- #1

issacnewton

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Suppose ##x \in \mathbb{Q}## and ##x > 1## and ## x^2 < 2##. I need to come up with some ##y \in \mathbb{Q}## such that ##x < y## and ## y^2 < 2##. Here is my attempt. Give that ##x > 1## and ## x^2 < 2##, I have ## (2-x^2) > 0## and ##4x > 0##. Also, ##2x >0##. Now define

$$ \alpha = \text{ min} \Bigl\{ \frac{(2-x^2)}{4x}, 2x \Bigl\} $$

It can be seen that ##\alpha > 0##. Also, I have

$$ \alpha \leqslant \frac{(2-x^2)}{4x} \cdots \cdots (1) $$

$$ \alpha \leqslant 2x \cdots \cdots (2)$$

Since ##x \in \mathbb{Q}##, it can be seen that ##\alpha \in \mathbb{Q}##. Since ##\alpha > 0##, there exists some ##\beta \in \mathbb{Q}## such that ## 0 < \beta < \alpha ##. From above equations, it follows that

$$ 0 < \beta < \frac{(2-x^2)}{4x} \cdots \cdots (3) $$

$$ 0 < \beta < 2x \cdots \cdots (4)$$

This simplifies to the following

$$ x^2 + 4x \beta < 2 $$

$$ \beta^2 < 2\beta x $$

Now consider ##(x + \beta)^2##

$$ (x + \beta)^2 = x^2 + 2\beta x + \beta^2 $$

Using above inequalities, I have

$$ (x + \beta)^2 = x^2 + 2\beta x + \beta^2 < x^2 + 2\beta x + 2\beta x $$

$$ x^2 + 2\beta x + 2\beta x = x^2 + 4 \beta x < 2 $$

Hence ## (x + \beta)^2 < 2 ##. Now, let ## y = x + \beta ##. Due to closure of ##\mathbb{Q}##, it can be seen that ##y \in \mathbb{Q}##. Also

## x < y## and finally ## y^2 < 2##.

Is this proof good enough ?

Thanks

$$ \alpha = \text{ min} \Bigl\{ \frac{(2-x^2)}{4x}, 2x \Bigl\} $$

It can be seen that ##\alpha > 0##. Also, I have

$$ \alpha \leqslant \frac{(2-x^2)}{4x} \cdots \cdots (1) $$

$$ \alpha \leqslant 2x \cdots \cdots (2)$$

Since ##x \in \mathbb{Q}##, it can be seen that ##\alpha \in \mathbb{Q}##. Since ##\alpha > 0##, there exists some ##\beta \in \mathbb{Q}## such that ## 0 < \beta < \alpha ##. From above equations, it follows that

$$ 0 < \beta < \frac{(2-x^2)}{4x} \cdots \cdots (3) $$

$$ 0 < \beta < 2x \cdots \cdots (4)$$

This simplifies to the following

$$ x^2 + 4x \beta < 2 $$

$$ \beta^2 < 2\beta x $$

Now consider ##(x + \beta)^2##

$$ (x + \beta)^2 = x^2 + 2\beta x + \beta^2 $$

Using above inequalities, I have

$$ (x + \beta)^2 = x^2 + 2\beta x + \beta^2 < x^2 + 2\beta x + 2\beta x $$

$$ x^2 + 2\beta x + 2\beta x = x^2 + 4 \beta x < 2 $$

Hence ## (x + \beta)^2 < 2 ##. Now, let ## y = x + \beta ##. Due to closure of ##\mathbb{Q}##, it can be seen that ##y \in \mathbb{Q}##. Also

## x < y## and finally ## y^2 < 2##.

Is this proof good enough ?

Thanks

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