# How to show that we can always choose a smaller number?

Suppose $A$ consists of all numbers $x \ge 0$ such that $x^2 \lt 2$. If the number $\sqrt{2}$ did not exist, there would not be a least number greater than all the numbers of $A$; for any $y > \sqrt{2}$ we chose, we could always choose a still smaller one.

I don't understand how we can choose a still smaller number than $y$.

If the number that could have been chosen is $c$, then to me this means:

$y>c>\sqrt{2}$

must be true.

But I am not sure why it must be true, because if we just select the next bigger number after $\sqrt{2}$ and call it $y$, how is it possible to select a number that is smaller than $y$ and not in $A$?

Thank you in advance for any help provided.

• Let $c=\frac{y+\sqrt{2}}{2}$. – Baby Dragon Aug 7 '13 at 19:15
• Also, you're missing a bit of context; while the statement is still true as written, it's almost certain from the specific topic under discussion that the numbers being considered are meant to be (at this stage) rational numbers, and that this argument shows the 'necessity' of completing the rationals to the reals. – Steven Stadnicki Aug 7 '13 at 19:16
• One property that the reals and the rationals share is density; for any two numbers (whether real or rational) $x, y$ with $x\lt y$ there is always a $z$ (for instance, $z=\frac{x+y}{2}$ works) with $x\lt z\lt y$. – Steven Stadnicki Aug 7 '13 at 19:19
• The reasoning in the grey box doesn't make sense. It first assumes $\sqrt 2$ doesn't exist, but then it uses the entity $\sqrt 2$ (which doesn't exist) in the inequality $y>\sqrt 2$. – Git Gud Aug 7 '13 at 19:22
• @StevenStadnicki You're probably right. It still is an horrible thing to write. – Git Gud Aug 7 '13 at 19:23

One property that real numbers and rational numbers share is density; for any $x, y$ with $x \lt y$, there's always a $z$ such that $x\lt z\lt y$ - just take $z=\frac{x+y}{2}$. This explains how, if we know $\sqrt{2}$ exists, we can conclude that there are no numbers next to it. But as pointed out in the comments, the argument here is somewhat circular; to run the simple density argument, we need to have a square root of $2$ in the first place.

Suppose instead that the only numbers we knew were rational; how could we conclude that for any number $y$ with $y^2\gt 2$, we can find a $z$ with $z\lt y$ but $z^2\gt 2$ still? One natural approach would be to run the averaging argument on the squares of $y$ and $z$, that is to choose $z$ such that $z^2 = \frac{2+y^2}2$; but if we can't find a square root of $2$ there's no reason to believe that we can find a square root of $\frac{2+y^2}{2}$.

Instead, let's take a different tack; if we know that $y^2$ is close to 2, then we know that $\frac2y$ is pretty close to $y$; what's more, we know that its square $\frac4{y^2}$ is on the 'other side' of $2$; $y^2\gt 2$, so $\frac4{y^2}\lt\frac42=2$. So let's try averaging $y$ and $\frac2y$; take $z=\frac12\left(y+\frac2y\right) = \frac{y^2+2}{2y}$. Since $y^2\gt 2$ then $\frac2y\lt y$, so their average $z$ must be $\lt y$; but $z^2 = \frac{(y^2+2)^2}{4y^2} = \frac{y^4+4y^2+4}{4y^2}$, and so $z^2-2 = \frac{y^4-4y^2+4}{4y^2} = \frac{(y^2-2)^2}{4y^2}\gt 0$.

This construction gives us a rational number $z$ (that is, a number that we know 'exists') that's less than y but whose square is still greater than $2$, and this $z$ disproves y's claim to be a least upper bound for the set $A$ ($=\{x: x^2\lt 2\}$).

The number $\dfrac{y^2+2}{2y}$ does the job. It is easy to check this is $\lt y$. To show that its square is greater than $2$, consider $$\left(\frac{y^2+2}{2y}\right)^2-2.$$ Simplification shows that this is $$\frac{(y^2-2)^2}{4y^2}.$$

Remarks: $1.$ "Newton's" Method is good for things other than computing solutions of equations to high accuracy! The technique for square roots was known at least $1500$ years before Newton was conceived. A similar technique was used by Islamic mathematician/astronomers many centuries before Newton.

$2.$ There is a more number-theoretic approach, connected with the theory of continued fractions. Given an overestimate $y$, we next use $$\frac{3y+4}{2y+3}.$$ Verification that this does the job is straightforward. There is reasonably quick convergence to $\sqrt{2}$, but not of the Newton Method level of performance.

• Too bad I couldn't upvote the original and the edited version. – Rick Decker Aug 7 '13 at 19:33