how do you proove this simple one?

[albertlee]

proove that between any two points a and b, a!=b, on the x axis, there are infinitely many rational points.

 

 

 

How do you solve the above? I know it's simple, but I just cant put down mathematically.

 

 

please help

thanks

[the tree]

This assumes a and b to be integers with a < b.

EDIT: it also assumes a>0 so is probably useless

 

[math]a< a+\frac{a}{b} < b[/math]

 

[math]a<a+\frac{a}{b\to\infty}<b[/math]

[neutrino86]

Which also needs to be proved:-)...rigorously...

 

I remember reading from Rudins Principles of Mathematical Analysis, a tricky inequality, which shows that between any 2 Rationals there can be infinitely many other rationals, by iterating the same inequality.

The nice part of it was that it killed all limits intrusions.

 

Unfortunately it avoids my memory...

 

 

Perhaps one must just get down to it.

[the tree]

Which also needs to be proved

Actually, it can't be, I was wrong. Damn.

[Holomorph]

It suffices to prove that one can find a rational (strictly) between any 2 distinct real numbers because then you can easily make an infinite sequence of rationals between the 2 (for example, if [math]x < y[/math] then we can find a rational [math]r_1[/math] between the 2 and reiterate the process with [math]x[/math] and [math]r_1[/math] or [math]r_1[/math] and [math]y[/math] to give a sequence [math](r_n)_n[/math] of rationals between [math]x[/math] and [math]y[/math]).

 

Here's a proof of this (we take [math]x < y[/math] and name [math]r[/math] the rational we want to find): suppose, without loss of generality, that [math]x[/math] and [math]y[/math] are positive (for if [math]x[/math] and [math]y[/math] are of different signs, we can pick the rational [math]r = 0[/math] and the case where they are both negative is analogous). Since [math]x-y > 0[/math], we can use the archimedian property of the real numbers (which is: given any positive real number [math]x[/math] and [math]y > 0[/math], one can find a positive integer [math]n[/math] such that [math]ny > x[/math]) to find a positive integer [math]b[/math] such that [math]b (x-y) > 1[/math], i.e. [math]x-y > \frac{1}{b}[/math]. Now let [math]a[/math] be the smallest integer such that [math]\frac{a}{b} > x[/math], then [math]r = \frac{a}{b}[/math] works since [math]r>x[/math] and also [math]r < y[/math] since otherwise we would have [math]r \geq y = x + (y - x) > \frac{1}{q} + x[/math] and hence [math]x < \frac{p-1}{q}[/math], contradicting the minimality of [math]p[/math].

 

Incidentally this is true not only for the reals but for any ordered field with the archimedian property (such a field always contains a sub-field isomorphic to [math]\mathbb{Q}[/math]).

[albertlee]

Well, I actually have my proof, but I dont know whether you can call it a proof or not.

 

Say, a>b,

 

(a+b)/2 must be the center point between a and b, right? let's call this point c.

 

(a+c)/2 is the point between a and c, which is also a point between a and (a+b)/2, so--> (a+[(a+b)/2])2, call this point d.

 

Then we can also find a point between a and d.

(a+d)/2, which is also (a+(a+[(a+b)/2])2)/2

 

etc etc, the above sequence can go as follow in simplfied:

 

1, (a+b)/2

 

2, (a+(a+b)/2)/2 --> (3a+b)/4

 

3, (a+(3a+b)/4)/2 --> (7a+b)/8

 

4, (15a+b)/16

 

5, (31a+b)/32

 

...

 

so, the formula of such sequence is:

((2ⁿ-1)a+b)/2ⁿ

 

therefore, there exists infinite rationals between any two points a and b.

 

 

Is this a good proof?

[cosine]

It suffices to prove that one can find a rational (strictly) between any 2 distinct real numbers because then you can easily make an infinite sequence of rationals between the 2 (for example' date=' if [math']x < y[/math] then we can find a rational [math]r_1[/math] between the 2 and reiterate the process with [math]x[/math] and [math]r_1[/math] or [math]r_1[/math] and [math]y[/math] to give a sequence [math](r_n)_n[/math] of rationals between [math]x[/math] and [math]y[/math]).

 

Here's a proof of this (we take [math]x < y[/math] and name [math]r[/math] the rational we want to find): suppose, without loss of generality, that [math]x[/math] and [math]y[/math] are positive (for if [math]x[/math] and [math]y[/math] are of different signs, we can pick the rational [math]r = 0[/math] and the case where they are both negative is analogous). Since [math]x-y > 0[/math], we can use the archimedian property of the real numbers (which is: given any positive real number [math]x[/math] and [math]y > 0[/math], one can find a positive integer [math]n[/math] such that [math]ny > x[/math]) to find a positive integer [math]b[/math] such that [math]b (x-y) > 1[/math], i.e. [math]x-y > \frac{1}{b}[/math]. Now let [math]a[/math] be the smallest integer such that [math]\frac{a}{b} > x[/math], then [math]r = \frac{a}{b}[/math] works since [math]r>x[/math] and also [math]r < y[/math] since otherwise we would have [math]r \geq y = x + (y - x) > \frac{1}{q} + x[/math] and hence [math]x < \frac{p-1}{q}[/math], contradicting the minimality of [math]p[/math].

 

Incidentally this is true not only for the reals but for any ordered field with the archimedian property (such a field always contains a sub-field isomorphic to [math]\mathbb{Q}[/math]).

 

I got confused with the proof right away when you said [math] x>y [/math] and then you said [math]x-y>0[/math] which implies by [math]x-y+y>0+y[/math] that [math]x>y[/math] which contradicts your first statement.

[cosine]

Well' date=' I actually have my proof, but I dont know whether you can call it a proof or not.

 

Say, a>b,

 

[b'](a+b)/2[/b] must be the center point between a and b, right? let's call this point c.

 

(a+c)/2 is the point between a and c, which is also a point between a and (a+b)/2, so--> (a+[(a+b)/2])2, call this point d.

 

Then we can also find a point between a and d.

(a+d)/2, which is also (a+(a+[(a+b)/2])2)/2

 

etc etc, the above sequence can go as follow in simplfied:

 

1, (a+b)/2

 

2, (a+(a+b)/2)/2 --> (3a+b)/4

 

3, (a+(3a+b)/4)/2 --> (7a+b)/8

 

4, (15a+b)/16

 

5, (31a+b)/32

 

...

 

so, the formula of such sequence is:

((2ⁿ-1)a+b)/2ⁿ

 

therefore, there exists infinite rationals between any two points a and b.

 

 

Is this a good proof?

 

This is what I would have thought, but this assumes that a and b are rational, which is not neccesarily the case. They could be irrational.

[Holomorph]

Ouch, I can't believe how careless I was when I wrote that, sorry ... I couldn't find how to edit that post so I'll just put the corrections here. Of course, it should actually be y-x instead of x-y in the whole post and in the last phrase of the second paragraph, p is actually a and q is actually b (this was because i wrote them as p and q on paper but then changed the notation when posting it here ) Also, in the bracketed phrase about the archimedian property it x can be any real number, not necessarily positive, only y needs to be positive (of course, these are not the same x and y as in the rest of the proof...). Hopefully it should make more sense now, sorry again...

 

Edit: Strange, I can edit this post but not the other ...

[jcarlson]

This is a pretty simple inductive proof for the case where a and b are both rational... I'm gunna have to think on the irrational case some.

 

 

Problem:

 

Assume a and b are two rational numbers, with [math]b > a[/math]. Show that there exists an infinate amount of rational numbers [math]t[/math] where [math]a < t < b[/math].

 

INDUCTIVE PROOF:

 

Let [math]f(n+1)[/math] be recursively defined as [math](f(n)+b)/2[/math] for all [math]n \geq 2[/math].

Let [math]f(1) = (a+b)/2[/math]

 

 

First we must show that every [math]f(k)[/math] where [math]1 \leq k \leq \infty [/math] is rational.

 

Part I:

 

Basis:

 

[math]f(1) = (a+b)/2[/math], where a and b are rational numbers.

 

A rational number by definition can be rewritten in the form [math]\frac{x}{y}[/math] where x and y are both integers. Thus we can rewrite a as [math]\frac{c}{d}[/math] and b as [math]\frac{e}{f}[/math], where c,d,e, and f are integers, resulting in:

 

[math]f(1) = \frac{\frac{c}{d}+\frac{e}{f}}{2}[/math]

 

This can then be written in the form

 

[math]f(1) = \frac{cf+ed}{2df}[/math]

 

Since the product of any two integers is an integer, and the sum of any two integers is also an integer, we can rewrite [math]f(1)[/math] as

 

[math]f(1) = \frac{g}{h}[/math]

 

where [math] g = cf+ed [/math] and [math] h = 2df [/math], both of which are integers. Therefore [math]f(1)[/math] is the ratio of two integers, and is by definition rational.

 

Inductive Hypothesis:

 

Assume that for some integer [math]k \geq 1[/math], [math]f(k)[/math] is rational. We need to show that [math]f(k+1)[/math] is rational.

 

By the definition of [math]f(n)[/math]:

 

[math]f(k+1) = \frac{f(k)+b}{2}[/math]

 

As in the case where k = 1 as shown above in the basis, since [math]f(k)[/math] (by the I.H.) and b are rational numbers, we can rewrite this as:

 

[math]f(k+1) = \frac{\frac{c}{d}+\frac{e}{f}}{2}[/math]

 

where c,d,e, and f are integers, and [math]f(k) = c/d[/math] and [math]b = e/f[/math].

 

This can then be written in the form

 

[math]f(k+1) = \frac{cf+ed}{2df}[/math]

 

Since the product of any two integers is an integer, and the sum of any two integers is also an integer, we can rewrite [math]f(k)[/math] as

 

[math]f(k+1) = \frac{g}{h}[/math]

 

where [math] g = cf+ed [/math] and [math] h = 2df [/math], both of which are integers. Therefore [math]f(k+1)[/math] is the ratio of two integers, and is by definition rational.

 

 

Thus we have shown [math]f(k)[/math] to be rational for every [math]k \geq 1[/math]. We must now show that [math]a < f(k) < b[/math] for all [math] k \geq 1 [/math]

 

Part II:

 

Basis:

 

[math]f(1) = (a+b)/2[/math]

[math]f(1) = \frac{a}{2} + \frac{b}{2}[/math]

 

[math]\frac{b}{2} + \frac{b}{2} = b[/math]

[math]\frac{a}{2} + \frac{a}{2} = a[/math]

 

since [math]a < b[/math],

 

[math]\frac{a}{2} < \frac{b}{2}[/math]

[math]\frac{a}{2} + \frac{b}{2} < b[/math]

[math]\frac{a}{2} + \frac{b}{2} > a[/math]

[math]a < f(1) < b[/math]

 

Inductive Hypothesis:

 

Assume for some value [math] k \geq 1 [/math], [math]a < f(k) < b[/math]. We need to show that [math]a < f(k+1) < b[/math]

 

By the definition of [math]f(n)[/math]:

 

[math]f(k+1) = \frac{f(k)+b}{2}[/math]

[math]f(k+1) = \frac{f(k)}{2} + \frac{b}{2}[/math]

 

[math]\frac{b}{2} + \frac{b}{2} = b[/math]

[math]\frac{a}{2} + \frac{a}{2} = a[/math]

 

since [math]a < f(k) < b[/math] (by I.H.),

 

[math]\frac{a}{2} < \frac{f(k)}{2} < \frac{b}{2}[/math]

[math]\frac{f(k)}{2} + \frac{b}{2} < b[/math]

[math]\frac{f(k)}{2} + \frac{b}{2} > a[/math]

[math]a < f(k+1) < b[/math]

 

 

Thus we have shown [math]a < f(k) < b[/math] for every [math]k \geq 1[/math].

 

Since there is an infinite number of values for k for which [math]k \geq 1[/math], it follows that there is an infinate number of values of [math]f(k)[/math] for which [math]k \geq 1[/math], and we have shown that every one of these values of [math]f(k)[/math] falls between the values of a and b, and that they are all rational.

 

 

EDIT: I've thought about it and the above proof could be easily extended to two irrational numbers a' and b' (and thus would be valid for all real numbers) if you show that there are always two distinct rational numbers a and b between the two distinct irrational numbers a' and b'. This can be proven by showing that there is at least one rational number b in between a' and b', and then that there is another rational number a in between a' and b. This is again, fairly simple to prove, but Its late and I'm tired, so I'm just gunna give you a link:

 

http://www.cut-the-knot.org/do_you_know/numbers.shtml

 

scroll down to the section "lemma 1" and you should find the proofs.

[jcarlson]

Edit: Strange' date=' I can edit this post but not the other ...[/quote']

 

Your posts become uneditable after a certain amount of time passes... couple hours I think.