MHB Proving the Existence of Irrational Numbers Between Two Reals

[OhMyMarkov]

Hello everyone!

I want to prove that between two reals, there exists an irrational. This is what I got:

$\forall x \in R$, $\exists m \in Z$ s.t. $m-1 < x < m$. In particular, $x\notin Q$.
$\exists a,b \in R$ s.t. $ax$ and $ax+b$ are irrational. Also, $a(m-1)<ax<am$, $a(m-1)+b<ax+b<am+b$.
End of proof.I also want to prove that between two reals, there are infinitely many rationals and irrationals. This is my proof:

(Using (1) the above theorem, and (2) the theorem that says that there exists a rational between every two reals)

Combining (1) and (2), between two reals, there exists a real. $\forall x,z \in R$ where $x<z$, $\exists y \in R$ s.t. $x<y<z$. In other words, $\forall x_0, x_1 \in R$ where $x_0 < x_1$, $\exists x_2 \in R$ s.t. $x_0<x_1<x_2$. Apply this again, $x_1<x_2<x_3$. Apply this N-times, $x_N<x_{n+1}<x_{N+2}$. If we let N grow indefinitely, we can say that there are infinitely many reals between two reals.Are these proofs correct? I am very new to establishing proofs in analysis. Any comments or criticism is highly appreciated. :)

 

[Sudharaka]

I want to prove that between two reals, there exists an irrational. This is what I got:

$\forall x \in R$, $\exists m \in Z$ s.t. $m-1 < x < m$. In particular, $x\notin Q$.
$\exists a,b \in R$ s.t. $ax$ and $ax+b$ are irrational. Also, $a(m-1)<ax<am$, $a(m-1)+b<ax+b<am+b$.
End of proof.

Hi OhMyMarkov, :)

You have finally obtained,

\[a(m-1)+b<ax+b<am+b\]

From this, I don't understand how you came to the final conclusion that between any two reals there is an irrational number. Can you please elaborate?

Kind Regards,
Sudharaka.

 

[OhMyMarkov]

Of course, $x$ is irrational, and so is $ax+b$ for choice of $ax+b$. Also, $a(m-1)+b$ and $am+b$ are real.

 

[Sudharaka]

Of course, $x$ is irrational, and so is $ax+b$ for choice of $ax+b$. Also, $a(m-1)+b$ and $am+b$ are real.

There is something incorrect in your proof. You are choosing specific values for \(a\), \(b\) and \(m\). Hence the numbers, $a(m-1)+b$ and $am+b$ are dependent upon your choice of \(x\). This is not what you need to prove. You should take any two real numbers and show that in between those two reals there is an irrational.

 

[tkhunny]

You may wish to familiarize your self with this: Modus ponens - Wikipedia, the free encyclopedia

You have rather magically declared x to be irrational. Are you SURE that leads to a significant result? Did you use its irrational properties or just your declaration to accomplish your proof?

Maybe its this one. Begging the question - Wikipedia, the free encyclopedia I can't decide.

 

[Plato2]

Hello everyone!
I want to prove that between two reals, there exists an irrational. This is what I got:
$\forall x \in R$, $\exists m \in Z$ s.t. $m-1 < x < m$. In particular, $x\notin Q$.
$\exists a,b \in R$ s.t. $ax$ and $ax+b$ are irrational. Also, $a(m-1)<ax<am$, $a(m-1)+b<ax+b<am+b$.
End of proof.

This problem depends on the theorem Between any two numbers there is a rational number.
With that theorem having been done this problem is trival.

If $a<b$ then $\sqrt2 a<\sqrt2 b$. So $\exists r\in\mathbb{Q}\setminus\{0\}$ such that $\sqrt2 a< r <\sqrt2 b$.

Now the irrational number $\dfrac{r}{\sqrt2}$ is between $a~\&~b~.$