# Terence Tao Analysis I Proposition 4.4.5

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 $$n \leq x < n+1$$.

• 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. Commented Jul 18, 2020 at 22:53
• 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 Commented Jul 18, 2020 at 23:02

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 $$(n\epsilon)^2 > 4 > 2$$.