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Numbers between Numbers

OhMyMarkov

Member
Mar 5, 2012
83
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Member
Mar 5, 2012
83
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Well-known member
MHB Math Helper
Jan 27, 2012
267

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
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 $a~\&~b~.$