivalent to $H \subseteq H^2$. Can anyone think of a counterexample?

Okay, we come to it at last:

Proposition 2. Let $H$ denote a square-root closed subgroup of $G$. Then: $$\mathop{\forall}_{x,y \in G} \left(x \in H^c \vee y \in H^c \rightarrow xy \in H^c \vee xy^{-1} \in H^c\right)$$

The proof is as above, and except that we use the square-root closedness of $H$ to finish it.

Returning to the subject of the rational numbers, note that since $\mathbb{Q}$ is being viewed as an additive subgroup of $\mathbb{R}$, hence to say that $\mathbb{Q}$ is square-root closed in the above sense is really just saying that $$\frac{1}{2} \mathbb{Q} \subseteq \mathbb{Q},$$ which is clearly true. Hence, we get


$$\mathop{\forall}_{x,y \in \mathbb{R}} \left(x \in \mathbb{Q}^c \vee y \in \mathbb{Q}^c \rightarrow x+y \in \mathbb{Q}^c \vee x-y \in \mathbb{Q}^c\right)$$

That is: "If either $x$ is irrational, or $y$ is irrational, then either $x+y$ is irrational, or $x-y$ is irrational."

This suggests an interesting question:

For which matrices $A \in \mathbb{R}^{2\times 2}$ is true that for all $x \in \mathbb{R}^2$, if one or both of the entries in $x$ fail to be rational, then one or both of the entries in $Ax$ will fail to be rational? We've just proved that the matrix $$\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}$$ has this property, but which others?

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Example: $\log_2(8) = 3$

can be rewritten as:

$\log_2(24) – \log_2(3) = 3$

It is easy to see that $\log_2(24)$ and $\log_2(3)$ are both irrational, due to the fact that $24$ and $3$ are not integer powers of $2$.

If you assume that there is some simplified rational number $\frac{a}{b}$ in which $a$ and $b$ are both integers, $b$ is not $1$, and $2^\frac{a}{b} = 3$, this implies that:

$$2^a = 3^b $$

which is impossible because raising $2$ to any integer power cannot magically yield $3$'s in its prime factorization, it is impossible for it to equal an integer power of $3$.

Similar reasoning can be applied to $24$, because $24$ does not contain all $2$'s in its prime factorization. A power of $2$ can never equal a power of $24$.

Prof. Yeager


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