Terence Tao Exercise 5.4.3: Integer part of $x$ proof.

Question

I am reading Terence Tao: Analysis 1. As you may be aware, certain objects are introduced bit by bit, so if i am not 'allowed' to use something yet, please understand.

Show that for every real number $x$ there is exactly one integer $N$ such that

$ N \leq x < N + 1$

If $x = 0$ then we can take $N = 0$. If $x > 0$ then $x$ is the formal limit of some Cauchy sequence $(a_n)$, of which is positively bounded away from zero. Since $(a_n)$ is Cauchy it is bounded by some rational $M$, this implies that $x \leq M$, if we take $M < N + 1$ for some integer $N$ then $x < N + 1$.

I seem to think this part is fine, but thinking is dangerous.

I now need to show that $ N \leq x $. I know that the sequence $(a_n)$ is bounded away from zero, so every term of the sequence $a_n > c$ for some rational $c$ But i don't think this gets me any further.

Surely i could use this as i know x is the formal limit of the sequence of rationals $(a_n)$ but i cant quite make the connection.

EDIT:

I know i have to prove that this is true when x < 0, but i have not completed from x > 0.

I know i can use Cauchy sequences, division algorithm... its pretty hard for me to say what i can't use because I am an undergraduate. I'm sorry if that makes it impossible to answer for some, it's very awkward i must admit. What i can tell you is the next chapter is on the least upper bound property.

Answer 1

Let $x=(x_1,x_2,\dots,x_n,\dots)$ be a Cauchy sequence of rational numbers, and let $\epsilon=\frac{1}{2}$. Then there exists an $M$ such that if $n,m\geq M$, $|x_n-x_m|<\epsilon$. That means, in particular, $x\in [x_M-\frac{1}{2},x_M+\frac{1}{2}]$.

Use the theorem for rational numbers, to show that there is an integer part $N_0$ of $x_M-\frac{1}{2}$, and that $N_0\leq x<N_0+2$. Then either $N_0\leq x<N_0+1$ or $N_0+1\leq x<N_0+2$.

Answer 2

If $x>0$ consider the set of natural numbers that are larger than $x$. It is non-empty by the Archimedean property, and it has a first element. Call that element $N+1$. We must have $N\leq x$ because otherwise $N+1$ wouldn't be the first element.

Your approach it leaving the chosen $N$ too lose because you are choosing it to be any $M<N+1$.

Answer 3

I propose the following proof which relies on the principle of infinite descent which Tao uses to prove “irrationality of $\sqrt{2}$” (cf. Proposition $4.4.4$).

First observe that it suffices to prove for positive real numbers.

For the sake of contradiction, assume that there exists a positive real number $x$ such that for all integers $n$, $x<n+1$ implies $x<n$. Now by Proposition $5.4.12$, there exists an integer $N$ such that $x<N=(N-1)+1$ and hence by our assumption, $x<N-1$. Note that both $N, N-1>0$ since $x>0$ and $N-1<N$. Now we can do the same process with $N-1$ and find a smaller positive integer $M<N-1$ such that $x<M$. Hence we can create a sequence of positive integers $N_0, N_1, N_2,\ldots$ such that each is less than $x$ and each one smaller than the previous. This contradicts the principle of infinite descent, and hence the proof follows.

Uniqueness is easy to prove.

Answer 4

The idea is that if your sequence does not limit to an integer, then eventually all of the terms in the sequence must have the same integer part. Let's see if we can formalize this.

Let $d = \inf\{|N-x|\ :\ N\in \mathbb{Z}\}$. This is a nonempty set of reals so it has an infimum. If $d = 0$, then $x\in\mathbb{Z}$ and you're done.

On reread, the above paragraph is suspect but I have to run and won't be able to fix it properly until later. Also if the next chapter is on the least upper bound property, you may not yet know that all sets of real numbers have infima and suprema.

Now write $x = \lim_m a_m$ where $a_m$ is a Cauchy sequence of rationals. Let $M$ be such that $m>M$ implies $|a_m - x|<d$. Observe (why?) that each $a_m$ may be written uniquely as $A_m + r_m$ where $A_m\in\mathbb{Z}$ and $r_m\in [0,1)$.

Now can you prove that $A_m = A_n$ for all $m,n>M$?

Answer 5

Given the sequence $(x_i)=x$ consider the sequences $(\lfloor x_i\rfloor)=\lfloor x \rfloor $and $(\lceil x_i\rceil)=\lceil x\rceil$. Where $\lfloor x_i\rfloor$ is the gratest integer $<x_i$ and $\lceil x_i\rceil$ is the lowest integer $>x_i$.

You can proof that if $(x_i)$ is Cauchy, than $\lfloor x \rfloor$ and $\lceil x\rceil$ are Cauchy sequences that define successive integers.

Answer 6

Since x is a real number it is the limit of some Cauchy sequence. Then this sequence is bounded by some M, where M is rational, so by the same propositon for rationals from Tao's book, we have that there exists an integer N such that N is less than or equal to M, and M is less than N + 1.

Since N is less than or equal to M we now that there exists some rational number which is the member of the sequence such that it is greater or equal to N and less than or equal to M.

Thus we have that N is less than or equal to x and x is less than N + 1, as desired.

Answer 7

First show that such an N exists, then show that it is unique.

$x$ can be exactly one of zero, positive or negative (Prop. 5.4.4).

1. For the easiest case, if $x=0$, we have $N = 0$.

2. Let x be positive. Suppose for the sake of contradiction that such an N does not exist. With this assumption we can prove by induction that $k \leq x$ for any integer $k\geq 0$. Use $k=0$ as the base case. $0 \leq x$ is true, as $x$ is positive. Assume that $k\leq x$ for some $k\geq 0$, then consider $k+1 \leq x$. By our assumption that $N$ does not exist, $k+1\leq x$ must be true. This closes the induction.

But we know the conclusion is false. There must exist an integer $M$ such that $M\gt x$, as reals are bounded by a rational (Prop. 5.4.12), and integers are interspersed by rationals (Prop. 4.4.1). We can, therefore, conclude that such an $N$ exists.

3. A similar induction can be constructed when x is negative.*

Now show that $N$ is unique. Let there be two integers $n_1$ and $n_2$ such that both $n_1 \leq x \lt n_1 + 1$ and $n_2 \leq x \lt n_2 + 1$. Let $n_1$ < $n_2$ (one must be less than the other since they cannot be equal).

$$ \begin{align} &n_1 \lt n_2 \\ &\implies n_1 + 1 \leq n_2 \\ &\implies n_1 + 1 \leq x && \text{as $n_2 \leq x$} \\ &\implies \text{false} &&\text{as we assumed $n_1 + 1 > x$} \end{align} $$

Thus, there exists a unique integer $N$ such that $N \leq x \lt N + 1$.

* Instead of breaking this proof into cases $x=0$, $x>0$ and $x<0$, you may wish to first prove that reals are bounded below by integers, a relatively simple extension of Prop. 5.4.12, then you may choose such an integer for the base case of the induction.