The sum of two irrational numbers is

Operations with Rational and Irrational Numbers

Alignments to Content Standards: N-RN.B.3

Task

Experiment with sums and products of two numbers from the following list to answer the questions that follow: $$ 5,\tfrac{1}{2},0,\sqrt{2},-\sqrt{2},\tfrac{1}{\sqrt{2}},\pi. $$

Based on the above information, conjecture which of the statements is ALWAYS true, which is SOMETIMES true, and which is NEVER true?

1. The sum of a rational number and a rational number is rational.
2. The sum of a rational number and an irrational number is irrational.
3. The sum of an irrational number and an irrational number is irrational.
4. The product of a rational number and a rational number is rational.
5. The product of a rational number and an irrational number is irrational.
6. The product of an irrational number and an irrational number is irrational.

IM Commentary

This task has students experiment with the operations of addition and multiplication, as they relate to the notions of rationality and irrationality. As such, this task perhaps makes most sense after students learn the key terms (rational and irrational numbers), as well as examples of each (e.g., the irrationality of $\sqrt{2}$, $\pi$, etc.), but before formally proving any of the statements to be discovered in this task. Discussion of such proofs is taken up in other tasks.

These conjectures are likely best discussed in small groups and/or with the whole class, and so is best used in instructional, rather than assessment-based, settings. The discussions generated by student conjectures will likely yield productive insights into the nature of sums and products of real numbers leading eventually to the explanations sought in content standard N.RN.3, preparing them for the formal statements of these results. Note that some of these decisions, e.g., the irrationality of $\pi+\sqrt{2}$, are well beyond the scope of high school mathematics, but this does not preclude students from being able to answer the always/sometimes/never questions being asked.

Solution

Solution to part (c):

1. The sum of a rational number and a rational number is rational.

   Always true.

2. The sum of a rational number and an irrational number is irrational.

   Always true.

3. The sum of an irrational number and an irrational number is irrational.

   Only sometimes true (for instance, the sum of additive inverses like $\sqrt{2}$ and $-\sqrt{2}$ will be 0).

4. The product of a rational number and a rational number is rational.

   Always true.

5. The product of a rational number and an irrational number is irrational.

   Not true -- but almost! This holds except when the rational number is zero.

6. The product of an irrational number and an irrational number is irrational.

   Only sometimes true (for instance, the product of multiplicative inverses like $\sqrt{2}$ and $\frac{1}{\sqrt{2}}$ will be 1).

Operations with Rational and Irrational Numbers

Experiment with sums and products of two numbers from the following list to answer the questions that follow: $$ 5,\tfrac{1}{2},0,\sqrt{2},-\sqrt{2},\tfrac{1}{\sqrt{2}},\pi. $$

Based on the above information, conjecture which of the statements is ALWAYS true, which is SOMETIMES true, and which is NEVER true?

1. The sum of a rational number and a rational number is rational.
2. The sum of a rational number and an irrational number is irrational.
3. The sum of an irrational number and an irrational number is irrational.
4. The product of a rational number and a rational number is rational.
5. The product of a rational number and an irrational number is irrational.
6. The product of an irrational number and an irrational number is irrational.

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:

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

Equivalently:

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

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

Corollary.

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

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?

Is the sum of two irrational numbers irrational?

Why is the sum of two irrational numbers rational?

What is the sum of two irrational numbers with example?

Are the sum of irrational numbers always irrational?