irrational number in the middle

[shahar]

There is a way to prove that between 2 irrational numbers there is another irrational number?

 

[topsquark]

There is a way to prove that between 2 irrational numbers there is another irrational number?

Sure. Given any two real numbers a and b the number $\dfrac{a + b}{2}$ will be between them.

If a + b is irrational, then we are done because half of that is also irrational.

If a + b is rational then the irrational parts of a and b are negatives of each other. So then take $a + \dfrac{b -a}{4}$.

-Dan

 

[Dr.Peterson]

Sure. Given any two real numbers a and b the number $\dfrac{a + b}{2}$ will be between them.

If a + b is irrational, then we are done because half of that is also irrational.

If a + b is rational then the irrational parts of a and b are negatives of each other. So then take $a + \dfrac{b -a}{4}$.

-Dan

How do you define "the irrational part of a number"?

 

[Cubist]

There is a way to prove that between 2 irrational numbers there is another irrational number?

Start with irrational numbers a and b (a<b). It's possible to find integer values $a_i \, ,\, b_i \,,\,n$ such that...

$\displaystyle a \lt \frac{a_i}{n} \lt \frac{b_i}{n} \lt b$

See spoiler if you can't work out how to do this. Next, find a new irrational number "c" such that...

$\displaystyle \frac{a_i}{n} \lt c \lt \frac{b_i}{n}$

OP, could you do this?

Spoiler: possible integer values to create two rationals

$$a \lt \frac{a_i}{n} \lt \frac{b_i}{n} \lt b$$
$$na \lt a_i \lt b_i \lt nb$$
The following values provide an answer (there are many more possible)...
$$n = \left\lceil \frac{3}{b-a} \right\rceil \\ a_i = \lfloor na + 1 \rfloor \\ b_i = \lceil nb - 1 \rceil$$

 

[topsquark]

How do you define "the irrational part of a number"?

I did say that wrong, didn't I? I suppose a better concept would be to use the fractional part. The concept I'm reaching for is that the non-integer part of the irrational numbers add to a rational number (between 0 and 1, but that isn't important.) So if we have two rational numbers a and b such that a + b is rational then we know that {a} + {b} = rational. But I guess we really don't need to comment on the possibility. All we need to do is find a rational number around b - a and add it to a. (As Cubit is saying above.)

-Dan

 

[Cubist]

I did say that wrong, didn't I? I suppose a better concept would be to use the fractional part. The concept I'm reaching for is that the non-integer part of the irrational numbers add to a rational number (between 0 and 1, but that isn't important.) So if we have two rational numbers a and b such that a + b is rational then we know that {a} + {b} = rational. But I guess we really don't need to comment on the possibility. All we need to do is find a rational number around b - a and add it to a. (As Cubit is saying above.)

-Dan

I loved your method/ post, and I immediately knew the kind of thing you intended by "the irrational part of a number" but I couldn't tie it down more formally myself

I've thought of an alternative (and simpler) method than post#4... Start with two different irrational numbers a and b (a<b, b>0). It's always possible to find an integer value $n, n>1$, such that...

$\displaystyle a \lt b \times \frac{n-1}{n} \lt b$

For example, this works if a=sqrt(2), b=pi, and n=2

$\displaystyle \sqrt{2} \lt \pi \times \frac{2-1}{2} \lt \pi$

1.414... < 3.141... * (1/2) < 3.141...
1.414... < 1.570... < 3.141...

I'm pretty sure that a rational times an irrational is always irrational. I'll leave it as a problem for OP to find a minimum acceptable value of n in terms of a and b

 

[darkyadoo]

We can prove it considering this proposition, For any real numbers $x,y$ with $x < y$ there exists a rational $t$ such that $x < t < y$.

Now we can prove your statement : Let $x,y$ two irrationals, regarding the proposition above there exists a rational $t$ such that $$\dfrac{x}{\sqrt{2}} < t < \dfrac{y}{\sqrt{2}}\iff x < t\sqrt{2} <y$$. And $t\sqrt{2}$ is an irrational.

It remains to prove the first statement : $$(\forall x,y\in \R)(x < y\implies \exists t\in \mathbb{Q} :x < t < y)$$

The idea is the following one: Imagine you are walking on the real number line in a direction of $x$ and $y$, what length must be your step to make sure that one of your steps will be between $x$ and $y$. (You will need to use the Archimedean property of $\R$).