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In the book the proof for

Proposition 4.4.5: For every rational number $\epsilon > 0$, there exists a non-negative rational number $x$ such that $x^2 < 2 < (x + \epsilon)^2$

Proof:

Let $\epsilon > 0$ be rational. Suppose for the sake of contradiction that there is no non-negative rational number $x$ for which $x^2 < 2 < (x + \epsilon)^2$. This means that whenever $x$ is non-negative and $x^2 < 2$, we must also have $(x + \epsilon)^2 < 2$ (note that $(x + \epsilon)^2$ cannot be equal 2 because no such rational exist according to Proposition 4.4.4). Since $0^2 < 2$, we thus have $\epsilon^2 < 2$, which then implies $(2\epsilon)^2 < 2$, and indeed a simple induction shows that $(n\epsilon)^2 < 2$ for every natural number $n$. But by Proposition 4.4.1 we can find an integer $n$ such that $n>2/\epsilon$, which implies that $(n\epsilon)*2 > 4 > 2$, contradicting the claim that $(n\epsilon)^2 < 2$ for every natural number $n$.

My question is that:

  1. When Tao says Since $0^2 < 2$, we thus have $\epsilon^2 < 2$, is he saying that because the assumption of non-existence of non-negative $x$ that satisfies the condition, so that $x^2 < 2$ when $x=0$ you have $0^2 < 2$, and since $x=0$, then $(x + \epsilon)^2 < 2$ becomes $(0 + \epsilon)^2 < 2$ and then $\epsilon^2 < 2$?

  2. How was the induction done to show that $(n\epsilon)^2 < 2$ for every natural number $n$ using the fact that $\epsilon^2 < 2$

  3. Why did Tao use an integer $n$ such that $n>2/\epsilon$?

Proposition 4.4.1 is (Interspersing of integers by rationals). Let $x$ be a rational number. Then there exists an integer $n$ such that $n \leq x < n+1$.

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  • $\begingroup$ The second sentence of the proof as you have transcribed it is incomplete: "This means that whenever ... [what]". I suspect "[what]" should be $(x + \epsilon)^2 < 2$. Please fix. $\endgroup$
    – Rob Arthan
    Commented Jul 18, 2020 at 22:53
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    $\begingroup$ 1. Yes 2. Use your previous question to do the "induction step" again. Since $\epsilon^2 < 2$ and there is no such $x$, then $(\epsilon + \epsilon)^2 < 2$ and so on... 3. You are supposing for a specific $\epsilon>0$, that there is no $x$ satisfying the assumption. In 2. you proved that if this $\epsilon$ exists, then $(n\epsilon)^2$ is also less than 2, $\forall n$. But the previous sentence can't be true, because at "some point" (that specific $n$) you will find a contradiction $\endgroup$
    – FormerMath
    Commented Jul 18, 2020 at 23:02

1 Answer 1

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  1. Yes. "Let $\epsilon > 0$ be rational. Suppose for the sake of contradiction that there is no non-negative rational number $x$ for which $x^2 < 2 < (x + \epsilon)^2$. This means that whenever $x$ is non-negative and $x^2 < 2$, we must also have $(x + \epsilon)^2 < 2$." Now take $x = 0$.

  2. Unter the above assumption, we want to prove that if $\epsilon > 0$, then $(n\epsilon)^2 < 2$ for all $n$. For $n=1$ this has been proved in 1. Now assume it is true for some $n \ge 1$, i.e. $(n\epsilon)^2 < 2$. Since $x= n\epsilon$ is a non-negative rational number such that $x^2 < 2$, we get $((n+1)\epsilon)^2 =(x + \epsilon)^2 < 2$. By the way, we could start iunduction with $n=0$ which is a trivial case. Then step 1. would be obsolete.

  3. To obtain a contradiction, we have to find $n$ such that $(n\epsilon)^2 \ge 2$. By 4.4.1 there is an integer $m$ such that $m \le 2/\epsilon < m+1$. Let $n = m+1$. Then $n\epsilon > 2$, thus $(n\epsilon)^2 > 4 > 2$.

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