# challenging proof(section idea needed)

• November 12th 2006, 09:47 PM
srinivas
challenging proof(section idea needed)
:cool: :cool: prove that between every two rationals there is another rational(do not use geometrical iterpretation )this is easy!!
also prove that between every two rationals there exists an irrational(this proof requires some thinking!!)
• November 13th 2006, 03:28 AM
ThePerfectHacker
Quote:

Originally Posted by srinivas
prove that between every two rationals there is another rational(do not use geometrical iterpretation )this is easy!!

a<b
Then, $\frac{a+b}{2}$ is between them and rational.

Quote:

also prove that between every two rationals there exists an irrational(this proof requires some thinking!!)
I am not familar with analysis on a serious level and this is what the proof requires.
However, I have my own little proof. (In fact 3)

Method 1)
You said between any two rationals,
$a,b$
So I can find a rational,
$\frac{a+b}{2}$
But we can transform this into an irrational number!
Since the that fraction is rational its continued fractional expansion terminates,
$[a_0;a_1,a_2,...,a_n]$
All you do know is attach ones to it,
$[a_0;a_1,a_2,....,a_n,1,1,1,1...]$
This will be an infinite continued fraction, which is irrational and be around the value of $(a+b)/2$

Method 2)
We can express any rational in terms of a decimal expansion (since the expansion is not necessarily unique we will prohibit the use the strings of nine).
So we have,
$a=.p_1p_2......$
$b=.q_1q_2......$
Let,
$p_1 if it exists (that is $p_1)[/tex])
Otherwise, $c_1=p_1=q_1$

Next go to $c_2$...
And so on...

Note the statment $c_k=p_k=q_k$ can not be always true for that would imply that $a=b$ which is false.

Thus there is a place where the fraction for $c_k$ is in between.
After that place write for the decimals,
12345678910111213.......
A number which is bound to be irrational
(Or use Louville's construction).

This will create an irrational number.

Method 3)
If that was not the case then the irrationals would be countable, an impossibility.
Hence there exists one such number, in fact infinitely many, in fact uncountable many.
• November 13th 2006, 04:52 AM
Plato
Here is another way. It requires a simple lemma that you my or not have.
LEMMA: If each of a and b is a real number and b-a>1 then there is integer N such that a<N<b.

If we assume that x and y are two real numbers and x<y then y-x is a positive real number. So there exists a positive integer K such K(y-x)>1. But this means that Ky-Kx>1. the by the lemma there is an integer J such that Kx<J<Ky or x<(J/K)<y.
Proof complete.
• November 13th 2006, 09:07 AM
srinivas
hey!!
i have different proof
it is based on division of rational no in to two parts by using a certain property
that is let P be a property rational no have or may not have
choose the property P such that we get two sets of rationals
one having the property and the other not having it an all rationals having the property are less than all others not having the property
the simplest of examples is x<a
so we have two idsjoint sets which cover all rational nos
one has all x<a and other x>=a
proving that the set x<a does not have a largest rational and x>=a does have a smallest rational we have formed a section of rationals
now rest can be proved by contradiction

irrational proof is using the idea that we know the existance of an irrational that is sqrt(2)
it is not very obvious proof!!
:D
• November 13th 2006, 09:27 AM
CaptainBlack
Quote:

Originally Posted by srinivas
prove that between every two rationals there exists an irrational(this proof requires some thinking!!)

Let a, b be two rationals a<b. The interval [a,b] is the image of [0,1] under
a 1-1 onto map. Therefore the cardinality of [a,b] is the same as that of [0,1].

If there were no irrationals in [a,b] then [a,b] would be enumerable, but its
not so there are.

RonL
• November 13th 2006, 09:32 AM
ThePerfectHacker
Quote:

Originally Posted by srinivas

irrational proof is using the idea that we know the existance of an irrational that is sqrt(2)
it is not very obvious proof!!
:D

There is a much simpler proof showing that $\sqrt{2}$ exists. (For non-math people, that is one of those times when we prove things that make us seem retarded).

I have posted this riddle here and supplied my own proof :cool: .
Using Intermediate value theorem of the continous function $f(x)=x^2-2$
• November 16th 2006, 07:50 AM
TriKri
Quote:

Originally Posted by srinivas
also prove that between every two rationals there exists an irrational(this proof requires some thinking!!)

What about taking a + (b-a) * x, where x is an irrational number between 0 and 1, for example sqrt(0.5)?

b-a is a rational number, hence (b-a)*x is an irrational number. And since a is a rational number a + (b-a) * x will be an irrational number.

supose b > a. It means b-a > 1. And since x > 1 too, a + (b-a)*x > a.

a + (b-a)*x = a + b*x - a*x = (1-x)*a + b*x = b - (b-a)*(1-x)

Since x < 1, 1-x > 0. And since b-a > 0, b - (b-a)*(1-x) < b.

So, if a and b are two rational numbers, a + (b-a) * x is an irrational number between a and b if x is an irrational number between 0 and 1.
• November 17th 2006, 10:06 AM
srinivas
ur proof is cyclic trikri
coz using what i gave to prove we get the proof that rational into irrational is irrational!!!
so that wont do
just see that we can find rationals as close to sqrt(2) as possible
and also we can find two rationals as close to each othe ras we wish
now take the difference between rationals on either side of sqrt(2) to be less han difference between two ratioals we have chosen
it can be seen easily that we can move the line between sqrt(2)+h and sqrt(2)-h between the two rationals we have chosen by moving it through a rational length!!
so it is proved