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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. – GNU Emacs Dec 9 '13 at 6:33

3 Answers 3

up vote 19 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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