# Real analysis question

1. May 22, 2013

### woundedtiger4

Hi all,

I am trying to understand this basic proof but I don't understand that where the equations (3) & (4) have come from?

[img=http://s9.postimg.org/8nwoy04vj/image.jpg]

p.s. sorry if I have posted this thread on wrong website.

2. May 22, 2013

### Fredrik

Staff Emeritus
Here's a link that works. http://s9.postimg.org/8nwoy04vj/image.jpg

(3) is a definition, and (4) is a simple consequence of (3). The reason why the definition (3) and the rewrite (4) are made is that they give us an easy way to prove the theorem. You probably understand all this, so I assume that what you're wondering about is how a person can come up with the idea to define q that way?

I don't know the answer to that. Maybe there's a simple motivation for it, but it's also possible that the person who came up with this just spent a week trying out different ideas and finally found one that works.

3. May 22, 2013

### Hypersphere

When it comes to the form of (3), I suppose it's just a clever choice, such that q gets the properties we want. However, we know that p^2<2 and p^2>2 give distinct cases, so a factor of p^2-2 somewhere in the expression is to be expected. Similarly, the term linear in p is suggested by the different p<q and p>q behaviours (it can be rewritten as q-p, after all). I don't have any argument for the scaling by 1/(p+2) though.

4. May 22, 2013

### micromass

Staff Emeritus
Here is some kind of way to find such things for yourself. Maybe it is what you want.

Basically, given a number x, you want to find a number f(x) (here, this will turn out to be $-\frac{x^2 - 2}{x+2}$ such that the following hold:

1) $x + f(x) < \sqrt{2}$ if $x< \sqrt{2}$.
2) If $x$ is rational, then $f(x)$ is rational.
3) $f(x)\geq 0$

Of course, there are many such functions, so let's try to identify such function. First, let's try to work with rational functions of the form

$$\frac{x + a}{ex + d}$$

If $(2)$ must be satisfied, then we will want $a$ and $b$ to be rational. If they are rational, then we have $(2)$ (this is the reason we pick a rational function).

The function is continuous and satisfies $x+ f(x)<\sqrt{2}$ for all $x<\sqrt{2}$. Thus we will have that $\sqrt{2} + f(\sqrt{2}) \leq \sqrt{2}$. Thus we must have that $f(\sqrt{2}) \leq 0$. It seems like a reasonable demand to ask for equality here. So we demand

4) $f(\sqrt{2}) = 0$

Now, if we pick a function of the form $\frac{x + a}{x + c}$ then this would satisfy $(4)$ only for $a=\sqrt{2}$. This is not a rational value, so we have a problem. So a function of this form will not work. Let's try to work with the next thing and work with functions

$$\frac{ax^2 + bx + c}{ex + d}$$

We want $\sqrt{2}$ to be a root, so it seems rather reasonable to ask that $x^2 - 2$ is the polynomial in the numerator. So we have the following function

$$\frac{x^2 - 2}{ex+d}$$

Now we must find $e$ and $d$ such that the other two conditions are satisfied. We want $x + f(x) - \sqrt{2} < 0$ if $x<\sqrt{2}$ and we have that $x + f(x) - \sqrt{2} = 0$ if $x = \sqrt{2}$. Now, if we could show that the function $g(x) = x + f(x) - \sqrt{2}$ is increasing, then the condition $(2)$ will be satisfied.

So we ask when $g$ would be increasing. We take the derivative and we get that $g^\prime(x)>0$ if and only if $f^\prime(x) > -1$

Now, if we could also get $f$ itself to be decreasing, then $f(\sqrt{2}) = 0$ would imply that $f(x)>0$ for $x<\sqrt{2}$. So condition $(3)$ would be satisfied. So we demand that $f^\prime(x) < 0$

If we work with the previous form we determined then we get that
$$0 > f^\prime(x) > -1$$

if and only if

$$0 > \frac{2x(ex + d) - e(x^2 - 2) }{(ex + d)^2}> -1$$

or

$$- (ex + d)^2 < x^2 e + 2xd + 2e < 0$$

We see from the above form that the problem will simplify if we take $e = -1$. So we get that

$$- (d - x)^2 < -x^2 + 2xd -2 < 0$$
and thus

$$- d^2 < -2 ~\text{and} ~x^2 + 2 > 2xd$$

The first condition will be true if $d^2$ is large enough. For example, for $d^2 = 4$. The last condition is true if $d$ is negative. So choosing $d= -2$ will satisfy all requirements.

5. May 22, 2013

### woundedtiger4

Many Thanks.
exactly, it is my question.

ok, is it possible for you to share something similar to (3)? if not then it's completely OK.

in the picture after equation (4) it says that "if p is in A then p^2 - 2 < 0 "WHY? ". how does (3) shows that q > p and (4) shows that q^2<2, finally q is in A.
for example, lets say that A={5} then here I can only see "p" where is q? I mean in the definition it says that for every p in A we can find rational q in A such that p<q OR should I consider a set with more than one p for example A={1,2,3,4,5} ?
Sorry if you find my question silly.

6. May 22, 2013

### Hypersphere

Well, the text says that "A is the set of ALL positive rationals p such that p^2<2". So a) p in A implies p^2-2<0 by definition and b) you can't say A={5} or something similar.

7. May 22, 2013

### woundedtiger4

Sir, thank you very much. I am relieved :)

In pic it says that "the purpose of the above discussion has been to show that rational numbers has certain gaps.........." what gaps?

8. May 22, 2013

### Useful nucleus

The gaps are the irrationals. Rudin showed that set A, for example, although bounded above, it does not have a least uper bound (which we know intuitively is √2). Reals will fill these gaps.