Tag: Irrationals

Prove there is no rational number which squares to 2.

by RoRi

July 1, 2015

Prove that there is no $r \in \mathbb{Q}$ such that $r^2 = 2$ .

Proof. Suppose otherwise, that there is some rational number $r \in \mathbb{Q}$ such that $r^2 = 2$ . Then, since $r \in \mathbb{Q}$ , we know there exist integers $a,b \in \mathbb{Z}$ not both even (I.3.12, Exercise #10 (e) such that $r = \frac{a}{b}$ . Then, since $r^2 = 2$ we have

$2 = \frac{a^2}{b^2} \quad \implies \quad 2b^2 = a^2.$

But, by I.3.12, Exercise #10 (d), we know $2b^2 = a^2$ implies both $a$ and $b$ are even, contradicting our choice of $a$ and $b$ not both even. Hence, there can be no such rational number $. \qquad \blacksquare$

Prove there is an irrational number between any two real numbers.

by RoRi

July 1, 2015

Let $x, y \in \mathbb{R}$ be given with $x < y$ . Prove that there exists an irrational number $z$ such that $x < z < y$ .

Note: To do this problem, I think we need to assume the existence of an irrational number. We will prove the existence of such a number (the $\sqrt{2}$ ) in I.3.12, Exercise #12.

Proof. Since the rationals are dense in the reals I.3.12, Exercise #6, we know that for $x, y \in \mathbb{R}$ with $x<y$ there exist $r,s \in \mathbb{Q}$ such that

$x < r < s < y.$

Now, assume the existence of an irrational number, say $w$ (see note preceding the proof about this). Since $w \in \mathbb{R}$ we know $-w \in \mathbb{R}$ and from the order axioms exactly one of $w$ or $-w$ is positive ( $w$ is nonzero since $0 \in \mathbb{Q}$ ). Without loss of generality, let $w > 0$ . Then, since $s-r > 0$ , we know there exists an integer $n$ such that

$n (s-r) > w \implies s > \frac{w}{n} + r$

Also, since $n,w > 0$ , we have $\frac{w}{n} > 0$ ; thus, $r < r + \frac{w}{n}$ .
Then, by I.3.12, Exercise #7 we have $\frac{w}{n}$ irrational and hence $r+\frac{w}{n}$ irrational.
Thus, letting $z = r + \frac{w}{n}$ , we have $x < z < y$ with $z$ irrational $. \qquad \blacksquare$

Sum and product of irrationals is not necessarily irrational.

by RoRi

July 1, 2015

Prove that the sum of two irrational numbers need not be irrational and the product of two irrational numbers need not be irrational.

Proof. Let $y$ be irrational. Then, by an argument we made in I.3.12, Exercise #7, we know that $-y$ and $y^{-1}$ are both irrational. However,

$y + (-y) = 0 \in \mathbb{Q} \qquad \text{and} \qquad y \cdot \left( \frac{1}{y} \right) = 1 \in \mathbb{Q}.$

Therefore, the sum and product of two irrationals need not be irrational $. \qquad \blacksquare$

(Note: we could also just say $\sqrt{2} \cdot \sqrt{2} = 2$ is not irrational, but we have not yet established that $\sqrt{2}$ is irrational. In fact, I’m not sure we’ve even established that an irrational number actually exists yet.)

Sums, differences, products and quotients of an irrational and a rational are irrational.

by RoRi

June 30, 2015

Prove that if $x \in \mathbb{Q}$ with $x \neq 0$ , and $y \in \mathbb{R} \smallsetminus \mathbb{Q}$ , then

$x+y,\quad x-y, \quad xy, \quad \frac{x}{y}, \quad \frac{y}{x}$

are all irrational, i.e., are all in $\mathbb{R} \smallsetminus \mathbb{Q}$ .

Proof. Since $x \in \mathbb{Q}$ , we know there are integers $m$ and $n$ such that $x = \frac{m}{n}$ . Now we consider each of the elements we wish to show are irrational.

Suppose otherwise, that $x+y \in \mathbb{Q}$ . Then, there exist $r,s \in \mathbb{Z}$ such that
$x+y = \frac{r}{s} \ \implies \ \frac{m}{n} +y = \frac{r}{s} \ \implies \ y = \frac{r}{s} - \frac{m}{n} \implies \ y = \frac{rn - ms}{sn} \ \implies \ y \in \mathbb{Q}.$

This contradicts our assumption that $y$ is irrational.
Since $y$ is irrational, so is $-y$ (by part (a), the sum of a rational and an irrational cannot be rational and since $y+(-y) = 0$ is rational, cannot have $-y$ rational). But then, $x-y = x+(-y)$ and by part (a) this sum must be irrational.
Suppose otherwise, that $xy \in \mathbb{Q}$ . Then there exist $r,s \in \mathbb{Z}$ such that $xy = \frac{r}{s}$ . Further, since $x \neq 0$ , we know $x^{-1} = \frac{n/m}$ exists.
$xy = \frac{r}{s} \ \implies \ \frac{m}{n} y = \frac{r}{s} \ \implies \ y = \frac{r}{s} \frac{n}{m} \ \implies \ y = \frac{rn}{sm} \in \mathbb{Q}.$

Contradicting our assumption that $y$ is irrational.
First, we since $y$ is irrational, we have $y \neq 0$ , and thus $y^{-1}$ exists. Further, since $yy^{-1} = 1$ is rational, and $y$ is irrational, by (c) we must have $y^{-1}$ irrational as well. Then by (c), since $\frac{x}{y} = xy^{-1}$ , we must have $xy^{-1}$ irrational.
By (d) we know $\frac{x}{y}$ is irrational, and since $\frac{x}{y} \frac{y}{x} = 1$ is rational, by (c), we must have $\frac{y}{x}$ irrational.