# Numbers between Numbers

#### OhMyMarkov

##### Member
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
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
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
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.

MHB Math Helper

#### Plato

##### Well-known member
MHB Math Helper
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 $a~\&~b~.$