5
$\begingroup$

I am reading a proof for the existence of $\sqrt 2$. The first half of the proof goes as follows:

Consider the set $T = \{t \in \mathbb{R} : t^2 \lt 2\}$. Let $\alpha = \sup T$.

Case 1: Show $\alpha^2 \lt 2$ is impossible by implying $\alpha$ is not an upper bound for $T$.

We want to find an element that is larger than $\alpha$. So

$(\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2 \alpha}{n} + \frac{1}{n^2}$

$\lt \alpha^2 + \frac{2 \alpha}{n} + \frac{1}{n} = \alpha^2 + \frac{2 \alpha + 1}{n}$

The author states:

But now assuming $\alpha^2 < 2$ gives us a little space in which to fit the $\frac{(2 \alpha + 1)}{n}$ term and keep the total less than 2. Specifically, choose $n_0 \in \mathbb{N}$ large enough so that

$ \frac{1}{n_0} \lt \frac{(2 - \alpha^2)}{2 \alpha + 1}$

This is where I get lost in the proof. I understand that his goal with this is that in order to preserve $a^2 \lt 2$ it must be the case the difference $2 - \alpha^2$ needs to be some fraction that doesn't push $a^2$ outside of $2$. That is, the "extra stuff" being added to $\alpha^2$ here can't push it over our assumed bound.

The confusion for me is understanding how he constructed $ \frac{1}{n_0} \lt \frac{(2 - \alpha^2)}{2 \alpha + 1}$ from the assumption $\alpha^2 < 2$. I had initially thought he rearranged the inequality so that $0 < 2 -\alpha^2$, but this isn't correct.

What exactly did he use to choose this? I took a look in the solution manual and he didn't really describe the methodology there either. It seems arbitrary - beneficial to the problem. I'm not exactly sure how it's constructed from the assumed inequality.

$\endgroup$
  • $\begingroup$ How are you defining $\mathbb{R}$? That looks like a Dedekind cut and hence a real by definition (if you use that definition). $\endgroup$ – badjohn Feb 3 at 10:53
9
$\begingroup$

This is just goal-oriented work: The author wants $$\tag1\left(\alpha+\frac1n\right)^2<2.$$ Equivalently, $$\alpha^2+\frac{2\alpha}n+\frac1{n^2}<2. $$ Rearrange to $$\frac{2\alpha}n+\frac1{n^2}<2-\alpha^2. $$ This looks promising because, by assumption, the right hand side is positive and we only have to find $n$ that makes the left (positive and) small enough. As $\frac1n\le\frac1{n^2}$, we can boldly strengthen our task to find $n$ that makes (even) $$\frac{2\alpha}n+\frac1{n}<2-\alpha^2. $$ (The advantage of this is that the dependence on $n$ is now simpler than with the square part).

Now multiply with the positive(!) $n$ $$ 2\alpha+1<(2-\alpha^2)n$$ and divide by the positive(!) $2-\alpha^2$ to arrive at $$ \frac{2\alpha+1}{2-\alpha^2}<n$$ as a sufficient (and readily fulfilled) condition for $n$ to make $(1)$ true.

$\endgroup$
  • 1
    $\begingroup$ Interesting, I am very weak in proofing (part of the reason I'm reading and really struggling with real analysis is because I want to get better). This makes sense to me. I wish the author had made this more clear. Sometimes math proofs are woefully brief. $\endgroup$ – CL40 Feb 3 at 10:36
  • $\begingroup$ @CL40 Yes, this is often a common issue when people first start analysis/proving inequalities. It is often better to get a feel for an inequality by starting with the final inequality, rearranging it to get something that is true, and then reversing steps. Do keep in mind however that reversing steps needs to be justified, i.e by iff statements. $\endgroup$ – rubikscube09 Feb 3 at 16:37
3
$\begingroup$

The author is trying to say there exists an sufficiently large $n_0$ s.t. $\alpha^2 + \frac{2 \alpha + 1}{n_0} < 2$, which

$\iff \frac{2 \alpha + 1}{n_0} < 2-\alpha^2$

$\iff \frac{2 \alpha + 1}{n_0} < 2-\alpha^2$

$\iff \frac{1}{n_0} < \frac{2-\alpha^2}{2\alpha-1}$

$\endgroup$
  • 1
    $\begingroup$ Note that if and only if in $\LaTeX$ is $\iff$ (\iff). $\endgroup$ – TheSimpliFire Feb 3 at 13:19
  • $\begingroup$ @TheSimpliFire, ha, thanks! updated. $\endgroup$ – iBG Feb 3 at 13:39
0
$\begingroup$

The idea is the the following. Assume that $\alpha^2+\frac{2\alpha +1}{n} \lt 2$ and rearrange for $1/n$. What do you get? $\frac{1}{n} \lt \frac{2-\alpha^2}{2\alpha+1} $. Now, this doesn't prove anything, but it gives us a hint as to how to solve the problem. This inequality certainly holds for some n, and the reason why is due to the Archimedean property. So, one can just choose an appropriate such n, and then from that, we have shown that for some n, $\alpha^2+\frac{2\alpha +1}{n} \lt 2$ holds. Now, we backtrack, to arrive at that for that particular n you have chosen, $(\alpha + \frac{1}{n} )^2 \lt 2$ . Again, I emphasise that this inequality does not hold for every n, but the fact that we have found some n that satisfies $(\alpha + \frac{1}{n} )^2 \lt 2$ is enough, because this proves that Case 1 is true: an $\alpha$ such that $\alpha^2 \lt 2$ cannot be an upper bound for T.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.