Prove that for every positive rational number $r$ where $r^2<2$, there must be another rational number $r+h$, such that $(r+h)^2<2$

Question

Question

Prove that for every positive rational number $r$ satisfying the condition $r^2<2$ one can always find a larger rational number $r+h$ ($h>0$) for which $(r+h)^2<2$.

The book's solution

We may assume $h<1$.

Then $h^2<h$ and $(r+h)^2<r^2+2rh+h$.

That is why it is sufficient to put $r^2+2rh+h = 2$, i.e., $h = \frac{2-r^2}{2r+1}$.

Note: As Ben in the comment section pointed out, this solution is wrong .

For example, when $r=\frac{1}{10}$, the value of h comes out to be $\frac{199}{120}$ and thus contradicts the assumption $h<1$.

My attempt

We have been given the following information:- $$r \in Q$$ $$r > 0$$ $$r^2<2$$

Let us first use the third piece of information to find the interval in which $r$ lies:-

$$r^2<2$$ $$r^2-2>0$$ $$(r-\sqrt2)(r+\sqrt2)<0$$

We can see that the value of the product is negative only in the interval between $-\sqrt2$ and $\sqrt2$.

But also $r$ is positive, therefore we get $$r \in (0,\sqrt2)$$

Now we have to prove that for every $r$, there exists a larger rational number $r+h$ which is bound by the following conditions:- $$(r+h)\in Q$$ $$(r+h) > 0$$ $$(r+h)^2 < 2$$

Again we can figure out the following:- $$ (r+h) \in (0,\sqrt2)$$

So for every positive rational number $r$ which satisfies the condition $r^2<2$, we can always find a larger number $r+h$ in the interval $(0,\sqrt2)$ that satisfies the condition $(r+h)^2<2$. This is because there are an infinite number of positive rational numbers in the interval $(0,\sqrt2)$, just like there are an infinite number of natural numbers.

Do you think this solution is right?

Answer 1

You can not assume that there exists a real number that you refer to as $\sqrt 2$. Namely you have no reason to presume that there exists a real number, $x$ so that $x^2 = 2$. That is precisely what this proof is supposed to establish.

If we know what irrational real numbers are, and if we know that $\sqrt 2$ actually exists, then this exercise simply follows by definitions. To be a real number, by definition, $\sqrt{2}$ is a least upper bound of a sequence of rational numbers all less than $\sqrt{2}$. So if any rational number, $r < \sqrt{2} = $ least upper bound of rational numbers less than $\sqrt{2}$ then there must be, by definition of least upper bound, a rational number $q$ (and $h= q-r$) so that $r < q=r+h \le \sqrt 2$. And as there is no rational number $q$ so that $q^2 = 2$ we know that $r < q=r+h < \sqrt 2$.

There was nothing to prove. It all followed by definition.

This exercise usually comes before the definition of the real numbers as an exercise to introduce the idea of a least upper bound. For any rational $r$ so that $r^2 < 2$, we must show that we can "get closer" to the idea of the square root of $2$ by finding a larger rational $q$ so that $r < q$ and $q^2 < 2$.

The book answer makes a few unstated assumptions (well, actually only one very obvious assumption) but it is correct (although it doesn't explicitly spell it out so it is hard to see). But here it is with more and correct details.

Preliminary: For any $0 < p < q$ it is true that $0< p^2 < q^2$. This is because $p< q$ and $p > 0$ so $p*p < q*p$. And $q > 0$ and $p < q$ we know that $p*q < q*q$ and so $p^2 < p*q < q^2$.

We can assume $r \ge 1$ because $1^2 = 1 < 2$. so if $r < 1$ we are done by letting $r+h = 1$. But we need to prove this is true for ALL rational $r$ where $r^2 < 2$.

So if $r \ge 1$ and IF such an $h$ exist we can assume $h < 1$. Because if $h \ge 1$ then $(r + h)^2 = r^2 + 2rh + h^2 \ge 1 + 2*1*1 + 1^2 = 4 > 2$.

Now we can do what your book does. .... Let's see; what does your book do?....

Let $h = \frac {2 - r^2}{2r + 1}$. We must prove that 1) $h> 0$ (and therefore $r < r+h$) 2) that $h$ is rational, and 3)that $(r+h)^2 < 2$.

1)Well $r^2 < 2$ so $2 -r^2 > 0$ and if we assume $r > 0$ we know $2r + 1 > 0$, so we have $h > 0$.

2) As $r$ is rational, so is $r^2, 2 -r^2, 2r+1$ and therefore $h =\frac {2-r^2}{2r+1}$ is rational.

And 3) is the real meat:

And as $r^2 \ge 1$ then $2-r^2 < 1$ and as $2r + 1 > 1$ we know that $h < 1$ and so $h^2 < 1$.

So

And $(r + h)^2 = r^2 + 2rh + h^2$

$ < r^2 + 2rh + h$

$ = r^2 + h(2r + 1) $

$ = r^2 + \frac {2-r^2}{2r+1}(2r+1)$

$= r^2 + 2 - r^2 = 2$.

And that completes the proof.

If $r < 1 $ then $h = 1 -r$ and $r+h = 1$ is one such rational rational number we want. (There are of course others.)

And if $r \ge 1$ and $r^2 < 2$ then $h =\frac {2-r^2}{2r+1}$ is one positive rational $h$ that we want so that $(r + h)^2 < 2$.

....

Now from there we know that we can create an infinite string of rational numbers $r < r+h =r_1 < r_q + h_1 = r_2 < r_3 < r_4 < ......... $ all so that $r^2 < r_1^2 < r_2^2 < r_3^2 < r_4^2 < ...... < 2$.

And once we define that the real numbers have the least upper bound property, we will know that some irrational real number, $x$, is the least upper bound of $r_i$. And we will be able to prove that $x^2 = ($ least upper bound $r_i)^2 = 2$. Thus THEN we will have proven that $\sqrt 2$ exists.