: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!!)

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- Nov 12th 2006, 09:47 PMsrinivaschallenging 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!!) - Nov 13th 2006, 03:28 AMThePerfectHacker
a<b

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

Quote:

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

However, I have my own little proof. (In fact 3)

Method 1)

You said between any two rationals,

$\displaystyle a,b$

So I can find a rational,

$\displaystyle \frac{a+b}{2}$

But we can transform this into an irrational number!

Since the that fraction is rational its continued fractional expansion terminates,

$\displaystyle [a_0;a_1,a_2,...,a_n]$

All you do know is attach ones to it,

$\displaystyle [a_0;a_1,a_2,....,a_n,1,1,1,1...]$

This will be an infinite continued fraction, which is irrational and be around the value of $\displaystyle (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,

$\displaystyle a=.p_1p_2......$

$\displaystyle b=.q_1q_2......$

Let,

$\displaystyle p_1<r_1<q_1$ if it exists (that is $\displaystyle p_1<q_1$)[/tex])

Otherwise, $\displaystyle c_1=p_1=q_1$

Next go to $\displaystyle c_2$...

And so on...

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

Thus there is a place where the fraction for $\displaystyle c_k$ 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. - Nov 13th 2006, 04:52 AMPlato
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. - Nov 13th 2006, 09:07 AMsrinivas
hey!!

good answer:cool: :cool: :cool:

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 - Nov 13th 2006, 09:27 AMCaptainBlack
- Nov 13th 2006, 09:32 AMThePerfectHacker
There is a much simpler proof showing that $\displaystyle \sqrt{2}$ 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 $\displaystyle f(x)=x^2-2$ - Nov 16th 2006, 07:50 AMTriKri
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. - Nov 17th 2006, 10:06 AMsrinivas
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